Trying to check if a Mysql user exists in a bash script

Question

I am trying to check if a MYSQL user exists. This is as far as I have got. I fall down on capturing the answer from the output.

 #!/bin/bash
echo -e "What is the MYSQL username called"
    read DBUSER
    if [ -z "$DBUSER" ]
    then
    exit

    mysql -uUSER -pPASS -e "SELECT EXISTS(SELECT 1 FROM mysql.user WHERE user = '$DBUSER')";

    if 
    yes
    do this
    else
    do this

this is the output I am getting

+-----------------------------------------------------+
| EXISTS(SELECT 1 FROM mysql.user WHERE user = 'bob') |
+-----------------------------------------------------+
|                                                   1 |
+-----------------------------------------------------+

Can any one help please

Rudi Kershaw · Accepted Answer · 2018-03-15 14:11:22Z

11

Thanks very much for your help. Here is the final result working.

It needs the -sse

RESULT_VARIABLE="$(mysql -uUSER -pPASS -sse "SELECT EXISTS(SELECT 1 FROM mysql.user WHERE user = '$DBUSER')")"

if [ "$RESULT_VARIABLE" = 1 ]; then
echo "TRUE"
else
  echo "FALSE"
fi

edited Mar 15, 2018 at 14:11

Rudi Kershaw

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answered Jan 23, 2015 at 14:55

Gary

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1 Comment

aequalsb Over a year ago

rather obvious, but just to be affirmative... this worked for me with shell_exec -->

$is_mysql_user = shell_exec('mysql --defaults-extra-file=MYPATHTOSTOREDCREDENTIALS -sse "SELECT EXISTS(SELECT 1 FROM mysql.user WHERE user = \'' . $user_name . '\')"');

fancyPants · Accepted Answer · 2015-01-23 14:58:12Z

3

Assigning the result to a variable can be done like this:

RESULT_VARIABLE="$(mysql -uUSER -pPASS -se "SELECT EXISTS(SELECT 1 FROM mysql.user WHERE user = '$DBUSER')")"

And you can also alias a column in MySQL, btw.

SELECT EXISTS(SELECT 1 FROM mysql.user WHERE user = '$DBUSER') AS does_it_exist

edited Jan 23, 2015 at 14:58

answered Jan 23, 2015 at 14:10

fancyPants

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1 Comment

Gary Over a year ago

Thanks very much. I am then trying to pick up the result. This gives me an error if [ $RESULT_VARIABLE = "1" ]; then echo "true" else echo "try again" fi

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