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