Php 如何在一个查询中从三个表中进行选择

我有以下数据库结构：

数据库配置列表

id

|

id\u配置

|

类型

db\u配置

id

|

id_列表

|

名称

|

值

db\u config\u lang

id

|

id_列表

|

id_lang

|

名称

|

值

在一个mySQL查询中，我想从

db\u config

和

db\u config\u lang

中选择内容，其中

db\u config.id\u list=db\u config\u list.id

和

db\u config\u lang.id\u list=db\u config\u list.id

我试过：

$sql = 'SELECT * FROM `'._DB_PREFIX_.$this->db_config_list.'` cl
        LEFT JOIN `'._DB_PREFIX_.$this->db_config.'` c
        ON (cl.id = c.id_list)
        LEFT JOIN `'._DB_PREFIX_.$this->db_config_lang.'` cll
        ON (cl.id = cll.id_list)
        WHERE cl.id_config = '.$this->c_id.'
        AND cl.type = "'.$this->c_type.'"
        AND cll.id_lang = "'.$this->default_lang.'"';

但它不能正常工作。当我在两个不同的查询中执行此操作时，它实际上是有效的：

$sql = 'SELECT * FROM `'._DB_PREFIX_.$this->db_config_list.'` cl
        LEFT JOIN `'._DB_PREFIX_.$this->db_config.'` c
        ON (cl.id = c.id_list)
        WHERE cl.id_config = '.$this->c_id.'
        AND cl.type = "'.$this->c_type.'"';

$sql = 'SELECT * FROM `'._DB_PREFIX_.$this->db_config_list.'` cl
        LEFT JOIN `'._DB_PREFIX_.$this->db_config_lang.'` cll
        ON (cl.id = cll.id_list)
        WHERE cll.id_lang = "'.$this->default_lang.'"';

但我想在一个查询中这样做。这可能吗

//编辑

我想我是想做一些不可能的事情。所以我选择了不同的解决方案

$sql = 'SELECT name,value FROM `'._DB_PREFIX_.$this->db_config_list.'` cl
        LEFT JOIN `'._DB_PREFIX_.$this->db_config.'` c
        ON (cl.id = c.id_list)
        WHERE cl.id_config = '.$this->c_id.'
        AND cl.type = "'.$this->c_type.'"';

$sql2 = 'SELECT name,value FROM `'._DB_PREFIX_.$this->db_config_list.'` cl
        LEFT JOIN `'._DB_PREFIX_.$this->db_config_lang.'` cll
        ON (cl.id = cll.id_list)
        WHERE cll.id_lang = "'.$this->default_lang.'"';

$query1 =  Db::getInstance()->executeS($sql);
$query2 =  Db::getInstance()->executeS($sql2);

$query = array_merge($query1, $query2);

//编辑2

以下是sqlfiddle.com的代码

创建代码：

CREATE TABLE IF NOT EXISTS `db_config_list` (
    `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
    `id_config` int(10) unsigned NOT NULL,
    `type` varchar(10) NOT NULL,
    PRIMARY KEY (`id`)
) DEFAULT CHARSET=UTF8;

CREATE TABLE IF NOT EXISTS `db_config` (
    `id_list` int(10) unsigned NOT NULL,
    `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
    `name` varchar(255) NOT NULL,
    `value` text NOT NULL,
    PRIMARY KEY (`id`)
) DEFAULT CHARSET=UTF8;


CREATE TABLE IF NOT EXISTS `db_config_lang` (
    `id_list` int(10) unsigned NOT NULL,
    `id_lang` int(10) unsigned NOT NULL,
    `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
    `name` varchar(255) NOT NULL,
    `value` text NOT NULL,
    PRIMARY KEY (`id`)
) DEFAULT CHARSET=UTF8;

ALTER TABLE `db_config` ADD INDEX(`id_list`);

ALTER TABLE `db_config_lang` ADD INDEX(`id_list`);

ALTER TABLE `db_config` ADD FOREIGN KEY (`id_list`) REFERENCES `db_config_list`(`id`) ON DELETE CASCADE ON UPDATE NO ACTION;

ALTER TABLE `db_config_lang` ADD FOREIGN KEY (`id_list`) REFERENCES `db_config_list`(`id`) ON DELETE CASCADE ON UPDATE NO ACTION;

