After reading the Bash man pages and with respect to this post, I am still having trouble understanding what exactly the eval command does and which would be its typical uses.
For example, if we do:
$ set -- one two three # Sets $1 $2 $3
$ echo $1
one
$ n=1
$ echo ${$n} ## First attempt to echo $1 using brackets fails
bash: ${$n}: bad substitution
$ echo $($n) ## Second attempt to echo $1 using parentheses fails
bash: 1: command not found
$ eval echo \${$n} ## Third attempt to echo $1 using 'eval' succeeds
one
What exactly is happening here and how do the dollar sign and the backslash tie into the problem?
eval takes a string as its argument, and evaluates it as if you'd typed that string on a command line. (If you pass several arguments, they are first joined with spaces between them.)
${$n} is a syntax error in bash. Inside the braces, you can only have a variable name, with some possible prefix and suffixes, but you can't have arbitrary bash syntax and in particular you can't use variable expansion. There is a way of saying “the value of the variable whose name is in this variable”, though:
echo ${!n}
one
$(…) runs the command specified inside the parentheses in a subshell (i.e. in a separate process that inherits all settings such as variable values from the current shell), and gathers its output. So echo $($n) runs $n as a shell command, and displays its output. Since $n evaluates to 1, $($n) attempts to run the command 1, which does not exist.
eval echo \${$n} runs the parameters passed to eval. After expansion, the parameters are echo and ${1}. So eval echo \${$n} runs the command echo ${1}.
Note that most of the time, you must use double quotes around variable substitutions and command substitutions (i.e. anytime there's a $): "$foo", "$(foo)". Always put double quotes around variable and command substitutions, unless you know you need to leave them off. Without the double quotes, the shell performs field splitting (i.e. it splits value of the variable or the output from the command into separate words) and then treats each word as a wildcard pattern. For example:
$ ls
file1 file2 otherfile
$ set -- 'f* *'
$ echo "$1"
f* *
$ echo $1
file1 file2 file1 file2 otherfile
$ n=1
$ eval echo \${$n}
file1 file2 file1 file2 otherfile
$eval echo \"\${$n}\"
f* *
$ echo "${!n}"
f* *
eval is not used very often. In some shells, the most common use is to obtain the value of a variable whose name is not known until runtime. In bash, this is not necessary thanks to the ${!VAR} syntax. eval is still useful when you need to construct a longer command containing operators, reserved words, etc.
Simply think of eval as "evaluating your expression one additional time before execution"
eval echo \${$n} becomes echo $1 after the first round of evaluation. Three changes to notice:
The \$ became $ (The backslash is needed, otherwise it tries to evaluate ${$n}, which means a variable named {$n}, which is not allowed)
$n was evaluated to 1
The eval disappeared
In the second round, it is basically echo $1 which can be directly executed.
So eval <some command> will first evaluate <some command> (by evaluate here I mean substitute variables, replace escaped characters with the correct ones etc.), and then run the resultant expression once again.
eval is used when you want to dynamically create variables, or to read outputs from programs specifically designed to be read like this. See Eval command and security issues for examples. The link also contains some typical ways in which eval is used, and the risks associated with it.
In my experience, a "typical" use of eval is for running commands that generate shell commands to set environment variables.
Perhaps you have a system that uses a collection of environment variables, and you have a script or program that determines which ones should be set and their values. Whenever you run a script or program, it runs in a forked process, so anything it does directly to environment variables is lost when it exits. But that script or program can send the export commands to standard output.
Without eval, you would need to redirect standard output to a temporary file, source the temporary file, and then delete it. With eval, you can just:
eval "$(script-or-program)"
Note the quotes are important. Take this (contrived) example:
# activate.sh
echo 'I got activated!'
# test.py
print("export foo=bar/baz/womp")
print(". activate.sh")
$ eval $(python test.py)
bash: export: `.': not a valid identifier
bash: export: `activate.sh': not a valid identifier
$ eval "$(python test.py)"
I got activated!
The eval statement tells the shell to take eval’s arguments as commands and run them through the command-line. It is useful in a situation like below:
In your script if you are defining a command into a variable and later on you want to use that command then you should use eval:
a="ls | more"
$a
Output:
bash: command not found: ls | more
The above command didn't work as ls tried to list file with name pipe (|) and more. But these files are not there:
eval $a
Output:
file.txt
mailids
remote_cmd.sh
sample.txt
tmp
Update: Some people say one should -never- use eval. I disagree. I think the risk arises when corrupt input can be passed to eval. However there are many common situations where that is not a risk, and therefore it is worth knowing how to use eval in any case. This stackoverflow answer explains the risks of eval and alternatives to eval. Ultimately it is up to the user to determine if/when eval is safe and efficient to use.
The bash eval statement allows you to execute lines of code calculated or acquired, by your bash script.
Perhaps the most straightforward example would be a bash program that opens another bash script as a text file, reads each line of text, and uses eval to execute them in order. That's essentially the same behavior as the bash source statement, which is what one would use, unless it was necessary to perform some kind of transformation (e.g. filtering or substitution) on the content of the imported script.
I rarely have needed eval, but I have found it useful to read or write variables whose names were contained in strings assigned to other variables. For example, to perform actions on sets of variables, while keeping the code footprint small and avoiding redundancy.
eval is conceptually simple. However, the strict syntax of the bash language, and the bash interpreter's parsing order can be nuanced and make eval appear cryptic and difficult to use or understand. Here are the essentials:
The argument passed to eval is a string expression that is calculated at runtime. eval will execute the final parsed result of its argument as an actual line of code in your script.
Syntax and parsing order are stringent. If the result isn't an executable line of bash code, in scope of your script, the program will crash on the eval statement as it tries to execute garbage.
When testing you can replace the eval statement with echo and look at what is displayed. If it is legitimate code in the current context, running it through eval will work.
The following examples may help clarify how eval works...
Example 1:
eval statement in front of 'normal' code is a NOP
$ eval a=b
$ eval echo $a
b
In the above example, the first eval statements has no purpose and can be eliminated. eval is pointless in the first line because there is no dynamic aspect to the code, i.e. it already parsed into the final lines of bash code, thus it would be identical as a normal statement of code in the bash script. The 2nd eval is pointless too, because, although there is a parsing step converting $a to its literal string equivalent, there is no indirection (e.g. no referencing via string value of an actual bash noun or bash-held script variable), so it would behave identically as a line of code without the eval prefix.
Example 2:
Perform var assignment using var names passed as string values.
$ key="mykey"
$ val="myval"
$ eval $key=$val
$ echo $mykey
myval
If you were to echo $key=$val, the output would be:
mykey=myval
That, being the final result of string parsing, is what will be executed by eval, hence the result of the echo statement at the end...
Example 3:
Adding more indirection to Example 2
$ keyA="keyB"
$ valA="valB"
$ keyB="that"
$ valB="amazing"
$ eval eval \$$keyA=\$$valA
$ echo $that
amazing
The above is a bit more complicated than the previous example, relying more heavily on the parsing-order and peculiarities of bash. The eval line would roughly get parsed internally in the following order (note the following statements are pseudocode, not real code, just to attempt to show how the statement would get broken down into steps internally to arrive at the final result).
eval eval \$$keyA=\$$valA # substitution of $keyA and $valA by interpreter
eval eval \$keyB=\$valB # convert '$' + name-strings to real vars by eval
eval $keyB=$valB # substitution of $keyB and $valB by interpreter
eval that=amazing # execute string literal 'that=amazing' by eval
If the assumed parsing order doesn't explain what eval is doing enough, the third example may describe the parsing in more detail to help clarify what is going on.
