CODEWITHSUNDEEP
mysqlbash
fricative
fricative

Reputation: 13

What is this bash script trying to do with mysql?

The person who wrote the follow bash script has left and I need to figure out the meaning of the follow bash script he wrote. This script is executed in a docker container before running some test cases which requires a running mysql instance.

I guess it is about starting a mysql server, but I am not exactly sure about the exact meaning of each statement in this script. 

echo -n "Loading [+"
( echo "SHOW TABLES;" | mysql mysql 2>/dev/null 1>/dev/null ) || \
    run-mysqld 2>/dev/null 1>/dev/null &

while ! ( echo "SHOW TABLES;" | mysql mysql 2>/dev/null 1>/dev/null ) ; 
do
    echo -n +
    sleep 1
done
echo "] Done."

I had to figure out this because our bitbucket pipeline recently gets stuck and timeout when running this script (previously it was fine). Thanks in advance.

Upvotes: 1

Views: 125

Answers (1)

codeforester
codeforester

Reputation: 42999

This sequence attempts to run SHOW TABLES through mysql, ignoring any output. If mysql fails (because mysqld isn't running), it starts mysqld to run in background.

( echo "SHOW TABLES;" | mysql mysql 2>/dev/null 1>/dev/null ) || \
    run-mysqld 2>/dev/null 1>/dev/null &

The second part of the code just waits for mysqld to start up, which is signaled by the following code exiting 0:

( echo "SHOW TABLES;" | mysql mysql 2>/dev/null 1>/dev/null )

If mysqld doesn't come up with a single attempt, the second part of the code would run forever. Looks like that's what happened.

The simplest way to make this code free from hanging is to put a limit on how long we sleep:

max_sleep=15
sleep=0
while ! ( echo "SHOW TABLES;" | mysql mysql 2>/dev/null 1>/dev/null ) ; 
do
    echo -n +
    sleep 1
    ((sleep++ > max_sleep)) && { echo "Failed to start mysqld] Error."; exit 1; }
done
echo "] Done."
exit 0

Upvotes: 3

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