INSERT INTO `db_config_list` (`id`, `id_config`, `type`) VALUES
(1, 1, 'global');

INSERT INTO `db_config` (`id_list`, `id`, `name`, `value`) VALUES
(1, 1, 'font_family', ''),
(1, 2, 'first_main_color', '#19BCE7'),
(1, 3, 'use_background_image', '1'),
(1, 4, 'background_color', '#F60'),
(1, 5, 'animated_tabs_carousel', '1'),
(1, 6, 'boxed_layout', '1'),
(1, 7, 'loading_animation', '1'),
(1, 8, 'smooth_scroll', '1'),
(1, 9, 'responsiveness', '1'),
(1, 10, 'sticky_header', '1');

INSERT INTO `db_config_lang` (`id_list`, `id_lang`, `id`, `name`, `value`) VALUES
(1, 1, 1, 'center_column_content', 'boo foo'),
(1, 1, 2, 'bottom_column_content', 'boo foo'),
(1, 2, 3, 'center_column_content', 'boo foo'),
(1, 2, 4, 'bottom_column_content', 'boo foo'),
(1, 3, 5, 'center_column_content', 'boo foo'),
(1, 3, 6, 'bottom_column_content', 'boo foo'),
(1, 4, 7, 'center_column_content', 'boo foo'),
(1, 4, 8, 'bottom_column_content', 'boo foo');

选择代码：

SELECT name,value FROM `db_config_list` cl
LEFT JOIN `db_config` c
ON (cl.id = c.id_list)
WHERE cl.id_config = 1
AND cl.type = "global";

SELECT name,value FROM `db_config_list` cl
LEFT JOIN `db_config_lang` cll
ON (cl.id = cll.id_list)
WHERE cll.id_lang = 1
AND cl.id_config = 1
AND cl.type = "global";

---

我认为发生的情况更可能是select是一个*属性，它只会为每个列名选择一个值，而不管有多少个表被联接。如果您可以发布此设置的详细信息，我可以帮助您修改查询。基本上，我相信您希望做以下事情：

$sql = 'SELECT *, c.name AS config_name, cll.name as config_lang_name
        FROM `'._DB_PREFIX_.$this->db_config_list.'` cl
        LEFT JOIN `'._DB_PREFIX_.$this->db_config.'` c
        ON (cl.id = c.id_list)
        LEFT JOIN `'._DB_PREFIX_.$this->db_config_lang.'` cll
        ON (cl.id = cll.id_list)
        WHERE cl.id_config = '.$this->c_id.'
        AND cl.type = "'.$this->c_type.'"
        AND cll.id_lang = "'.$this->default_lang.'"';

编辑： 执行两个单独的查询，然后组合结果集将比调整查询以返回所需的结果集慢得多。这个查询是否更接近您想要的？（以上只是一个例子，既然你已经展示了你的最终目标，我相信这应该是可行的。）

编辑2：

---

看起来您正在使用此查询的框架。我真的建议使用一些参数绑定，而不是将值直接放入查询。

这是可能的。我会核实你的想法。谢谢回复，@WalterWhite，如果米克尔的回答能解决你的问题，记得接受它。@Mikel Bitson你让我大开眼界了。问题解决了，我更新了帖子。谢谢你。@Spencer Doak，但如果这不是实际的解决方案，但有帮助，我是否应该接受他的回答？@WalterWhite，这取决于你。如果你想要一个更好的答案，你当然可以修改你的帖子，把问题留着。接受对你有帮助的答案只是StackOverflow的标准程序。我只是提醒你接受这个答案，这样其他人就能看到什么对你有用。

$sql = 'SELECT cl.name AS config_list_name, cl.value AS config_list_value,
        cll.name AS config_lang_name, cll.value AS config_lang_value
        FROM `'._DB_PREFIX_.$this->db_config_list.'` cl
        LEFT JOIN `'._DB_PREFIX_.$this->db_config.'` c
        ON (cl.id = c.id_list)
        LEFT JOIN `'._DB_PREFIX_.$this->db_config_lang.'` cll
        ON (cl.id = cll.id_list)
        WHERE cl.id_config = '.$this->c_id.'
        AND cl.type = "'.$this->c_type.'"
        AND cll.id_lang = "'.$this->default_lang.'"';