在同一張桌子上的多對多關係

一個gabarit可以有很多Gabarits（稱爲選項）
一個gabarit可以來自許多Gabarits（稱爲OptionsFrom）

這裏是我的schema.yml：

Gabarit: 
    actAs: [Attachable] 
    columns: 
    libelle: { type: string, size: 255 } 
    description: { type: clob } 
    relations: 
    Options: 
     class: Gabarit 
     refClass: Gabarit2Gabarit 
     local: gabarit_id 
     foreign: fils_id 
     foreignAlias: OptionsFrom 

Gabarit2Gabarit: 
    columns: 
    fils_id: { type: integer, primary: true } 
    gabarit_id: { type: integer, primary: true } 
    relations: 
    Gabarit: 
     class: Gabarit 
     local: gabarit_id 
     alias: Gabarit 
     foreignAlias: Gabarit2Gabarits 
    Fils: 
     class: Gabarit 
     local: fils_id 
     alias: Fils 
     foreignAlias: Gabarit2Gabarits

生成的代碼是：

$this->hasMany('Gabarit as Options', array(
     'refClass' => 'Gabarit2Gabarit', 
     'local' => 'gabarit_id', 
     'foreign' => 'fils_id')); 

    $this->hasMany('Gabarit as OptionsFrom', array(
     'refClass' => 'Gabarit2Gabarit', 
     'local' => 'fils_id', 
     'foreign' => 'gabarit_id'));

看來不錯，根據http://www.doctrine-project.org/projects/orm/1.2/docs/manual/defining-models/en#relationships:join-table-associations:self-referencing-nest-relations:non-equal-nest-relations

從邏輯上講，它會給我這樣的：對於getOptionsFrom（）

SELECT gabarit.id AS gabarit__id, gabarit.libelle AS gabarit__libelle, gabarit.description AS gabarit__description, gabarit2_gabarit.fils_id AS gabarit2_gabarit__fils_id, gabarit2_gabarit.gabarit_id AS gabarit2_gabarit__gabarit_id FROM gabarit INNER JOIN gabarit2_gabarit ON gabarit.id = gabarit2_gabarit.fils_id WHERE gabarit.id IN (SELECT fils_id FROM gabarit2_gabarit WHERE gabarit_id = '507') ORDER BY gabarit.id ASC

生成的SQL：

foreach($gabarit_>getOptions() as $option) 
{ 
    in_array($gabarit->getId(), $options->getOptionsFrom()->getPrimaryKeys()); // should be true but returns false !! 
}

生成getOptions）SQL（

SELECT gabarit.id AS gabarit__id, gabarit.libelle AS gabarit__libelle, gabarit.description AS gabarit__description, gabarit2_gabarit.fils_id AS gabarit2_gabarit__fils_id, gabarit2_gabarit.gabarit_id AS gabarit2_gabarit__gabarit_id FROM gabarit INNER JOIN gabarit2_gabarit ON gabarit.id = gabarit2_gabarit.gabarit_id WHERE gabarit.id IN (SELECT gabarit_id FROM gabarit2_gabarit WHERE fils_id = '529') ORDER BY gabarit.id ASC

你有什麼想法爲什麼Gabarit :: getOptio nsFrom返回一個空集合？

在此先感謝，弗洛裏安。

來源

2010-06-23 Florian

你是否缺少'Gabarit'的主鍵？有'description'和'libelle'但沒有定義的主鍵。 – DrColossos 2010-06-23 12:37:04

嗨，當省略主鍵時，教條自動生成它。所以我Gabarit有一個ID BIGINT AUTO_INCREMENT（mysql）我覺得問題不在這裏。 – Florian 2010-06-23 12:51:27

這將極有可能是事實，你使用的是相同的別名

foreignAlias：Gabarit2Gabarits

你Gabarit2Gabarit應該是這樣的：

Gabarit2Gabarit: 
    columns: 
    fils_id: { type: integer, primary: true } 
    gabarit_id: { type: integer, primary: true } 
    relations: 
    Gabarit: 
     class: Gabarit 
     local: gabarit_id 
     alias: Gabarit 
     foreignAlias: Gabarit2GabaritsGabarit 
    Fils: 
     class: Gabarit 
     local: fils_id 
     alias: Fils 
     foreignAlias: Gabarit2GabaritsFils

這可能會幫助？

來源

2010-06-24 12:47:05 johnwards

令人驚歎！有用！ :) 非常感謝你，即使現在看來邏輯，我想我也不會想到它... 謝謝！弗洛裏安。 – Florian 2010-06-25 11:12:45

在同一張桌子上的多對多關係

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