Example 4:
Discover whether vars, whose names are contained in strings, themselves contain string values.
a="User-provided"
b="Another user-provided optional value"
c=""
myvarname_a="a"
myvarname_b="b"
myvarname_c="c"
for varname in "myvarname_a" "myvarname_b" "myvarname_c"; do
eval varval=\$$varname
if [ -z "$varval" ]; then
read -p "$varname? " $varname
fi
done
In the first iteration:
varname="myvarname_a"
Bash parses the argument to eval, and eval sees literally this at runtime:
eval varval=\$$myvarname_a
The following pseudocode attempts to illustrate how bash interprets the above line of real code, to arrive at the final value executed by eval. (the following lines descriptive, not exact bash code):
1. eval varval="\$" + "$varname" # This substitution resolved in eval statement
2. .................. "$myvarname_a" # $myvarname_a previously resolved by for-loop
3. .................. "a" # ... to this value
4. eval "varval=$a" # This requires one more parsing step
5. eval varval="User-provided" # Final result of parsing (eval executes this)
Once all the parsing is done, the result is what is executed, and its effect is obvious, demonstrating there is nothing particularly mysterious about eval itself, and the complexity is in the parsing of its argument.
varval="User-provided"
The remaining code in the example above simply tests to see if the value assigned to $varval is null, and, if so, prompts the user to provide a value.
I originally intentionally never learned how to use eval, because most people will recommend to stay away from it like the plague. However I recently discovered a use case that made me facepalm for not recognizing it sooner.
If you have cron jobs that you want to run interactively to test, you might view the contents of the file with cat, and copy and paste the cron job to run it. Unfortunately, this involves touching the mouse, which is a sin in my book.
Lets say you have a cron job at /etc/cron.d/repeatme with the contents:
*/10 * * * * root program arg1 arg2
You cant execute this as a script with all the junk in front of it, but we can use cut to get rid of all the junk, wrap it in a subshell, and execute the string with eval
eval $( cut -d ' ' -f 6- /etc/cron.d/repeatme)
The cut command only prints out the 6th field of the file, delimited by spaces. Eval then executes that command.
I used a cron job here as an example, but the concept is to format text from stdout, and then evaluate that text.
The use of eval in this case is not insecure, because we know exactly what we will be evaluating before hand.
I've recently had to use eval to force multiple brace expansions to be evaluated in the order I needed. Bash does multiple brace expansions from left to right, so
xargs -I_ cat _/{11..15}/{8..5}.jpg
expands to
xargs -I_ cat _/11/8.jpg _/11/7.jpg _/11/6.jpg _/11/5.jpg _/12/8.jpg _/12/7.jpg _/12/6.jpg _/12/5.jpg _/13/8.jpg _/13/7.jpg _/13/6.jpg _/13/5.jpg _/14/8.jpg _/14/7.jpg _/14/6.jpg _/14/5.jpg _/15/8.jpg _/15/7.jpg _/15/6.jpg _/15/5.jpg
but I needed the second brace expansion done first, yielding
xargs -I_ cat _/11/8.jpg _/12/8.jpg _/13/8.jpg _/14/8.jpg _/15/8.jpg _/11/7.jpg _/12/7.jpg _/13/7.jpg _/14/7.jpg _/15/7.jpg _/11/6.jpg _/12/6.jpg _/13/6.jpg _/14/6.jpg _/15/6.jpg _/11/5.jpg _/12/5.jpg _/13/5.jpg _/14/5.jpg _/15/5.jpg
The best I could come up with to do that was
xargs -I_ cat $(eval echo _/'{11..15}'/{8..5}.jpg)
This works because the single quotes protect the first set of braces from expansion during the parsing of the eval command line, leaving them to be expanded by the subshell invoked by eval.
There may be some cunning scheme involving nested brace expansions that allows this to happen in one step, but if there is I'm too old and stupid to see it.
You asked about typical uses.
One common complaint about shell scripting is that you (allegedly) can't pass by reference to get values back out of functions.
