Question

我正在努力在Yii中创建正确的查询，我相信我正在取得进展。

我需要检查一个相关的表，只想返回相关表中没有记录的记录。这在这里得到了解答 - Yii determining existence of related models

使这种情况复杂化的原因是我不确定如何克服它，多个用户可以在此相关表中拥有记录。因此，完整的要求是返回不存在相关记录的记录，但仅记录登录用户的记录。

两个相关对象如下─ SurveyQuestion AnsweredQuestion

SurveyQuestion HAS_MANY AnsweredQuestion

AnsweredQuestion表包含以下列 -

id - survey_question_id - user_id

survey_question_id是SurveyQuestion表的外键。

到目前为止，我的方法是尝试使用关系定义将记录限制为与登录用户相关的记录 -

public function relations()
{
    // NOTE: you may need to adjust the relation name and the related
    // class name for the relations automatically generated below.
    return array(
        'survey_answer'=>array(self::HAS_MANY,'SurveyAnswer','survey_question_id'),
        'answered_questions' => array(self::HAS_MANY, 'AnsweredQuestion', 'question_id',
            'condition'=>'answered_questions.user_id = '.Yii::app()->user->id,
            'joinType'=>'LEFT JOIN',
            ),
    );
}

要将查询限制在父表中的记录而子表中没有相关的记录，我在findAll函数中使用了一个条件，如下所示 -

            $questions = SurveyQuestion::model()->with(array(
                                            'survey_answer',
                                            'answered_questions'=>array(

                                                    'select'=>false,

                                                    'joinType'=>'LEFT JOIN',
                                                    'condition'=>'`answered_questions` . `id` is NULL'
                                                    ),))->findAll();

即使子表被清除，这两段代码也不会返回任何结果。

任何人都可以在方法或执行中发现我出错的地方吗？

非常感谢，

尼克

更新

根据要求，这是运行的sql语句。这是第二个相关的Join，第一个Join收集多项选择答案。

SELECT `t`.`id` AS `t0_c0`, `t`.`area_id` AS `t0_c1`,
`t`.`question_text` AS `t0_c2`, `t`.`date_question` AS `t0_c3`,
`survey_answer`.`id` AS `t1_c0`, `survey_answer`.`survey_question_id` AS
`t1_c1`, `survey_answer`.`answer_text` AS `t1_c2`, `survey_answer`.`tag_id`
AS `t1_c3` FROM `tbl_survey_questions` `t`  LEFT OUTER JOIN
`tbl_survey_answers` `survey_answer` ON
(`survey_answer`.`survey_question_id`=`t`.`id`)  LEFT JOIN
`tbl_answered_questions` `answered_questions` ON
(`answered_questions`.`question_id`=`t`.`id`)  WHERE
((answered_questions.user_id = 2) AND (`answered_questions` . `id` is
NULL))

Answer 1

在发布关于直观地运行查询的评论后，我发现了。

我认为您需要将user_id的条件放在关系的on子句中，而不是condition子句中。因为它只返回子行中NULL id和user_id为2的父行。这显然不会发生。但是你需要它符合JOIN标准。所以它应该是：

'answered_questions' => array(self::HAS_MANY, 'AnsweredQuestion', 'question_id',
    'on'=>'answered_questions.user_id = '.Yii::app()->user->id,
    'joinType'=>'LEFT JOIN',
),

Yii查询返回没有结果

1 个答案: