I am new to bash scripting but after trying several syntax approaches and researching, I am a bit stuck storing the result of an external script call in my bash script. $r has no visible value when I echo it...
From the command line, it works as expected:
./external-prog 23334
echo $?
2
#!/bin/bash
# build the command
c="./external-prog 23334"
# invoke command that returns an integer value
eval "$c"
#collect result in $r
r=$(eval "$?")
#see result
echo $r
It seems you just want to run the command and retrieve it's returned value, so there is no need far eval:
#!/bin/bash
# run the command
./external-prog 23334
#collect result in $r
r=$?
#see result
echo $r
You can do it as below, storing the external script result in a variable :
c=$(./external-prog 23334)
echo $c
This sequence $?will return the error code of the last process not the output.
Check this out
$ echo ok; echo $?
ok
0
First echo printed 'ok' and the second one printed 0, which means command succeeded. And a non 0 code means some kind of error occured.
So this
./external-prog 23334
echo $?
2
Means that external-prog failed and the error code is 2 whatever it means.
And to catch the output of some command to a var you need this
var=$(echo ok)
$ echo $var
ok
