I'm sure I'm missing something stupid. I want to pass a full path variable to a perl script, where I do some work on it and then pass it back. So I have:
echo "Backing up: $f ";
$write_file="$(perl /home/spider/web/foo.com/public_html/gen-path.cgi $f)";
echo "WRITE TO: $write_file \n";
However, this gives me:
Backing up: /home/spider/web/foo.com/public_html/websites-uk/uk/q/u
backup-files-all.sh: line 7: =backup-uk-q-u.tar.gz: command not found
WRITE TO: \n
I can't work out why its not saving the output into $write_file. I must be missing something (bash isn't my prefered language, which is why I'm passing to Perl as I'm a lot more fluent in that :))
Unless your variable write_file already exists, the command $write_file="something" will translate to ="something"(1).
When setting a variable, leave off the $ - you only need it if you want the value of the variable.
In other words, what you need is (note no semicolons needed):
write_file="$(perl /home/spider/web/foo.com/public_html/gen-path.cgi $f)"
(1) It can be even hairier if it is set to something. For example, the code:
write_file=xyzzy
$write_file="something"
will result in something being placed into a variable called xyzzy, not write_file :-)
Related
I'm defining a variable as a composition of other variables and some text, and I'm trying to get this variable to not expand its containing variables on the assigning. But I want it to expand when called later. That way I could reuse the same template to print different results as the inner variables keep changing. I'm truing to avoid eval as much as possible as I will be receiving some of the inner variables from third parties, and I do not know what to expect.
My use case, as below, is to have some "calling stack" so I can log all messages with the same format and keep a record of the script, function, and line of the logged message in some format like this: script.sh:this_function:42.
My attempted solution
called.sh:
#!/bin/bash
SCRIPT_NAME="`basename "${BASH_SOURCE[0]}"`"
CURR_STACK="${SCRIPT_NAME}:${FUNCNAME[0]}:${LINENO[0]}"
echo "${SCRIPT_NAME}:${FUNCNAME[0]}:${LINENO[0]}"
echo "${CURR_STACK}"
echo
function _func_1 {
echo "${SCRIPT_NAME}:${FUNCNAME[0]}:${LINENO[0]}"
echo "${CURR_STACK}"
}
_func_1
So, I intend to get the same results while printing the "${CURR_STACK}" as when printing the previous line.
If there is some built-in or other clever way to log this 'call stack', by all means, let me know! I'll gladly wave my code good-bye, but I'd still like to know how to prevent the variables from expanding right away on the assigning of CURR_STACK, but still keep them able to expand further ahead.
Am I missing some shopt?
What I've tried:
Case 1 (expanding on line 4):
CURR_STACK="${SCRIPT_NAME}:${FUNNAME[0]}:${LINENO[0]}"
CURR_STACK="`echo "${SCRIPT_NAME}:${FUNCNAME[0]}:${LINENO[0]}"`"
CURR_STACK="`echo "\${SCRIPT_NAME}:\${FUNCNAME[0]}:\${LINENO[0]}"`"
called.sh::7 <------------------| These are control lines
called.sh::4 <---------------. .------------| With the results I expect to get.
X
called.sh:_func_1:12 <---´ `-------| Both indicate that the values expanded
called.sh::4 <-------------------------| on line 4 - when CURR_STACK was set.
Case 2 (not expanding at all):
CURR_STACK="\${SCRIPT_NAME}:\${FUNNAME[0]}:\${LINENO[0]}"
CURR_STACK=\${SCRIPT_NAME}:\${FUNCNAME[0]}:\${LINENO[0]}
CURR_STACK="`echo '${SCRIPT_NAME}:${FUNCNAME[0]}:${LINENO[0]}'`"
called.sh::7
${SCRIPT_NAME}:${FUNNAME[0]}:${LINENO[0]} <-------.----| No expansion at all!...
