[SOLVED] Bash Shell Script - Store a variable as a string not an integer

RML1992 · 09-12-2012, 07:02 AM

Is it possible to store a variable as a string rather than an integer? For instance a " date +'%d%m%y' " so as to keep the leading zero and avoid the "value is too great for base" error when the value begins in either 08 or 09.

414N · 09-12-2012, 07:16 AM

bash already considers all variables as strings, not ints.
If you're missing leading zeroes then it's likely you're using some kind of operator which translates the string from the "date" command into an int.

kabamaru · 09-12-2012, 07:23 AM

You might want to have a look at this. Bash sees the leading zeros and thinks this is an octal value. That's why 08 and 09 are "too great for base".

Also see this about Bash variables.

RML1992 · 09-12-2012, 07:58 AM

Thank you very much for your replies!

I am setting the variable as follows:

backupDate=$(date +'%d%m%y');

Does it automatically convert it into a string? As it still seems to bring up the same error that it is 'too great for base'. I have also tried emitting the leading zero and then adding it back onto the front of a separate variable calling in the first to follow, with no such luck, the same error is produced.

Is it at all possible for it to not be recognised as Octal so that the leading zero will remain?

suicidaleggroll · 09-12-2012, 08:10 AM

That works for me, what system are you on?

RML1992 · 09-12-2012, 08:28 AM

Thanks!

The system I'm running is Cent 5.6. Currently the script reads as follows:

#!/bin/bash

dow=$(date +'%a');

function determineBackup {
if [ "$dow" == "Wed" ]; then
let backupDate=$(date +'%d%m%y' -d "3 days ago");
else
let backupDate=$(date +'%d%m%y' -d "1 day ago");
fi
}

determineBackup;

echo Backup Date: "$backupDate";

exit

As 3 (and 4) days ago are 08 or 09 then it kicks up the error 'too great for base'.

414N · 09-12-2012, 08:41 AM

Why are you using let?
Simply define the backupDate variable in a global scope (right after dow, for example) and you'll be able to access it both from inside determineBackup and outside of it:

Code:

#!/bin/bash

dow=$(date +'%a')
backupDate=""

function determineBackup {
if [ "$dow" = "Wed" ]; then
	backupDate=$(date +'%d%m%y' -d "3 days ago")
else
	backupDate=$(date +'%d%m%y' -d "1 day ago")
fi
}

determineBackup

echo Backup Date: "$backupDate"

Please use code tags ([code]insert code here[/code]) to post source code or commands output, as it makes it more readable.
Note that all those ";" I removed are useless, as the exit at the end of the script with no value.

suicidaleggroll · 09-12-2012, 08:44 AM

Quote:

Originally Posted by 414N

Why are you using let?
Simply define the backupDate variable in a global scope (right after dow, for example) and you'll be able to access it both from inside determineBackup and outside of it:

Code:

#!/bin/bash

dow=$(date +'%a')
backupDate=""

function determineBackup {
if [ "$dow" = "Wed" ]; then
	backupDate=$(date +'%d%m%y' -d "3 days ago")
else
	backupDate=$(date +'%d%m%y' -d "1 day ago")
fi
}

determineBackup

echo Backup Date: "$backupDate"

Please use code tags ([code]insert code here[/code]) to post source code or commands output, as it makes it more readable.
Note that all those ";" I removed are useless, as the exit at the end of the script with no value.

Correct. The problem is not the leading 0 in your date output, the problem is your syntax. Get rid of the let's and the useless semicolons and it should work (does for me).

RML1992 · 09-12-2012, 09:19 AM

Thank you for your help! Works brilliantly now