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php - 如何在一个查询中从三个表中进行选择

转载 作者:行者123 更新时间:2023-11-29 01:55:25 25 4
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我有以下数据库结构:

db_config_list编号 | id_config | 类型

数据库配置编号 | id_list | 名称 |

db_config_lang编号 | id_list | id_lang | 名称 |

在一个 mySQL 查询中,我想从 db_configdb_config_lang 中选择内容,其中 db_config.id_list = db_config_list.id db_config_lang.id_list = db_config_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执行此操作。这可能吗?

//编辑

我想我曾试图实现不可能的事情。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

这是 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";

最佳答案

我认为正在发生的事情更可能只是因为 select 是一个 * 属性——它只会为每个列名选择一个值,而不管连接了多少个表。如果你可以发布 SQL Fiddle对于此设置,我可以帮助您修改查询。从本质上讲,我相信您希望做类似的事情:

$sql = 'SELECT *, c.name AS config_name, cll.name as config_lang_name
FROM `'._DB_PREFIX_.$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。这个查询更接近你想要的吗? (以上只是一个例子,既然你已经展示了你的最终目标,我相信这应该可行。)

$sql = 'SELECT cl.name AS config_list_name, cl.value AS config_list_value,
cll.name AS config_lang_name, cll.value AS config_lang_value
FROM `'._DB_PREFIX_.$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.'"';

编辑 2:看起来您正在为此查询使用框架。我真的建议使用一些参数绑定(bind),而不是将值直接放入查询中。

关于php - 如何在一个查询中从三个表中进行选择,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30488829/

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