Bash combining two for loops

Question

Im trying to create a dynamic cronjob which executes a PHP script which needs a DB argument for specific directorys.

The Output should look like this:

/usr/bin/php /var/www/html/test.php db_test

I got following:

    #!/bin/bash
    DIR="/var/www/html/*"
    PHP="/usr/bin/php"
    ACCOUNTS=$(ls -d $DIR)
    DBPREFIX="db_"
    DB=$( find /var/www/html/ -mindepth 1 -maxdepth 1 -type d | awk -F'/' '{print $5}' )

    for CUSTOMER in $ACCOUNTS
    do
            echo $PHP $CUSTOMER $DBPREFIX$DB
    done

This outputs following:

/usr/bin/php /var/www/html/test.php db_test test2 test3 test4

The Problem is that the DBPREFIX and DB vars need also to go trough a loop so that i get for every single db a complete command.

How am I able to do so?

Don't use uppercase for your private variables; uppercase variable names are reserved for system use. — tripleee, Dec 07 '17 at 10:08
And hard-coding a path to PHP is just silly. Amend the `PATH` if you have to, but allow your users to override the system location if they need to (good for testing, for example!) — tripleee, Dec 07 '17 at 10:09
Annnnd [don't use `ls` in scripts](http://mywiki.wooledge.org/ParsingLs). — tripleee, Dec 07 '17 at 10:09
Probably you're looking for https://stackoverflow.com/questions/28725333/looping-over-pairs-of-values-in-bash — tripleee, Dec 07 '17 at 10:10

score 1 · Accepted Answer · answered Dec 07 '17 at 10:15

1

You seem to be looking for something like

for dir in /var/www/html/*/; do
    base=${dir#/var/www/html/}
    db=${base%/}
    echo php "$db" "db_$db"   # or maybe php "${dir%/}" "db_$db"?
done

You could probably use something like find -printf "..." to accomplish much the same thing, but the -printf option to find is a GNU extension.

answered Dec 07 '17 at 10:15

tripleee

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Thank you very much thats what I was searching for. Could you explain what the var base and db are doing? – patrick_ Dec 07 '17 at 10:29
I'm discarding the directory name with `${var#prefixtoremove}` and the trailing slash with `${var%suffixtoremove}`. – tripleee Dec 07 '17 at 10:42
Read about [parameter expansions](http://wiki.bash-hackers.org/syntax/pe#substring_removal) for further explanation. – PesaThe Dec 07 '17 at 11:21

Bash combining two for loops

1 Answers1