Recursive function in AIX bash

Question

Overview:

I want to code a function that will keep on executing till the value is equal to 0 and if the value is greater than 0, I'm writing the value into file and break-ing the loop.

Problem:

If the function is executed 8 times then while exiting the function in case the value is greater than 0 then 8 times the value is written into the file instead of 1 time. Kindly help me with the issue I'm facing.

PFB the code used,

#!/usr/bin/ksh

    fx_Running()
    {
        v_line=$1
        v_SRC_NM=$2

        v_VAR =`echo $v_line|awk -F',' '{print $1}'`

        v_STATUS=`wc -l $v_VAR | awk '{print $1}`

        if [ $v_STATUS == 0 ]
        then
            fx_Running $v_line $v_SRC_NM 
        fi

        if [ $v_STATUS == 1 -o $v_STATUS == 2 ]
            then
                ${v_line} >> ${v_COMPLETED_LIST}
                break
        fi

        if [ $v_JOB_STATUS == 3 ]
        then        
            ${v_line} >> ${v_FAILED_LIST}
            break
        fi

    }

Why does your title mention bash if your script starts with #!/usr/bin/ksh? — Tom Fenech
– Tom Fenech, Commented Apr 22, 2016 at 11:05

agc · Accepted Answer · 2021-09-14 20:20:01Z

1

Using break is incorrect, use return instead. Specifically break exits a loop, whereas return exits a function.

Here's a demo of the problem:

foo(){ break ; } ; for f in `seq 10` ; do echo $f ; foo ; done

Output:

edited Sep 14, 2021 at 20:20

answered Apr 23, 2016 at 0:03

agc

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