I am trying to dynamically create alias' from the output of another command line tool.
For example:
> MyScript
blender="/opt/apps/blender/blender/2.79/blender"
someOtherAlias="ls -l"
I am trying the following code:
MyScript | {
while IFS= read -r line;
do
`echo alias $line`;
done;
}
But when I run this, I get the following error:
bash: alias: -l": not found
Just trying to run this command by itself gives me the same error:
> `echo 'alias someOtherAlias="ls -l"'`
bash: alias: -l": not found
But obviously the following command does work:
alias someOtherAlias="ls -l"
I've tried to find someone else who may have done this before, but none of my searches have come up with anything.
I would appreciate any and all help. Thanks!
See how bash (and posix shells) command parsing and quoting works and see difference between syntax and literal argument: for example '.."..' "..'.." are litteral quotes in an argument whereas " or ' are shell syntax and are not part of argument
also, enabling tacing with set -x may help to understand :
set -x
`echo 'alias someOtherAlias="ls -l"'`
++ echo 'alias someOtherAlias="ls -l"'
+ alias 'someOtherAlias="ls' '-l"'
bash: alias: -l": not found
bash sees 3 words : alias, someOtherAlias="ls and -l".
and alias loops over its arguments if they contain a = it create an alias otherwise it displays what alias argument is as -l" is not an alias it shows the error.
Note also as backquotes means command is run in a subshell (can be seen with mutiple + in trace) it will have no effect in current shell.
eval may be use to reinterpret literal as bash syntax (or to parse again a string).
So following should work, but be careful using eval on arbitrary arguments (from user input) can run arbitrary command.
eval 'alias someOtherAlias="ls -l"'
Finally also as bash commands after pipe are also run in subshell.
while IFS= read -r line;
do
`echo alias $line`;
done <MyScript
Related
In one of my Bash scripts, there's a point where I have a variable SCRIPT which contains the /path/to/an/exe, and what the script ultimately needs to do, is executing that executable. Therefore the last line of the script is
$($SCRIPT)
so that $SCRIPT is expanded to /path/to/an/exe, and $(/path/to/an/exe) executes the executable.
However, running shellcheck on the script generates this error:
In setscreens.sh line 7:
$($SCRIPT)
^--------^ SC2091: Remove surrounding $() to avoid executing output.
For more information:
https://www.shellcheck.net/wiki/SC2091 -- Remove surrounding $() to avoid e...
Is there a way I can rewrite that $($SCRIPT) in a more appropriate way? eval does not seem to be of much help here.
$($SCRIPT) indeed does not do what you think it does.
The outer $() will execute any commands inside the parenthesis and execute the result string.
The inner $SCRIPT will expand to the value of the SCRIPT variable and execute this string while splitting words on spaces/
If you want to execute the command contained into the SCRIPT variable, you just write as an example:
SCRIPT='/bin/ls'
"$SCRIPT" # Will execute /bin/ls
Now if you also need to handle arguments with your SCRIPT variable command call:
SCRIPT='/bin/ls'
"$SCRIPT" -l # Will execute /bin/ls -l
To also store or build arguments dynamically, you'd need an array instead of a string variable.
Example:
SCRIPT=(/bin/ls -l)
"${SCRIPT[#]}" # Will execute /bin/ls -l
SCRIPT+=(/etc) # Add /etc to the array
"${SCRIPT[#]}" # Will execute /bin/ls -l /etc
It worked for me with sh -c:
$ chrome="/opt/google/chrome/chrome"
$ sh -c "$chrome"
Opening in existing browser session.
It also passed the ShellCheck without any issues.
with bash, just use $SCRIPT:
cat <<'EOF' > test.sh
SCRIPT='echo aze rty'
$SCRIPT
EOF
bash test.sh
produce:
aze rty
I am trying to write shell script (sh), Where I am getting below error
variable i contains:
test.txt
code:
echo "${i/.txt/}"
Error:
just.sh: 16: just.sh: Bad substitution
expected output string :
text
Reproduce steps
Create file:
touch text.txt
Create file test.sh contents using any of editor
code:
#!/bin/sh
for i in `find *.txt`
do
echo "$i"
echo "${i/.txt/}"
done
How to run:
sh test.sh
sh is not bash. Fix your shebang (the 1st line) as #!/bin/bash first.
References
Difference between sh and bash, search "expansion" in the thread
Bash features a rich set of expanded non-standard parameter expansions such as ${substring:1:2}, ${variable/pattern/replacement}, case conversion, etc.
I am trying to run a script on multiple lists of files while also passing arguments in parallel. I have file_list1.dat, file_list2.dat, file_list3.dat. I would like to run script.sh which accepts 3 arguments: arg1, arg2, arg3.
For one run, I would do:
sh script.sh file_list1.dat $arg1 $arg2 $arg3
I would like to run this command in parallel for all the file lists.
