Question

我的问题与此Undefined index: inverseJoinColumns while trying to define ManyToMany relationship between two entities

密切相关

我试图找到所有没有部分的书。 - ＆GT; findBy（阵列（＆＃34;节＆＃34; = GT;空）

这是我的实体配置：的 EditedBooks

    /**
 * @var \Doctrine\Common\Collections\Collection
 *
 * @ORM\ManyToMany(targetEntity="Sections", inversedBy="editedBook")
 * @ORM\JoinTable(name="edited_book_has_section",
 *   joinColumns={
 *     @ORM\JoinColumn(name="edited_book_id", referencedColumnName="id")
 *   },
 *   inverseJoinColumns={
 *     @ORM\JoinColumn(name="section_id", referencedColumnName="id")
 *   }
 * )
 */ 
private $section;

这是我的实体配置：的节

/**
 * Bidirectional (INVERSE SIDE)
 *
 * @ORM\ManyToMany(targetEntity="EditedBooks", mappedBy="section")
 */       
private $editedBook;

我得到的错误

未定义的索引：BasicEntityPersister.php中的joinColumns

我只用

尝试了它

    /**
 * @var \Doctrine\Common\Collections\Collection
 *
 * @ORM\ManyToMany(targetEntity="Sections", inversedBy="editedBook")
 * @ORM\JoinTable(name="edited_book_has_section")
 */

或切换定义但我收到此错误

You cannot search for the association field '\Entity\EditedBooks#section', because it is the inverse side of an association. Find methods only work on owning side associations.

解决方法： 今天我试着用我编辑的BooksRepository

中的querybuilder修复我的问题

        $query = $qb->select('books')
        ->from('Bundle:EditedBooks', 'books')
        ->leftJoin('books.section', 'sec')
        ->addSelect('COUNT(sec.id) AS sec_count')
        ->andWhere('sec_count = 0');

但我收到了以下错误：

执行＆＃39; SELECT e0_.id AS id0，e0_.doi时发生异常   AS doi1，e0_.isbn_print AS isbn_print2，e0_.isbn_electronic AS   isbn_electronic3，e0_.publication_date AS publication_date4，   e0_.price_print AS price_print5，e0_.price_electronic AS   price_electronic6，e0_.summary AS summary7，e0_.title AS title8，   e0_.ongoing AS ongoing9，e0_.pages AS pages10，e0_.illustrations AS   illustrations11，e0_.entry_date AS entry_date12，e0_.google_id AS   google_id13，e0_.specialIssue_comment AS specialIssue_comment14，   e0_.deleted AS deleted15，e0_.specialIssue_id AS specialIssue_id16，   COUNT（s1_.id）AS sclr17 ，e0_.book_series_id AS book_series_id18，   e0_.copyright_id AS copyright_id19，e0_.publisher_id AS publisher_id20   FROM edited_books e0_ LEFT JOIN edited_book_has_section e2_ ON e0_.id   = e2_.editedbooks_id LEFT JOIN部分s1_ ON s1_.id = e2_.sections_id WHERE sclr17 = 0＆＃39;：

SQLSTATE [42S22]：未找到列：1054未知列＆＃39; sclr17 ＆＃39;在   ＆＃39; where where＆＃39;

但是对我而言sclr17似乎存在，我错过了什么吗？

丑陋的解决方法 我知道这不是正确的事情，但有时男人必须要做一个男人必须做的事情：

    $noSection = new Sections();      
    $noSection->setTitle("Sectionless");
    //add all the books 
    $noSection->setEditedBook(new ArrayCollection($books));
    //remove the assigned ones
    foreach($sections as $section){
        $sBooks = $section->getEditedBook();
        foreach($sBooks as $b){
            $noSection->removeEditedBook($b);
        }
    }

使用脏修复它现在正在工作，但我很高兴任何其他解决方案。

Answer 1

这是一个工作示例

    class Audio
    {
      /**
      * @ORM\ManyToMany(targetEntity="Acme\MyBundle\Entity\Groupe",inversedBy="audios")
      */
      private $groupes;

  class Groupe
  {

    /**
    * @ORM\ManyToMany(targetEntity="Acme\MyBundle\Entity\Audio",mappedBy="groupes")
    */
    private $audios;

注意mappedby和inversedby以及属性如何具有＆＃34; s＆＃34;在末尾。（Groupe =＆gt; private $ groupe s ）。也许它会帮助你

Answer 2

你能不能用null检查？

$query = $qb->select('books')
    ->from('Bundle:EditedBooks', 'books')
    ->leftJoin('books.section', 'sec')
    ->where('sec IS NULL');

Doctrine n：m relation未定义的索引：BasicEntityPersister.php中的joinColumns

2 个答案: