PHP数据库解决方法

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
------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);
}

------解决方案--------------------

$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);