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首页 > 解决方案 > 有人可以建议一种使用带有多个“where”子句的 Doctrine (QueryBuilder) 进行此查询的方法吗？

php - 有人可以建议一种使用带有多个“where”子句的 Doctrine (QueryBuilder) 进行此查询的方法吗？

问题描述

我是 QueryBuilder 的新手，我正在尝试执行一个 POST 请求（使用 JSON）来检索我的数据库中的一些信息。

我使用数组是因为每个属性都可以有多个值。这是我当前发送的 JSON：

{
    "name":["Martin"],
    "state":["Ohio", "Texas"],
    "job":["Photographer", "Reporter"]
}

这是我的数据库：

ID  |   NAME        |   FIRST_NAME  |   STATE       |   JOB             |   SPEAK_FRENCH
1   |   Martin      |   John        |   Ohio        |   Photographer    |   1
2   |   Martin      |   Max         |   Ohio        |   Reporter        |   1
3   |   Martin      |   Sophie      |   Texas       |   Model           |   1
4   |   Alexander   |   David       |   Kansas      |   Author          |   0
5   |   Archie      |   Kira        |   Maine       |   Photographer    |   1
6   |   Lushen      |   Albert      |   Nevada      |   Pilot, Model    |   1
7   |   Wilkins     |   Minnie      |   Utah        |   Tailor          |   0
8   |   Martin      |   Thomas      |   Texas       |   Reporter        |   1
9   |   Patino      |   Stephen     |   Virginia    |   Pilot, Reporter |   1
10  |   Ting        |   Maria       |   Nevada      |   Dentist         |   0
11  |   Brown       |   Barbara     |   Virginia    |   Reporter        |   1
12  |   Martin      |   William     |   Texas       |   Photographer    |   1
13  |   Zachary     |   Thomas      |   Virginia    |   Telephonist     |   1

我想要的请求：

SELECT * FROM 'application'
WHERE SPEAK_FRENCH = 1
AND NAME = "Martin"
AND STATE = "Ohio"
AND JOB LIKE "%Photographer%"
OR SPEAK_FRENCH = 1
AND NAME = "Martin"
AND STATE = "Ohio"
AND JOB LIKE "%Reporter%"
OR SPEAK_FRENCH = 1
AND NAME = "Martin"
AND STATE = "Texas"
AND JOB LIKE "%Photographer%"
OR SPEAK_FRENCH = 1
AND NAME = "Martin"
AND STATE = "Texas"
AND JOB LIKE "%Reporter%"

而我目前在 Symfony 中所做的事情并没有按照我想要的方式工作：

$repository = $this->getDoctrine()->getRepository(Application::class);
$query = $repository->createQueryBuilder('request');
$temp_name = 0;
$temp_state = 0;
$temp_job = 0;
foreach ($app->getName() as $name) {
    $temp_name = $temp_name + 1;

    $query = $query->orWhere('request.speakFrench = 1')
        ->andWhere('request.name LIKE :JSONname' . strval($temp_name))
        ->setParameter('JSONname' . strval($temp_name), $name);

    foreach ($app->getState() as $state) {
        $temp_state = $temp_state + 1;

        $query = $query->andWhere('request.state LIKE :JSONstate' . strval($temp_state))
            ->setParameter('JSONstate' . strval($temp_state), $state);

        foreach ($app->getJob() as $job) {
        $temp_job = $temp_job + 1;

        $query = $query->andWhere('request.job LIKE :JSONjob' . strval($temp_job))
            ->setParameter('JSONjob' . strval($temp_job), '%' . $job . '%');
        }
    }
}

我的目标是得到这个结果：

ID  |   NAME        |   FIRST_NAME  |   STATE       |   JOB             |   SPEAK_FRENCH
1   |   Martin      |   John        |   Ohio        |   Photographer    |   1
2   |   Martin      |   Max         |   Ohio        |   Reporter        |   1
8   |   Martin      |   Thomas      |   Texas       |   Reporter        |   1
12  |   Martin      |   William     |   Texas       |   Photographer    |   1

我的代码运行不正常，它不会发送任何错误，而只会返回ID : 8。

我正在寻找一种方法来执行仅返回ID : 1, 2, 8 and 12.

非常感谢您未来的回答。

PS：我正在运行 Symfony 4.3.11。

标签： phpmysqlsqlsymfonydoctrine

解决方案

以下似乎是符合您要求的有效查询：

SELECT * 
  FROM application
 WHERE SPEAK_FRENCH = 1
    AND NAME = "Martin"
    AND STATE = IN("Ohio","Texas")
    AND (JOB LIKE "%Photographer%" OR JOB LIKE "%Reporter%");

另外，请参阅在数据库列中存储分隔列表真的那么糟糕吗？

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