Search Results

The definition of $! is "Expands to the process ID of the most recently executed background (asynchronous) command."

But since I assume your while loop runs after using MyProg without &, then...

myVar=$((myVar+1)) #note that there is no $ sign in myVar
myVar=`expr $myVar + 1` #note that there has to be $ sign in myVar

That's the variable scoping. The loop runs in subshell and hence they doesn't return the value back to the parent shell.

You can force everything to be executed in the current shell.


| {...

The while loop, in this case runs in another subshell with its own scope of variables, which are created and destroy as the while loop starts and finishes.

Some more...

There are different ways to achive this.
Try this for one:
$ MYBUILDPATH=/randomstuff/myBuildRoot/data_tables
$ echo $MYBUILDPATH | sed 's~data_tables[/]*$~~'

/randomstuff/myBuildRoot/
$...