gpt4 book ai didi

php - 使用 union 创建 zf2 查询 'not in' 子句

转载 作者:行者123 更新时间:2023-11-29 21:04:55 27 4
gpt4 key购买 nike

我想使用联合查询结果 Not in在ZF2中。将返回该查询的结果

Query: $selectLevelTwoPointFive->combine($selectLevelAll, 'UNION');
Result: ( SELECT `assigned`.`item_id` AS `item_id` FROM `assign_items_level_twopointfive` AS `assigned` ) UNION ( SELECT `assigned`.`item_id` AS `item_id` FROM `assign_items` AS `assigned` )

现在我想在“not in”子句中使用此查询的结果。

$select = $sql->select()->from(array(
"items" => "cu_items"
))->columns(array('item_name'=> new \Zend\Db\Sql\Expression(" group_concat(`items`.`item_name`)") ));

$select->join(array(
"ca" => "cu_areas"
), new Expression(" ca.area_id = items.area_id ")
, array(
'area_name'
), $selectLevelAll::JOIN_INNER);
$select->where->addPredicate(new \Zend\Db\Sql\Predicate\Expression('items.item_id NOT IN (?)',
array($selectLevelTwoPointFive)));

结果是:

SELECT group_concat(`items`.`item_name`) AS `item_name`, `ca`.`area_name` AS `area_name` FROM `cu_items` AS `items` INNER JOIN `cu_areas` AS `ca` ON ca.area_id = items.area_id WHERE `items`.`is_opted` = 'yes' AND `ca`.`is_opted` = 'yes' AND items.item_id NOT IN ((( SELECT `assigned`.`item_id` AS `item_id` FROM `assign_items_level_twopointfive` AS `assigned` ) UNION ( SELECT `assigned`.`item_id` AS `item_id` FROM `assign_items` AS `assigned` )))

它显示 mysql 错误: Error Code: 1064. You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'UNION ( SELECT已分配. item_id AS item_id FROM分配项目AS受让人'位于第 1 行`

我需要的查询是:

SELECT group_concat(items.item_name) AS item_name, ca.area_name AS area_name FROM cu_items
AS items INNER JOIN cu_areas AS ca ON ca.area_id = items.area_id
WHERE items.is_opted = 'yes' AND ca.is_opted = 'yes' AND items.item_id NOT IN
(
(
SELECT assigned.item_id AS item_id FROM assign_items_level_twopointfive AS assigned
UNION ( SELECT assigned.item_id AS item_id FROM assign_items AS assigned )
)
)

实际上,当我使用combine()时,它会添加圆括号(第一个查询)并集(第二个查询)但是in not我需要第一个查询联合第二个查询。

请帮忙。如果有人建议mysql改变它也可以。

最佳答案

感谢上帝,找到该页面。我们需要重写setSpecification()这是答案的链接。 we can override the query output

关于php - 使用 union 创建 zf2 查询 'not in' 子句,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36905594/

27 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com