CODEWITHSUNDEEP
bashshell
user265330
user265330

Reputation: 2553

bash script - exit status of previous command different if checked in a function

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.

Upvotes: 2

Views: 971

Answers (1)

camh
camh

Reputation: 42458

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=$?

Upvotes: 7

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