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PHP数据库解决方法

时间:2021-07-01 10:21:17 帮助过:17人阅读

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);

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