R. *_*kan 7 php sql symfony doctrine-orm
我必须将相同的字段与不同的参数相关联.
多对多:
/**
* @ORM\ManyToMany(targetEntity="BRCN\TaxonomyBundle\Entity\Taxonomy", mappedBy="genderCategories")
*/
private $genders;
/**
* @ORM\ManyToMany(targetEntity="BRCN\TaxonomyBundle\Entity\Taxonomy", inversedBy="genders")
* @ORM\JoinTable(name="menu_relations",
* joinColumns={@ORM\JoinColumn(name="gender_id", referencedColumnName="id")},
* inverseJoinColumns={@ORM\JoinColumn(name="category_id", referencedColumnName="id")}
* )
*/
private $genderCategories;
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这些变量在同一个表"id"中使用相同的列名.
当我查看"menu_relations"表时,我看到两个与我的参数关联的主键.
当我尝试插入新记录时,我收到此错误:
使用参数[94,1]执行'INSERT INTO menu_relations(gender_id,category_id)VALUES(?,?)'时发生异常:
SQLSTATE [23000]:完整性约束违规:1062键'PRIMARY'的重复条目'94 -1'
插入值:
if($request->isXmlHttpRequest()) {
$categories = $request->request->get('category');
foreach($categories as $key => $value) {
/** @var $gender \BRCN\TaxonomyBundle\Entity\Taxonomy */
$gender = $em->find('BRCNTaxonomyBundle:Taxonomy', $key);
if($gender) {
foreach($value as $category) {
$category = $em->find('BRCNTaxonomyBundle:Taxonomy', $category);
$gender->addGenderCategory($category);
$em->persist($gender);
}
$em->flush();
}
}
return JsonResponse::create([
'status' => true,
'message' => '?li?kilendirme Ba?ar?l?'
]);
}
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您可以利用Doctrine的索引关联来避免添加重复的关系.
下面将指数的代码$genderCategories通过category_id,所以它永远不会包含多个类别具有相同的ID.
请注意注释中的indexBy属性.
/**
* @ORM\ManyToMany(targetEntity="BRCN\TaxonomyBundle\Entity\Taxonomy", inversedBy="genders", indexBy="id")
* @ORM\JoinTable(name="menu_relations",
* joinColumns={@ORM\JoinColumn(name="gender_id", referencedColumnName="id")},
* inverseJoinColumns={@ORM\JoinColumn(name="category_id", referencedColumnName="id")}
* )
*/
private $genderCategories;
public addGenderCategory($category) {
$this->genderCategories[$category->getId()] = $category;
}
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边注
我建议将persist呼叫拉出你的循环以提高性能.
foreach($value as $category) {
$category = $em->find('BRCNTaxonomyBundle:Taxonomy', $category);
$gender->addGenderCategory($category);
}
$em->persist($gender);
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