From: Michael Dykman
Date: October 3 2012 7:41pm
Subject: Re: passing shell variable to the SET data type in parentheses
List-Archive: http://lists.mysql.com/mysql/228301
Message-Id: <CANLyif_B8U3QdgW0n+tkzg5M=rac+dM4iGENQoCzLaeaLhr-xg@mail.gmail.com>
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What is the result if you echo that line instead of running it?  ie:

echo mysql -u $user -p${password} --skip-column-names -e 'ALTER TABLE
' $table' MODIFY '$kolom' SET(" '$var' ");' $database ;

I'm not clear exactly what the text is of the command you are trying to run.

 - michael dykman

On Wed, Oct 3, 2012 at 9:35 AM, Morning Star
<morning.star.crew@stripped> wrote:
> Hi guys,
> i have a problem when trying to pass shell variable to the SET data
> type in parentheses.
> i have a variable like this:
>
> $ echo $var
> "value1","value2","value3"
>
> what i did:
> mysql -u $user -p${password} --skip-column-names -e 'ALTER TABLE
> '$table' MODIFY '$kolom' SET(" '$var' ");' $database ;
>
> the result:
> ERROR 1064 (42000) at line 1: You have an error in your SQL syntax;
> check the manual that corresponds to your MySQL server version for the
> right syntax to use near 'value1","value2","value3' at line 1
>
> what do i have to do? please help me.
>
> Greetings,
>
> Marco
>
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-- 
 - michael dykman
 - mdykman@stripped

 May the Source be with you.