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.c...@gmail.com> 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
- mdyk...@gmail.com
May the Source be with you.
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