PHP数据库

数组:Array ( [0] => Array ( [id] => 42 ) [1] => Array ( [id] => 49 ) [2] => Array ( [id] => 50 ) [3] => Array ( [id] => 51 ) [4] => Array ( [id] => 52 ) ) ?，这里为数据表里的id字段，怎样根据这些id，修改每一个id对应的另一个字段count？第一个id的count加1，第二个id对应的加2，第三个加3.。。。。这样循环下去

回复讨论(解决方案)

$arr = array ( 0 => array ( 'id' => 42 ) ,1 => array ( 'id' => 49 ), 2 => array ( 'id' => 50 ), 3 => array ( 'id' => 51 ), 4 => array ( 'id' => 52 ) );foreach ($arr as $key => $value) {	$sql = "update tableName set count = count + ".($key+1)." where id = ".$value['id'];	mysql_query($sql);}

$ar = array ( 0 => array ( 'id' => 42 ) ,1 => array ( 'id' => 49 ), 2 => array ( 'id' => 50 ), 3 => array ( 'id' => 51 ), 4 => array ( 'id' => 52 ) );$s = join(',', array_map('current', $ar));$sql = "update tbl_name set count=count+find_inset(id,'$s')";mysql_query($sql);

版主大大的代码 看不懂！

版主好强！不过好像是find_in_set