From: Morning Star
Date: October 4 2012 1:17am
Subject: Re: passing shell variable to the SET data type in parentheses
List-Archive: http://lists.mysql.com/mysql/228312
Message-Id: <CAD9kJ0PJ=Jr4RY+93GUBHSVkN01w-RfFcayRJ__M5hP3CPBMgA@mail.gmail.com>
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Thanks guys for all the corrections.
I fix it by executing:

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

:)

Greetings,

Marco

On Thu, Oct 4, 2012 at 4:49 AM,  <hsv@stripped> wrote:
>>>>> 2012/10/03 20:35 +0700, Morning Star >>>>
> $ 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.
> <<<<<<<<
> If your "var" really has three pairs of double-quotes, with the double-quotes within the single-quotes, you get that this
> ... SET(" "value1","value2","value3" ");
> is passed to MySQL client, which is not valid.
>
>
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