My attempt:
Ncores=4
ls file_list*.dat | xargs -P "$Ncores" -n 1 [sh script.sh [$arg1 $arg2 $arg3]]
This results in the error: invalid number for -P option. I think the order of this command is wrong.
My 2nd attempt:
echo $arg1 $arg2 $arg3 | xargs ls file_list*.dat | xargs -P "$Ncores" -n 1 sh script.sh
But this results in the error: xargs: ls: terminated by signal 13
Any ideas on what the proper syntax is for passing arguments to a bash script with xargs?
I'm not sure I understand exactly what you want to do. Is it to execute something like these commands, but in parallel?
sh script.sh $arg1 $arg2 $arg3 file_list1.dat
sh script.sh $arg1 $arg2 $arg3 file_list2.dat
sh script.sh $arg1 $arg2 $arg3 file_list3.dat
...etc
If that's right, this should work:
Ncores=4
printf '%s\0' file_list*.dat | xargs -0 -P "$Ncores" -n 1 sh script.sh "$arg1" "$arg2" "$arg3"
The two major problems in your version were that you were passing "Ncores" as a literal string (rather than using $Ncores to get the value of the variable), and that you had [ ] around the command and arguments (which just isn't any relevant piece of shell syntax). I also added double-quotes around all variable references (a generally good practice), and used printf '%s\0' (and xargs -0) instead of ls.
Why did I use printf instead of ls? Because ls isn't doing anything useful here that printf or echo or whatever couldn't do as well. You may think of ls as the tool for getting lists of filenames, but in this case the wildcard expression file_list*.dat gets expanded to a list of files before the command is run; all ls would do with them is look at each one, say "yep, that's a file" to itself, then print it. echo could do the same thing with less overhead. But with either ls or echo the output can be ambiguous if any filenames contain spaces, quotes, or other funny characters. Some versions of ls attempt to "fix" this by adding quotes or something around filenames with funny characters, but that might or might not match how xargs parses its input (if it happens at all).
But printf '%s\0' is unambiguous and predictable -- it prints each string (filename in this case) followed by a NULL character, and that's exactly what xargs -0 takes as input, so there's no opportunity for confusion or misparsing.
Well, ok, there is one edge case: if there aren't any matching files, the wildcard pattern will just get passed through literally, and it'll wind up trying to run the script with the unexpanded string "file_list*.dat" as an argument. If you want to avoid this, use shopt -s nullglob before this command (and shopt -u nullglob afterward, to get back to normal mode).
Oh, and one more thing: sh script.sh isn't the best way to run scripts. Give the script a proper shebang line at the beginning (#!/bin/sh if it uses only basic shell features, #!/bin/bash or #!/usr/bin/env bash if it uses any bashisms), and run it with ./script.sh.
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 would like to invoke alias from ruby code so as to test the alias which I programmatically inserted into the dotfile. Say for example, the alias is the following:
alias something="echo somethingelse"
I searched the web and found the solution for bash:
#solution for bash
system %(
source ~/.bash_profile
shopt -s expand_aliases
something
)
However, this does not work for zsh.
I tried to invoke the alias using the following code (and a combination of other commands) but to no avail.
system %(
exec zsh #this seems to source .zshrc
something #this does not work
)
I would like it to work for zsh too. How can I get it working for zsh? Does anyone have a suggestion? Thanks in advance!
I have also tested the following but they don't work:
system %(
# exec /bin/zsh #this causes the subsequent lines to not run.
source ~/.zshrc #this causes the error lines to be printed
# setopt aliases #don't think it helps
something #trying to invoke this which is already in zshrc
)
The error messages:
/Users/ytbryan/.zprezto/init.zsh: line 14: autoload: command not
found
/Users/ytbryan/.zprezto/init.zsh: line 15: print: command not
found
/Users/ytbryan/.zshrc: line 42: `#': not a valid identifier
One approach is to run Zsh code via zsh -c. Aliases are not expanded when run from zsh -c, but the builtin aliases array is still accessible, so one can still expand aliases by manually retrieving expansions from the array and manually performing word splitting. This should cover most of the commonly seen aliases. For more advanced aliases (that involves process substitution, parameter expansion, command substitution, arithmetic expansion, brace expansion, filename expansion or filename generation, or that is more than a simple command), one might need to use eval (but one needs to be very cautious when using eval, and never ever use it when input comes from an untrusted source, or from a trusted but possibly tempered-with source).
Example code that could be embedded in Ruby system calls:
> zsh -c 'alias foo="print bar"; ${=aliases[foo]}'
bar
> zsh -c 'alias foo=print; ${=aliases[foo]} $#' -- 1 2 3
1 2 3
> zsh -c 'alias foo="print a b c | grep -o a"; ${=aliases[foo]}' # simple case where naive expansion fails
a b c | grep -o a
> zsh -c 'alias foo="print a b c | grep -o a"; eval "$aliases[foo]"' # eval comes to rescue, but be extra careful
a
Note that source works in zsh -c, so the alias definitions above could be sourced from any file just fine.