But actually, via "eval", you can pass by reference. The callee can pass back a list of variable assignments to be evaluated by the caller. It is pass by reference because the caller can allowed to specify the name(s) of the result variable(s) - see example below. Error results can be passed back standard names like errno and errstr.
Here is an example of passing by reference in bash:
#!/bin/bash
isint()
{
re='^[-]?[0-9]+$'
[[ $1 =~ $re ]]
}
#args 1: name of result variable, 2: first addend, 3: second addend
iadd()
{
if isint ${2} && isint ${3} ; then
echo "$1=$((${2}+${3}));errno=0"
return 0
else
echo "errstr=\"Error: non-integer argument to iadd $*\" ; errno=329"
return 1
fi
}
var=1
echo "[1] var=$var"
eval $(iadd var A B)
if [[ $errno -ne 0 ]]; then
echo "errstr=$errstr"
echo "errno=$errno"
fi
echo "[2] var=$var (unchanged after error)"
eval $(iadd var $var 1)
if [[ $errno -ne 0 ]]; then
echo "errstr=$errstr"
echo "errno=$errno"
fi
echo "[3] var=$var (successfully changed)"
The output looks like this:
[1] var=1
errstr=Error: non-integer argument to iadd var A B
errno=329
[2] var=1 (unchanged after error)
[3] var=2 (successfully changed)
There is almost unlimited band width in that text output! And there are more possibilities if the multiple output lines are used: e.g., the first line could be used for variable assignments, the second for continuous 'stream of thought', but that's beyond the scope of this post.
In the question:
who | grep $(tty | sed s:/dev/::)
outputs errors claiming that files a and tty do not exist. I understood this to mean that tty is not being interpreted before execution of grep, but instead that bash passed tty as a parameter to grep, which interpreted it as a file name.
There is also a situation of nested redirection, which should be handled by matched parentheses which should specify a child process, but bash is primitively a word separator, creating parameters to be sent to a program, therefore parentheses are not matched first, but interpreted as seen.
I got specific with grep, and specified the file as a parameter instead of using a pipe. I also simplified the base command, passing output from a command as a file, so that i/o piping would not be nested:
grep $(tty | sed s:/dev/::) <(who)
works well.
who | grep $(echo pts/3)
is not really desired, but eliminates the nested pipe and also works well.
In conclusion, bash does not seem to like nested pipping. It is important to understand that bash is not a new-wave program written in a recursive manner. Instead, bash is an old 1,2,3 program, which has been appended with features. For purposes of assuring backward compatibility, the initial manner of interpretation has never been modified. If bash was rewritten to first match parentheses, how many bugs would be introduced into how many bash programs? Many programmers love to be cryptic.
As clearlight has said, "(p)erhaps the most straightforward example would be a bash program that opens another bash script as a text file, reads each line of text, and uses eval to execute them in order". I'm no expert, but the textbook I'm currently reading (Shell-Programmierung by Jürgen Wolf) points to one particular use of this that I think would be a valuable addition to the set of potential use cases collected here.
For debugging purposes, you may want to go through your script line by line (pressing Enter for each step). You could use eval to execute every line by trapping the DEBUG signal (which I think is sent after every line):
trap 'printf "$LINENO :-> " ; read line ; eval $line' DEBUG
I like the "evaluating your expression one additional time before execution" answer, and would like to clarify with another example.
var="\"par1 par2\""
echo $var # prints nicely "par1 par2"
function cntpars() {
echo " > Count: $#"
echo " > Pars : $*"
echo " > par1 : $1"
echo " > par2 : $2"
if [[ $# = 1 && $1 = "par1 par2" ]]; then
echo " > PASS"
else
echo " > FAIL"
return 1
fi
}
# Option 1: Will Pass
echo "eval \"cntpars \$var\""
eval "cntpars $var"
# Option 2: Will Fail, with curious results
echo "cntpars \$var"
cntpars $var
The curious results in option 2 are that we would have passed two parameters as follows:
First parameter: "par1
Second parameter: par2"
How is that for counter intuitive? The additional eval will fix that.
