在一個SQL查詢中更新多個表

Question

我做了一些研究，發現不可能通過一個查詢來完成，而是可以通過事務來完成。 但是我無法使SQL查詢正常工作。 如果我一個接一個地使用這些語句，那么它們都很好。

我該怎么辦？

目標是使用來自一種表單的信息來更新兩個表。

$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;

";

Answer 1

Mysqli具有處理事務的方法，例如http://php.net/manual/en/mysqli.begin-transaction.php 。

代碼可以是：

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