I'm writing a shell script using #!/bin/bash instead of #!/bin/sh because of the substitution: ${!variable}, which won't work with sh. My main problem is the following (just a summarized example, the script is much more complex):
Thank you!
---------- Post updated at 02:00 PM ---------- Previous update was at 12:32 PM ----------
At least someone to tell me the POSIX alternative for ${!variable}, please:
Last edited by teresaejunior; 08-28-2010 at 02:12 PM..
Reason: You could paste this now in a terminal
I have this situation in my script (simplified):
A=C
C=10
I need to get number 10 using just A variable.
I tried with :
echo $`echo $A` - but i get $C string (i need number)
Thanks very much for any help! (1 Reply)
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################################################
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OK, I'm striving to abide by all the rules this time.
Here is a fragment of my windows10/cygwin64/bash script:
export BUPLOG=$(BackupRecords --log "$src")
robocopy $(BackupRecords -mrbd "$src" --path "$src") $(BackupRecords --appSwitches "$src") "$src" "$dst" $(BackupRecords --fileSwitches... (0 Replies)
OK, I'm striving to abide by all the rules this time.
Here is a fragment of my windows10/cygwin64/bash script:
export BUPLOG=$(BackupRecords --log "$src")
robocopy $(BackupRecords -mrbd "$src" --path "$src") $(BackupRecords --appSwitches "$src") "$src" "$dst" $(BackupRecords --fileSwitches... (15 Replies)