On 2/22/2012 9:47 AM, Steven Staples wrote:
Good [insert time of day here] all!
I am trying to reorder my auto-inc field in my database, and I have
successfully done it with my "front end" that I use (SQLYog) with the
following code:
SET @var_name = 0;
UPDATE `my_database`.`my_table` SET `id` = (@var_name := @var_name +1);
Now, when I try this within PHP... I can't get it to work at all. I assume
that the SET is the issue, but I am not 100% sure.
I use the PEAR MDB2 class, and I have tried it in 1 statement, but it
failed, and I tried it with 2 statements, it didn't puke on it, but it
didn't work either.
<?php
# db connection is already set up #
echo $db->exec('SET @var_name = 0;')
echo '<br />';
echo $db->exec('UPDATE `my_database`.`my_table` SET `id` =
(@var_name:= @ var_name +1);');
exit;
?>
Does anyone know how I can do this within PHP? Worst case scenario, is that
I just write a php shell() command instead, but I would rather avoid that if
at all possible.
The manual warns us not to rely on repeat user var assignments, but your
approach works for me:
$conn=mysql_connect( ... );
mysql_select_db("test");
mysql_query( "drop table if exists t" ) or exit(mysql_error());
mysql_query( "create table t (id int)" ) or exit(mysql_error());
for( $i=0; $i<10; $i++ ) mysql_query( "insert into t values(0)" ) or
exit(mysql_error());
mysql_query( "set @var=0" ) or exit(mysql_error());
mysql_query( "update t set id=(@var:=@var+1)" ) or exit(mysql_error());
$res = mysql_query( "select id from t" ) or exit(mysql_error());
while( $row = mysql_fetch_row( $res )) echo $row[0], " ";
Output: 1 2 3 4 5 6 7 8 9 10
PB
-----
Thanks in advance!
Steven Staples
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