在一个SQL查询中更新多个表

Posted 2021-04-05

我已经做了一些研究，并发现通过一个查询不可能做到，但可以通过例如事务完成。但我无法使SQL查询正常工作。如果我逐一使用这些陈述，那么他们的工作一切顺利。

我该怎么办？

目标是使用一个表单中的信息更新两个表。

$query = "

START TRANSACTION;

UPDATE projects

SET

projects.project_name = '".$mysqli->real_escape_string($_POST['project_name'])."',

projects.project_group = '".$mysqli->real_escape_string($_POST['project_group'])."',

projects.project_notes = '".$mysqli->real_escape_string($_POST['project_notes'])."',

projects.project_created = '".$mysqli->real_escape_string($_POST['project_created'])."',

projects.project_start = '".$mysqli->real_escape_string($_POST['project_start'])."',

projects.project_delivery  = '".$mysqli->real_escape_string($_POST['project_delivery'])."',

projects.project_orderdetails = '".$mysqli->real_escape_string($_POST['project_orderdetails'])."',

projects.project_owner = '".$mysqli->real_escape_string($_POST['project_owner'])."'

where projects.project_id = '".$mysqli->real_escape_string($_REQUEST['id'])."';

INSERT INTO hours

hours.userhours_id = '".$mysqli->real_escape_string($_POST['project_owner'])."',

hours.projecthours_id = '".$mysqli->real_escape_string($_REQUEST['id'])."',

hours.user_hours = '".$mysqli->real_escape_string($_POST['user_hours'])."'

where projects.project_id = '".$mysqli->real_escape_string($_REQUEST['id'])."';

COMMIT;

";

答案

Mysqli有处理交易的方法，例如http://php.net/manual/en/mysqli.begin-transaction.php。

代码可以是：

$mysqli->begin_tansaction();
$mysqli->query('query1');
$mysqli->query('query2');
$mysqli->commit();