/
called.sh::12 /
${SCRIPT_NAME}:${FUNNAME[0]}:${LINENO[0]} <----´
Shell variables are store plain inert text(*), not executable code; there isn't really any concept of delayed evaluation here. To make something that does something when used, create a function instead of a variable:
print_curr_stack() {
echo "$(basename "${BASH_SOURCE[1]}"):${FUNCNAME[1]}:${BASH_LINENO[0]}"
}
# ...
echo "We are now at $(print_curr_stack)"
# Or just run it directly:
print_curr_stack
Note: using BASH_SOURCE[1] and FUNCNAME[1] gets info about context the function was run from, rather than where it is in the function itself. But for some reason I'm not clear on, BASH_LINENO[1] gets the wrong info, and BASH_LINENO[0] is what you want.
You could also write it to allow the caller to specify additional text to print:
print_curr_stack() {
echo "$#" "$(basename "${BASH_SOURCE[1]}"):${FUNCNAME[1]}:${BASH_LINENO[0]}"
}
# ...
print_curr_stack "We are now at"
(* There's an exception to what I said about variables just contain inert text: some variables -- like $LINENO, $RANDOM, etc -- are handled specially by the shell itself. But you can't create new ones like this except by modifying the shell itself.)
Are you familiar with eval?
$ a=this; b=is; c=a; d=test;
$ e='echo "$a $b $c $d"';
$ eval $e;
this is a test
$ b='is NOT'; # modify one of the variables
$ eval $e;
this is NOT a test
$ f=$(eval $e); # capture the value of the "eval" statement
$ echo $f;
this is NOT a test
I have recently just made this script:
if test -s $HOME/koolaid.txt ; then
Billz=$(grep / $HOME/koolaid.txt)
echo $Billz
else
Billz=$HOME/notkoolaid
echo $Billz
fi
if test -d $Billz ; then
echo "Ok"
else touch $Billz
fi
So basically, if the file $HOME/koolaid.txt file does NOT exist, then Billz will be set as $HOME/koolaid.txt. It then sucesfully creates the file.
However, if I do make the koolaid.txt then I get this
mkdir: cannot create directory : No such file or directory
Any help would be appreciated
Here is a difference between content of a variable and evaluated content...
if your variable contains a string $HOME/some - you need expand it to get /home/login/same
One dangerous method is eval.
bin=$(grep / ~/.rm.cfg)
eval rbin=${bin:-$HOME/deleted}
echo "==$rbin=="
Don't eval unless you're absolutely sure what you evaling...
Here are a couple things to fix:
Start your script with a "shebang," such as:
#!/bin/sh
This way the shell will know that you want to run this as a Bourne shell script.
Also, your conditional at the top of the script doesn't handle the case well in which .rm.cfg exists but doesn't contain a slash character anywhere in it. In that case the rbin variable never gets set.
Finally, try adding the line
ls ~
at the top so you can see how the shell is interpreting the tilde character; that might be the problem.
I'm dealing with a pipeline of predominantly shell and Perl files, all of which pass parameters (paths) to the next. I decided it would be better to use a single file to store all the paths and just call that for every file. The issue is I am using awk to grab the files at the beginning of each file, and it's turning out to be a lot of repetition.
My question is: I do not know if there is a way to store key-value pairs in a file so shell can natively do something with the key and return the value? It needs to access an external file, because the pipeline uses many scripts and a map in a specific file would result in parameters being passed everywhere. Is there some little quirk I do not know of that performs a map function on an external file?
You can make a file of env var assignments and source that file as need, ie.
$ cat myEnvFile
path1=/x/y/z
path2=/w/xy
path3=/r/s/t
otherOpt1="-x"
Inside your script you can source with either . myEnvFile or the more versbose version of the same feature sourc myEnvFile (assuming bash shell) , i.e.
$cat myScript
#!/bin/bash
. /path/to/myEnvFile
# main logic below
....
