I'm debugging an old shell script; I want to check the values of all the variables used, it's a huge ugly script with approx more than 140 variables used. Is there anyway I can extract the variable names from the script and put them in a convenient pattern like:
#!/bin/sh
if [ ${BLAH} ....
.....
rm -rf ${JUNK}.....
to
echo ${BLAH}
echo ${JUNK}
...
Try running your script as follows:
bash -x ./script.bash
Or enable the setting in the script:
set -x
You can dump all interested variables in one command using:
set | grep -w -e BLAH -e JUNK
To dump all the variables to stdout use:
set
or
env
from inside your script.
You can extract a (sub)list of the variables declared in your script using grep:
grep -Po "([a-z][a-zA-Z0-9_]+)(?==\")" ./script.bash | sort -u
The expression given will match string followed by an egal sign (=) and a double quote ("). So if you don't use syntax such as myvar="my-value" it won't work.
But you got the idea.
grep Options-P --perl-regexp: Interpret PATTERN as a Perl regular expression (PCRE, see below) (experimental) ;-o --only-matching: Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.I'm using a positive lookahead: (?==\") to require an egal sign followed by a double quote.
In bash, but not sh, compgen -v will list the names of all variables assigned (compare this to set, which has a great deal of output other than variable names, and thus needs to be parsed).
Thus, if you change the top of the script to #!/bin/bash, you will be able to use compgen -v to generate that list.
That said, the person who advised you use set -x did well. Consider this extension on that:
PS4=':$BASH_SOURCE:$LINENO+'; set -x
This will print the source file and line number before every command (or variable assignment) which is executed, so you will have a log not only of which variables are set, but just where in the source each one was assigned. This makes tracking down where each variable is set far easier.
