I have the following unix shell script, in which i have two integer variables namely a and b.
If a is greater then or equal to b then shell script should exit with returning 0.
Else it should exit with returning 1.
Script: ConditionTest.sh
#!/bin/sh
a=10
b=20
if [ $a -ge $b ]
then
exit 0
else
exit 1
fi
....
....
....
Running Script:
$ ./ConditionTest.sh
$
Note: I am not getting any return value after executing the file.
The shell puts the exit status of the last command in the variable ?.
You could simply inspect it:
mycommand
echo $?
... or you could use it to do something else depending on its value:
mycommand && echo "ok" || echo "failed"
or alternatively, and slightly more readable:
if mycommand; then
# exit with 0
echo "ok"
else
# exit with non-zero
echo "failed"
if
$? expands to the value, not the variable. The variable is called ?.
Your script looks fine; you did everything right.
#!/bin/sh
a=10
b=20
if [ $a -ge $b ]
then
exit 0
else
exit 1
fi
So here's where we run it and check the return value:
$ sh test.sh
$ echo $?
1
$
10 is not greater than or equal to 20.
Another way to test it would be like this:
$ sh test.sh && echo "succeeded" || echo "failed"
failed
As noted in the comments, you should also quote your variables, always:
if [ $a -ge $b ]
Should be:
if [ "$a" -ge "$b" ]
To add to the previous answers, the key idea you should understand is that every program provides a number when exiting. That number is used as a way to report if the command has completed its operation successfully, and if not, what type of error has occurred.
Like mentioned, the exit code of the last command executed can be accessed with $?.
The reason nothing was printed by your script, is that your script returned 1, but the exit code of a command is not printed. (This is analogous to calling a function, you get a return value from the function but it's not printed)
$?. Doecho $?after running the script. If you want to see the value as part of the script output then print it:echo 0; exit 0andecho 1; exit 1.