兽人学说的问题

Question

我有这样的东西：

$languages = ["English", "German", "Spanish"];

作业实体：

/**
* @FlowEntity
*/
class Job {
/**
* The name of first language for the job (M:1 unidirectional)
*
* @var PATHLanguage
* @ORMManyToOne
*/
protected $language1;
/**
* The name of second language for the job (M:1 unidirectional)
*
* @var PATHLanguage
* @ORMManyToOne
*/
protected $language2;
}

和语言实体：

/**
* @FlowEntity
*/
class Language
{
/**
* The language name
*
* @var string
* @FlowIdentity
* @FlowValidate(type="Text")
* @FlowValidate(type="NotEmpty")
* @FlowValidate(type="StringLength", options={ "minimum"=1, "maximum"=80})
* @ORMColumn(length=80)
*/
protected $name;
}

如何在$languages中获得所有具有语言1或语言2的工作？ 我尝试了以下方法，但它不起作用...

我得到空的结果。

$queryBuilder
->resetDQLParts()
->select("job")
->from("Job", "job")
->andWhere(
$queryBuilder->expr()->orX(
$queryBuilder
->innerJoin('job.language1', 'language1')
->andWhere($queryBuilder->expr()->in("language1.name", $languages)),
$queryBuilder
->innerJoin('job.language2', 'language2')
->andWhere($queryBuilder->expr()->in("language2.name", $languages))
)
);

有什么想法吗？

Answer 1

我不完全理解你写的查询，这肯定不是我过去使用QueryBuilder的方式。我已经重写了如何使用它。

语言字段没有说明它们引用的实体应该是：

/** @ORMManyToOne(targetEntity='AppBundleEntityLanguage')

假设表结构为：

ID | Job Name | Language1 | Language2
1  | Job1     | French    | English
2  | Job2     | English   | Spanish

该 SQL 将是：

SELECT * FROM `jobs` 
WHERE `language1` IN ("English", "French") 
OR    `language2` IN ("English", "French");

翻译过来就是：

$this->createQueryBuilder('job')            
->where('job.language1 IN (:languages)')
->orWhere('job.language2 IN (:languages)')
->setParameter("languages", $languages)
->getQuery()
->getResult();

但是，表结构为：

-- Jobs
ID | Job Name | Language1 | Language2
1  | Job1     | 1         | 2
2  | Job2     | 2         | 3
-- Languages
ID | Language
1  | French
2  | English
3  | Spanish

你需要这样的东西：

$this->createQueryBuilder('job')
->leftJoin('job.language1', 'language1')
->leftJoin('job.language2', 'language2')
->where('language1.language IN (:languages)')
->orWhere('language2.language IN (:languages)')
->setParameter("languages", $languages)
->getQuery()
->getResult();

Answer 2

我也遇到了不同的问题，这就是为什么 Dougs 解决方案对我不起作用的原因。在解决了这个问题和他的解决方案的帮助下，我想出了这个：

$queryBuilder
->leftJoin('PathLanguage', 'language1', JOIN::LEFT_JOIN,  "job.language1 = language1")
->leftJoin('PathLanguage', 'language2', JOIN::LEFT_JOIN,  "job.language2 = language2")
->andWhere(
$queryBuilder->expr()->orX(
$queryBuilder->expr()->in("language1.name", ":languages"),
$queryBuilder->expr()->in("language2.name", ":languages")
)
)->setParameter("languages", $languages);