3

我知道这个问题已经被问过很多次了，但是普遍接受的答案对我没有帮助。但很偶然，我偶然发现了一个答案。

这是设置：我有大量查询（主要是 CREATE TABLE）通过连接。但是 CREATE TRIGGER 不断抛出可怕的 2014 错误。这与打开的游标无关，因为即使它是程序中唯一的命令也会发生。例如，这失败了：

<?php
$db = new PDO ($cnstring, $user, $pwd);
$db->setAttribute (PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$db->setAttribute (PDO::ATTR_EMULATE_PREPARES, false);
$db->setAttribute (PDO::MYSQL_ATTR_USE_BUFFERED_QUERY, true);

$st = $db->query ("CREATE TRIGGER `CirclesClosureSync` AFTER INSERT ON Circles
FOR EACH ROW BEGIN
    INSERT INTO CirclesClosure (ancestor, descendant) 
    SELECT ancestor, NEW.ID from CirclesClosure WHERE descendant=NEW.Parent;
    INSERT INTO CirclesClosure (ancestor, descendant) values (NEW.ID, NEW.ID);
END;");

$st->closeCursor();
?>

这似乎类似于涉及创建存储过程的其他问题。

这是 php 5.4.5、MySql 5.5、Windows XP（尽管它在其他 Windows 上也失败了）

4

2 回答 2

6

花了点功夫，但我发现当我把 ATTR_EMULATE_PREPARES=false 拿出来（默认是模拟）时，它起作用了：

<?php
$db = new PDO ($cnstring, $user, $pwd);
$db->setAttribute (PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
//$db->setAttribute (PDO::ATTR_EMULATE_PREPARES, false);
$db->setAttribute (PDO::MYSQL_ATTR_USE_BUFFERED_QUERY, true);

$st = $db->query ("CREATE TRIGGER `CirclesClosureSync` AFTER INSERT ON Circles
FOR EACH ROW BEGIN
    INSERT INTO CirclesClosure (ancestor, descendant) 
    SELECT ancestor, NEW.ID from CirclesClosure WHERE descendant=NEW.Parent;
    INSERT INTO CirclesClosure (ancestor, descendant) values (NEW.ID, NEW.ID);
END;");

$st->closeCursor();
?>

希望这可以帮助某人

于 2013-06-10T19:29:08.360 回答
2

即使 PDO::ATTR_EMULATE_PREPARES = false，也可以通过 PDO::exec() 创建触发器：

<?php
$db = new PDO ($cnstring, $user, $pwd);
$db->setAttribute (PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$db->setAttribute (PDO::ATTR_EMULATE_PREPARES, false);
$db->setAttribute (PDO::MYSQL_ATTR_USE_BUFFERED_QUERY, true);

$db->exec("CREATE TRIGGER `CirclesClosureSync` AFTER INSERT ON Circles
FOR EACH ROW BEGIN
    INSERT INTO CirclesClosure (ancestor, descendant) 
    SELECT ancestor, NEW.ID from CirclesClosure WHERE descendant=NEW.Parent;
    INSERT INTO CirclesClosure (ancestor, descendant) values (NEW.ID, NEW.ID);
END;");
于 2015-05-18T08:39:36.027 回答