how can i pass argument to bash in exec command?

Asked Jun 04 '23 at 11:41

Active Jun 04 '23 at 11:48

Viewed 54 times

I have a bash script :

#!/bin/bash
./buind/env/hue shell -i python -c 'exec open("my.py").read,{"pars":[$1,$2]})'

the program gives error $1,$2 invalid syntax it doesnt take my args.( i try ${1},${2} or with double quotes)

When i echo my params in sh it shows my params correct.

i run sh : ./my.sh "mypar1" "mypar2"

How can i solve this problem?

edited Jun 04 '23 at 11:48

asked Jun 04 '23 at 11:41

CompEng

"'$1'" works thanks – CompEng Jun 04 '23 at 11:47
It doesn't work _safely_; you have serious security problems if you go that route and your arguments come from somewhere less trusted than your source code. – Charles Duffy Jun 04 '23 at 15:42
If you use `python -c '...' "$@"` your arguments will be available in `sys.argv`. – Charles Duffy Jun 04 '23 at 15:43
(you can use `__import__("sys").argv[1:]` in place of `[$1,$2]` if you want to keep it all one command inside the Python interpreter; run `python -c 'print(repr(__import__("sys").argv))' one two three` to understand how/why this works). – Charles Duffy Jun 04 '23 at 15:52
1

Are you actually using Python 2? `exec` is a function in Python 3, not a statement: `exec('...')`. – chepner Jun 04 '23 at 17:50
You are in a context of single quotes here. Therefore, the shell does not expand $1 and $2. – user1934428 Jun 05 '23 at 06:20
no I use python 3 @chepner – CompEng Jun 05 '23 at 08:13
@CompEng If you are, then `exec open...` should be a (different) syntax error, because Python 3 doesn't have an `exec` statement. – chepner Jun 05 '23 at 13:34

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