将连接关系对象排除在Propel中的结果集中的方法？

Question

我有4个表，它们表示一个简单的调查数据库设置：

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

下面是我用来提取所有相关问题和答案的所有调查的查询：

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

然而，随着我增加更多的调查，特别是如果调查有共同的问题的话，多对多的'SurveyToSurveyQuestion‘关系会变得臃肿。

是否有方法将'**SurveyToSurveyQuestion**‘数据从结果数据集中排除？

编辑：

附加数据库架构。

<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完成了我的工作。谢谢你。

Answer 1

您不需要使用->joinwith()来使用->useSurveyToSurveyQuestion()

您可以排除导致联接对象不存在的联接调用。

此外，我对每项调查都有自己的方法，得到他们各自的问题和答案。

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

这样做无疑会增加对DB的访问，但我认为总体效率会好得多。