It was adapted from another answer on How can I reference a file for variables using Bash?
In my program I need to know the maximum number of process I can run. So I write a script. It works when I run it in shell but but when in program using system("./limit.sh"). I work in bash.
Here is my code:
#/bin/bash
LIMIT=\`ulimit -u\`
ACTIVE=\`ps -u | wc -l \`
echo $LIMIT > limit.txt
echo $ACTIVE >> limit.txt
Anyone can help?
Why The Original Fails
Command substitution syntax doesn't work if escaped. When you run:
LIMIT=\`ulimit -u\`
...what you're doing is running a command named
-u`
...with the environment variable named LIMIT containing the value
`ulimit
...and unless you actually have a command that starts with -u and contains a backtick in its name, this can be expected to fail.
This is because using backticks makes characters which would otherwise be syntax into literals, and running a command with one or more var=value pairs preceding it treats those pairs as variables to export in the environment for the duration of that single command.
Doing It Better
#!/bin/bash
limit=$(ulimit -u)
active=$(ps -u | wc -l)
printf '%s\n' "$limit" "$active" >limit.txt
Leave off the backticks.
Use modern $() command substitution syntax.
Avoid multiple redirections.
Avoid all-caps names for your own variables (these names are used for variables with meaning to the OS or system; lowercase names are reserved for application use).
Doing It Right
#!/bin/bash
exec >limit.txt # open limit.txt as output for the rest of the script
ulimit -u # run ulimit -u, inheriting that FD for output
ps -u | wc -l # run your pipeline, likewise with output to the existing FD
You have a typo on the very first line: #/bin/bash should be #!/bin/bash - this is often known as a "shebang" line, for "hash" (#) + "bang" (!)
Without that syntax written correctly, the script is run through the system's default shell, which will see that line as just a comment.
As pointed out in comments, that also means only the standardised options available to the builtin ulimit command, which doesn't include -u.
I have a variable containing some command output, and I want to count the number of lines in that output. I am trying to do this using pure bash (instead of piping to wc like in this question).
The best I have come up with seems kind of cumbersome:
function count_lines() {
num_lines=0
while IFS='' read -r line; do
((num_lines++))
done <<< "$1"
echo "$num_lines"
}
count_lines "$my_var"
Is there a cleaner or shorter way to do this?
count_lines () (
IFS=$'\n'
set -f
set -- $1
echo $#
)
Some tricks used:
The function is defined using a subshell instead of a command group to localize the change to IFS and the -f shell option. (You could be less fancy and use local IFS=$'\n' instead, and running set +f at the end of the function).
Disable filename generation to avoid any metacharacters in the argument from expanding and interfering with the line count.
Once IFS is changed, set the positional parameters for the function using unquoted parameter expansion for the argument.
Finally, output the number of positional parameters; there is one per line in the original argument.
IFS=$'\n'
set -f
x=( $variable )
Now ${#x[#]} will contain the number of 'lines'. (Use "set +f" to undo "set -f".)
Another possible solution I came up with:
export x=0;while read i;do export x=$(($x+1));done < /path/to/your/file;echo $x
I have a bash script that I wish to read from a file to get it's arguments set. Basically my script reads arguments positionally ($1, $2, $3, etc.)
while test $# -gt 0; do
case $1 in
-h | --help)
echo "Help cruft"
exit 0
;;
esac
shift
done
One of the options I was hoping could be a config file that reads in arguments (for simple and easy config) so I was hoping the set -- command would work (-- to over ride the arguments). However, since they are defined in a file I have to read it in and use xargs to pass them:
-c | --config)
cat $2 | xargs set --
continue
;;
The trouble is that xargs buggers up the -- so I don't know how to accomplish this.
Note: I realize I could use source config_file and have it set variable; might be the final option. I wanted to know if I could do it like above and simplify the documentation.
