时间:2021-07-01 10:21:17 帮助过:43人阅读
//附加 id,变形为
$ar = array(
3 => array(13 => '张三'),
4 => array(14 => '李四'),
);
foreach($ar as $id=>$v) {
list($name, $aeg) = each($v);
$sql = "update tbl_name set aeg=$aeg, name='$name' where id=$id";
//执行数据库操作
}