Array
(
[0] => Array
(
[id] => 1
)
[1] => Array
(
[id] => 16
)
[2] => Array
(
[id] => 21
)
[3] => Array
(
[id] => 22
)
[4] => Array
(
[id] => 2
)
)
上面的数组是$id
$grp_id="";
foreach($id as $gid)
{
$grp_id .= " LIKE %".$gid['id']."% AND rj.applyto";
}
my query is
$sql="SELECT * FROM `table` j INNER JOIN `table2` rj ON j.eid = rj.eid WHERE rj.applyto LIKE $grp_id";
我写错了,请建议我改正
你真的需要在这里使用LIKE
吗?1
和2
太常见了,我不认为这是你想要的。
假设这不是你真正想要的,就使用IN()
。更加简单。
// Create comma separated list of IDs
$ids = implode(',', $id);
// Use IN() in our WHERE clause
$sql="SELECT * FROM `table` j INNER JOIN `table2` rj ON j.eid = rj.eid WHERE rj.applyto IN($ids);
你只需要用逗号分隔你的id并使用FIND_IN_SET
$ids = implode(',', $id_array);
$sql="SELECT * FROM `table` j INNER JOIN `table2` rj ON j.eid = rj.eid WHERE FIND_IN_SET(rj.applyto, $ids)";
试试这个!