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Home/ Questions/Q 7872903
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Editorial Team
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Editorial Team
Asked: June 3, 20262026-06-03T02:21:13+00:00 2026-06-03T02:21:13+00:00

I don’t get this – if I check the exit status of a command

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I don’t get this – if I check the exit status of a command in a function and store in a local variable, I always get the answer 0. From outside the function, I get the correct exit status.

#!/bin/bash

function check_mysql()
{
    local output=`service mysql status`
    local mysql_status=$?

    echo "local output=$output"
    echo "local status=$mysql_status"
}

check_mysql

g_output=`service mysql status`
g_mysql_status=$?

echo "g output=$g_output"
echo "g status=$g_mysql_status"

Output is:

local output=MySQL is running but PID file could not be found..failed
local status=0
g output=MySQL is running but PID file could not be found..failed
g status=4

The status of 4 is the correct one.

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  1. Editorial Team
    Editorial Team
    2026-06-03T02:21:15+00:00Added an answer on June 3, 2026 at 2:21 am

    The local command is run after the service mysql status command in your function. It is that which is returning 0. You are losing the return status of the service command.

    Split the local statement into two:

    local output
    local mysql_status
    
    output=`service mysql status`
    mysql_status=$?
    
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