A simplified example script:
# foo.sh
echo "x y z" | xargs set --
echo $*
# Command line
$ bash foo.sh a b c
xargs: set: No such file or directory
a b c
xargs can't execute set because:
set is a shell built-in, not an external command. xargs only knows how to execute commands. (Some shell built-ins shadow commands with the same name, such as printf, true, and [. So xargs can execute those commands, but the semantics might not be identical to the built-in.)
Even if xargs could execute set, it would have no effect because xargs does not run inside of the shell's environment; every command executed by xargs is a separate process. So you will get no error if you do this:
echo a b c | xargs bash -c 'set -- "${#}"' _
But it also won't do anything useful. (Substitute set with echo and you'll see that it does actually invoke the command.)
How to read arguments from a file.
First, you need to answer the question: what does it mean to have arguments in a file? Are they individual whitespace-separated words with no mechanism to include whitespace in any argument? (That would also be required for xargs to work in its default mode, so it is not a totally unreasonable assumption, although it is almost certainly going to get you into trouble at some point.)
In that case you don't need xargs at all; you can just use command substitution:
set -- $(<file)
While that will work fine, this won't:
echo a b c | set -- $(</dev/stdin)
because the pipeline (created by the | operator) causes the processes on either side to be run in subshells, and consequently the set doesn't modify the current shell's environment variables.
A more robust solution
Suppose that each argument is in a single line in the file, which makes it possible to include whitespace in an argument, but not a newline. Then we could use the useful mapfile built-in to read the arguments into an array, and set the positional arguments from the array. (Or just use the array directly, but that would be a different question.)
mapfile -t args < file
set -- "${args[#]}"
Again, watch out for piping into mapfile; it won't work, for the same reason that it didn't work with set.
I'm facing a small problem here, I want to pass a string containing whitespaces , to another program such that the whole string is treated as a command line argument.
In short I want to execute a command of the following structure through a bash shell script:
command_name -a arg1 -b arg2 -c "arg with whitespaces here"
But no matter how I try, the whitespaces are not preserved in the string, and is tokenized by default. A solution please,
edit: This is the main part of my script:
#!/bin/bash
#-------- BLACKRAY CONFIG ---------------#
# Make sure the current user is in the sudoers list
# Running all instances with sudo
BLACKRAY_BIN_PATH='/opt/blackray/bin'
BLACKRAY_LOADER_DEF_PATH='/home/crozzfire'
BLACKRAY_LOADER_DEF_NAME='load.xml'
BLACKRAY_CSV_PATH='/home/crozzfire'
BLACKRAY_END_POINT='default -p 8890'
OUT_FILE='/tmp/out.log'
echo "The current binary path is $BLACKRAY_BIN_PATH"
# Starting the blackray 0.9.0 server
sudo "$BLACKRAY_BIN_PATH/blackray_start"
# Starting the blackray loader utility
BLACKRAY_INDEX_CMD="$BLACKRAY_BIN_PATH/blackray_loader -c $BLACKRAY_LOADER_DEF_PATH/$BLACKRAY_LOADER_DEF_NAME -d $BLACKRAY_CSV_PATH -e "\"$BLACKRAY_END_POINT\"""
sudo time $BLACKRAY_INDEX_CMD -a $OUT_FILE
#--------- END BLACKRAY CONFIG ---------#
You're running into this problem because you store the command in a variable, then expand it later; unless there's a good reason to do this, don't:
sudo time $BLACKRAY_BIN_PATH/blackray_loader -c $BLACKRAY_LOADER_DEF_PATH/$BLACKRAY_LOADER_DEF_NAME -d $BLACKRAY_CSV_PATH -e "$BLACKRAY_END_POINT" -a $OUT_FILE
If you really do need to store the command and use it later, there are several options; the bash-hackers.org wiki has a good page on the subject. It looks to me like the most useful one here is to put the command in an array rather than a simple variable:
BLACKRAY_INDEX_CMD=($BLACKRAY_BIN_PATH/blackray_loader -c $BLACKRAY_LOADER_DEF_PATH/$BLACKRAY_LOADER_DEF_NAME -d $BLACKRAY_CSV_PATH -e "$BLACKRAY_END_POINT")
sudo time "${BLACKRAY_INDEX_CMD[#]}" -a $OUT_FILE
This avoids the whole confusion between spaces-separating-words and spaces-within-words because words aren't separated by spaces -- they're in separate elements of the array. Expanding the array in double-quotes with the [#] suffix preserves that structure.
