Should have included an example:
echo $var
"value1","value2","value3"
echo mysql -u $user -p${password} --skip-column-names -e 'ALTER TABLE
' $table' MODIFY '$kolom' SET(" '$var' ");' $database ;
mysql -u -p --skip-column-names -e ALTER TABLE
MODIFY SET(" "value1","value2","value3" ");
echo mysql -u $user -p${password} --skip-column-names -e 'ALTER TABLE
' $table' MODIFY '$kolom' SET(" $var ");' $database ;
mysql -u -p --skip-column-names -e ALTER TABLE
MODIFY SET(" $var ");
echo mysql -u $user -p${password} --skip-column-names -e 'ALTER TABLE
' $table' MODIFY '$kolom' SET( '$var' );' $database ;
mysql -u -p --skip-column-names -e ALTER TABLE
MODIFY SET( "value1","value2","value3" );
I believe that this is what Michael was eluding to as well...
[email protected]
________________________________
From: Garot Conklin <[email protected]>
To: Michael Dykman <[email protected]>; Morning Star
<[email protected]>
Cc: "[email protected]" <[email protected]>
Sent: Wednesday, October 3, 2012 3:50 PM
Subject: Re: passing shell variable to the SET data type in parentheses
are you trying to get the value of var encapsulated with ' ' marks?
Typical shell expansion while within " " will output the literal:
var=10
echo "'$var'"
'10'
Have you tried removing the single quotes? The shell can be funny with ' and "
[email protected]
________________________________
From: Michael Dykman <[email protected]>
To: Morning Star <[email protected]>
Cc: [email protected]
Sent: Wednesday, October 3, 2012 3:41 PM
Subject: Re: passing shell variable to the SET data type in parentheses
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
<[email protected]> 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
>
> --
> MySQL General Mailing List
> For list archives: http://lists.mysql.com/mysql
> To unsubscribe: http://lists.mysql.com/mysql
>
--
- michael dykman
- [email protected]
May the Source be with you.
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe: http://lists.mysql.com/mysql
