是否有一种简单的方法来获得查询生成器的contain
方法的逆?
我使用belongsToMany关联和一个连接表来关联两个模型。
EmailsTable $this->belongsToMany('Issues');
IssuesTable $this->belongsToMany('Emails');
所以我不能这样做:
$unparsed_emails = $this->Emails->find('all')->where(['issue_id is null']);
似乎下面的反向将提供没有相关问题的电子邮件:
$unparsed_emails = $this->Emails->find->contain(['Issues']); //need inverse of this
我想我一定是从cakephp 3 ORM/Querybuilder文档中缺少一些东西,但我找不到它。
我还没有那么多进入查询生成器,所以我不确定是否有更简单的方法,但左连接应该做到这一点。
$this->Emails
->find('all')
->leftJoin('emails_issues', 'emails_issues.email_id = Emails.id')
->where('emails_issues.id IS NULL');
这应该转换成像
这样的查询SELECT
Emails.id AS `Emails__id`, ...
FROM
emails AS Emails
LEFT JOIN
emails_issues emails_issues ON emails_issues.email_id = Emails.id
WHERE
emails_issues.id IS NULL