PDO在准备语句中不能使用整数来比较mysql ENUM [英] PDO cannot compare mysql ENUM using integers in prepared statements

问题描述

我正在使用PDO和准备的语句，但是当将ENUM字段与整数进行比较时，我似乎无法获得任何结果。

I am using PDO and prepared statements, but i cannot seem to get any results when comparing an ENUM field with an integer.

示例：

$db = new PDO('mysql:host=localhost;dbname=****', '***', '***');
$s = $db->prepare('SELECT id FROM t2 WHERE lang = ?');

$s->execute(array('en')); // Works              
print_r($s->fetchAll());

$s->execute(array(2)); // Does not work            
print_r($s->fetchAll());

我正在测试此表：

DROP TABLE IF EXISTS t2;
CREATE TABLE t2 (
  id int(10) NOT NULL AUTO_INCREMENT,
  lang enum('no','en','fr') NOT NULL,
  PRIMARY KEY (id)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8;
INSERT INTO  t2 (id, lang) VALUES (NULL , 'en');

有关如何让这项工作有任何想法？

Any idea on how to get this to work?

我正在转换为PDO，我不想在我的应用程序中重写所有的常量，枚举和查询：（

I am converting to PDO, and I'd prefer not to rewrite all constants, enums, and queries in my app :(

推荐答案

2 不是一个有效的ENUM元素，很简单。

2 is not a valid ENUM element. It's as simple as that.

在原始MySQL中工作，当您执行 lang = 2 时，它是暴露了列表的底层存储机制（基本上它只是一个标准化的值，但标准化是从你隐藏的通过ENUM列）

The reason it works in raw MySQL when you do lang = 2, is that it's exposing the underlying storage mechanism for the list (basically it's just a normalized value, but the normalization is hid from you by the ENUM column).

我建议不要这样做，MySQL隐藏了一个原因的实现，意识到使用 2 进行比较只不过是一个魔术数字。 ..

I'd suggest not trying to do this. MySQL hides the implementation for a reason. Realize that using 2 for the comparison is nothing more than a magic number...

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