Php 如何从“推进”中的结果集中排除关联关系对象？

我有4个表，代表一个简单的测量数据库设置：

table name: (delimited columns)
---------------------------------------------------------------
survey: (id, title)
surveyQuestion: (id, title)
surveyAnswer: (id, surveyQuestionID, title, sortOrder)
surveyToSurveyQuestion: (surveyID, surveyQuestionID, sortOrder)

下面是一个查询，我使用它来获取所有调查以及所有相关的问题和答案：

$query = SurveyQuery::create()
    ->joinWith('SurveyToSurveyQuestion')
    ->useSurveyToSurveyQuestionQuery()
        ->orderBySurveyId()
        ->orderBySortOrder()
        ->joinWith('SurveyQuestion')
        ->useSurveyQuestionQuery()
            ->joinWith('SurveyAnswer')
            ->useSurveyAnswerQuery()
                ->orderBySortOrder()
            ->endUse()
        ->endUse()
    ->endUse();

然而，当我添加更多调查时，多对多“

SurveyToSurveyQuestion

”关系变得臃肿，特别是当调查共享问题时

是否有办法从结果数据集中排除'

SurveyToSurveyQuestion

'数据？

编辑：

附加数据库模式

<table name="survey" idMethod="native" phpName="Survey">
    <column name="id" phpName="Id" type="INTEGER" primaryKey="true" autoIncrement="true" required="true"/>
    <column name="statusID" phpName="StatusID" type="INTEGER" required="true"/>
    <column name="code" phpName="Code" type="VARCHAR" required="true"/>
    <column name="createdDate" phpName="CreatedDate" type="TIMESTAMP" default="0000-00-00 00:00:00"/>
    <index name="INDEX_StatusID">
        <index-column name="statusID"/>
    </index>
    <foreign-key foreignTable="status">
        <reference local="statusID" foreign="id"/>
    </foreign-key>
    <vendor type="mysql">
        <parameter name="Engine" value="InnoDB"/>
    </vendor>
</table>
<table name="surveyQuestion" idMethod="native" phpName="SurveyQuestion">
    <column name="id" phpName="Id" type="INTEGER" primaryKey="true" autoIncrement="true" required="true"/>
    <column name="surveyQuestionTypeID" phpName="SurveyQuestionTypeID" type="INTEGER" required="true"/>
    <column name="code" phpName="Code" type="VARCHAR" required="true"/>
    <index name="INDEX_SurveyQuestionTypeID">
        <index-column name="surveyQuestionTypeID"/>
    </index>
    <foreign-key foreignTable="surveyQuestionType">
        <reference local="surveyQuestionTypeID" foreign="id"/>
    </foreign-key>
    <vendor type="mysql">
        <parameter name="Engine" value="InnoDB"/>
    </vendor>
</table>
<table name="surveyQuestionType" idMethod="native" phpName="SurveyQuestionType">
    <column name="id" phpName="Id" type="INTEGER" primaryKey="true" autoIncrement="true" required="true"/>
    <column name="title" phpName="Title" type="VARCHAR" required="true"/>
    <vendor type="mysql">
        <parameter name="Engine" value="InnoDB"/>
    </vendor>
</table>
<table name="surveyAnswer" idMethod="native" phpName="SurveyAnswer">
    <column name="id" phpName="Id" type="INTEGER" primaryKey="true" autoIncrement="true" required="true"/>
    <column name="surveyQuestionID" phpName="SurveyQuestionID" type="INTEGER" required="true"/>
    <column name="code" phpName="Code" type="VARCHAR" required="true"/>
    <column name="sortOrder" phpName="SortOrder" type="INTEGER"/>
    <index name="INDEX_SurveyQuestionID">
        <index-column name="surveyQuestionID"/>
    </index>
    <foreign-key foreignTable="surveyQuestion">
        <reference local="surveyQuestionID" foreign="id"/>
    </foreign-key>
    <vendor type="mysql">
        <parameter name="Engine" value="InnoDB"/>
    </vendor>
</table>
<table name="surveyToSurveyQuestion" idMethod="native" phpName="SurveyToSurveyQuestion">
    <column name="id" phpName="Id" type="INTEGER" primaryKey="true" autoIncrement="true" required="true"/>
    <column name="surveyID" phpName="SurveyID" type="INTEGER" required="true"/>
    <column name="surveyQuestionID" phpName="SurveyQuestionID" type="INTEGER" required="true"/>
    <column name="sortOrder" phpName="SortOrder" type="INTEGER"/>
    <index name="INDEX_SurveyID">
        <index-column name="surveyID"/>
    </index>
    <index name="INDEX_SurveyQuestionID">
        <index-column name="surveyQuestionID"/>
    </index>
    <foreign-key foreignTable="survey">
        <reference local="surveyID" foreign="id"/>
    </foreign-key>
    <foreign-key foreignTable="surveyQuestion">
        <reference local="surveyQuestionID" foreign="id"/>
    </foreign-key>
    <vendor type="mysql">
        <parameter name="Engine" value="InnoDB"/>
    </vendor>
</table>

---

编辑：使用

join

代替

joinWith

为我完成了这项工作。谢谢@kriple.

使用

->useSurveyToSurveyQuestion（）

不需要使用

->joinwith（）

可以排除导致连接对象不存在的连接调用

此外，我会在每次调查中提供相应的问题和答案的方法

$surveys = SurveyQuery::create()->filterByXXX()->find()
foreach($survey as $s){
  //Grab each survey's respective data
  $QandA = $s->getQuestionsAndAnswers();
  // do something with the data.
  $otherClass->doStuffWithData($QandA);
}

---

这样做无疑会增加到DB的行程，但我认为总体效率会更好

跳过

joinwith（）

，+1的好处很好。但是，如果删除joinWith（'SurveyToSurveyQuestion'）行，则查询将失败。还有其他想法吗？查询失败，没有结果？或者某种php错误？错误是模型、别名或表“Survey”上的未知列“SurveyId”

joinWith（）

方法

向查询中添加一个JOIN子句并对相关对象进行水合物化，试着使用just
JOIN（）
@Kriple:再次运行并注意到您最后的注释。谢谢