# references to defined var
if [[ -d $path2 ]] ; then
cd $path2
else
echo "no pa4h2=$path2 found, can't continue" 1>&1
exit 1
fi
Based on how you've described your problem this should work well, and provide a-one-stop-shop for all of your variable settings.
IHTH
In bash, there's mapfile, but that reads the lines of a file into a numerically-indexed array. To read a whitespace-separated file into an associative array, I would
declare -A map
while read key value; do
map[$key]=$value
done < filename
However this sounds like an XY problem. Can you give us an example (in code) of what you're actually doing? When I see long piplines of grep|awk|sed, there's usually a way to simplify. For example, is passing data by parameters better than passing via stdout|stdin?
In other words, I'm questioning your statement "I decided it would be better..."
I have written a small script with which I take the name of a File.
#objectname
echo "objectname"
read ON
Can't get simpler.
I do some processing with the file I get.
gpg -c --no-use-agent "$ON"
For example if I have a file a.exe --> It will encrypt it and give me a file with a different md5 and an extension. Now, the file looks this way a.exe.gpg
Now, if I give it a bind the name of the file directly.
like this for example:
Taken from : this link
# This works
fileName='a.exe.gpg'
md5sum=$(md5sum ${fileName})
echo $md5sum
it returns it properly.
What if I want to do it dynamically.
This is what I tried:
#does not work
gpg -c --no-use-agent "$ON"
fileName= `$ON.gpg`
md5sum= $(md5sum ${fileName})
echo $md5sum
I get this bug here: upload.sh: 1: upload.sh: Fire.exe.gpg: not found and the program does not exit.
May I ask where exactly is the mistake I am doing?
The error is here:
fileName= `$ON.gpg`
There should be no space after =. (Also look at the next line.)
You used back-quotes, which execute $ON.gpg rather than simply evaluating it. Back-quotes are the same as $(...) but less elegant. Use double-quotes for this.
Read Greg's wiki entry on quotes for an ultra-detailed explanation with opinionated commentary. :-)
Be careful when making assignments in shell script. Don't use spaces in any sides of the operator=. Try the following:
fileName="$ON.gpg"
md5sum=$(md5sum ${fileName})
Note that the variable and the assignment operator= are together with no space.
Also, when you use backticks as `expression`, it will be executed by shell like using $(expression), as pointed by user ghoti.
You goofed on fixing the filename.
fileName="$ON.gpg"
I have one loop that creates a group of variables like DISK1, DISK2... where the number at the end of the variable name gets created by the loop and then loaded with a path to a device name. Now I want to use those variables in another loop to execute a shell command, but the variable doesn't give its contents to the shell command.
for (( counter=1 ; counter<=devcount ; counter++))
do
TEMP="\$DISK$counter"
# $TEMP should hold the variable name of the disk, which holds the device name
# TEMP was only for testing, but still has same problem as $DISK$counter
eval echo $TEMP #This echos correctly
STATD$counter=$(eval "smartctl -H -l error \$DISK$counter" | grep -v "5.41" | grep -v "Joe")
eval echo \$STATD$counter
done
Don't use eval ever, except maybe if there is no other way AND you really know what you are doing.
The STATD$counter=$(...) should give an error. That's not a valid assignment because the string "STATD$counter" is not a valid variable name. What will happen is (using a concrete example, if counter happened to be 3 and your pipeline in the $( ) output "output", bash will only expand that line as far as "STATD3=output" so it will try to find a command named "STATD3=output" and run it. Odds are this is not what you intended.
It sounds like everything you want to do can be accomplished with arrays instead. If you are not familiar with bash arrays take a look at Greg's Wiki, in particular this page or the bash man page to find out how to use them.
For example, in the loop you didn't post in your question: make disk (not DISK: don't use all upper case variable names) an array like so
disk+=( "new value" )
or even
disk[counter]="new value"
Then in the loop in your question, you can make statd an array as well and assign it with values from disk by
statd[counter]="... ${disk[counter]} ..."
It's worth saying again: avoid using eval.