(BTW, another option would be to use escaped quotes rather like you're doing, then run the command with eval. Don't do this; it's a good way to introduce weird parsing bugs.)
Edit:
Try:
BLACKRAY_END_POINT="'default -p 8890'"
or
BLACKRAY_END_POINT='"default -p 8890"'
or
BLACKRAY_END_POINT="default\ -p\ 8890"
or
BLACKRAY_END_POINT='default\ -p\ 8890'
and
BLACKRAY_INDEX_CMD="$BLACKRAY_BIN_PATH/blackray_loader -c $BLACKRAY_LOADER_DEF_PATH/$BLACKRAY_LOADER_DEF_NAME -d $BLACKRAY_CSV_PATH -e $BLACKRAY_END_POINT"
Original answer:
Is blackray_loader a shell script?
Here is a demonstration that you have to deal with this issue both when specifying the parameter and when handling it:
A text file called "test.txt" (include the line numbers):
1 two words
2 two words
3 two
4 words
A script called "spacetest":
#!/bin/bash
echo "No quotes in script"
echo $1
grep $1 test.txt
echo
echo "With quotes in script"
echo "$1"
grep "$1" test.txt
echo
Running it with ./spacetest "two--------words" (replace the hyphens with spaces):
No quotes in script
two words
grep: words: No such file or directory
test.txt:1 two words
test.txt:2 two words
test.txt:3 two
With quotes in script
two words
2 two words
You can see that in the "No quotes" section it tried to do grep two words test.txt which interpreted "words" as a filename in addition to "test.txt". Also, the echo dropped the extra spaces.
When the parameter is quoted, as in the second section, grep saw it as one argument (including the extra spaces) and handled it correctly. And echo preserved the extra spaces.
I used the extra spaces, by the way, merely to aid in the demonstration.
I have a suggestion:
# iterate through the passed arguments, save them to new properly quoted ARGS string
while [ -n "$1" ]; do
ARGS="$ARGS '$1'"
shift
done
# invoke the command with properly quoted arguments
my_command $ARGS
probably you need to surround the argument by double quotes (e.g. "${6}").
Following OP comment it should be "$BLACKRAY_END_POINT"
Below is my example of restarting a script via exec su USER or exec su - USER. It accommodates:
being called from a relative path or current working directory
spaces in script name and arguments
single and double-quotes in arguments, without crazy escapes like: \\"
#
# This script should always be run-as a specific user
#
user=jimbob
if [ $(whoami) != "$user" ]; then
exec su -c "'$(readlink -f "$0")' $(printf " %q" "$#")" - $user
exit $?
fi
A post on other blog saved me for this whitespaces problem: http://logbuffer.wordpress.com/2010/09/23/bash-scripting-preserve-whitespaces-in-variables/
By default, whitespaces are trimed:
bash> VAR1="abc def gh ijk"
bash> echo $VAR1
abc def gh ijk
bash>
"The cause of this behaviour is the internal shell variable $IFS (Internal Field Separator), that defaults to whitespace, tab and newline.
To preserve all contiguous whitespaces you have to set the IFS to something different"
With IFS bypass:
bash> IFS='%'
bash> echo $VAR1
abc def gh ijk
bash>unset IFS
bash>
It works wonderfully for my command case:
su - user1 -c 'test -r "'${filepath}'"; ....'
Hope this helps.