CakePHP在每个表上查找具有条件的连接记录

Question

我想查看连接中的所有记录,在连接的WHERE每一侧设置条件.

例如,我有LOAN和BORROWER(加入borrower.id = loan.borrower_id).我想要记录LOAN.field = 123和BORROWER.field = 'abc'.

这里的答案(例如这一个)似乎说我应该使用Containable.

我试过了.这是我的代码:

$stuff = $this->Borrower->find('all', array(
    'conditions' => array(
        'Borrower.email LIKE' => $this->request->data['email'] // 'abc'
    ),
'contain'=>array(
    'Loan' => array(
        'conditions' => array('Loan.id' => $this->request->data['loanNumber']) // 123
        )
    )
)); 

我期望得到一个结果,因为在我的数据中,只有一个连接记录具有这两个条件.相反,我得到两个结果,

结果1是{Borrower: {field:abc, LOAN: {field: 123} }//正确的

{Borrower: {field:abc, LOAN: {NULL} }// 结果2 //错误

当我查看CakePHP使用的SQL时,我没有看到连接.我看到的是两个单独的查询:

问题1: SELECT * from BORROWER // (yielding 2 IDs),

查询2: SELECT * FROM LOAN WHERE borrower_id in (IDs)

这不是我想要的.我想加入表格,然后应用我的条件.我可以很容易地编写SQL查询,但是我尝试以Cake方式执行它,因为我们已经采用了该框架.

可能吗？

Answer 1

尝试做这样的事情:

    $options['conditions'] = array(
           'Borrower.email LIKE' => $this->request->data['email'] // 'abc',
           'loan.field' => '123' )

    $options['joins'] = array(
        array('table' => 'loans',
              'alias' => 'loan',
              'type' => 'INNER',
              'conditions' => array(
                    'borrower.id = loan.borrower_id')
                )
            );

    $options['fields'] = array('borrower.email', 'loan.field');

    $test = $this->Borrower->find('all', $options);

你应该看到一个SQL语句,如:

SELECT borrower.email, loan.field
FROM borrowers AS borrower
INNER JOIN loans AS loan
    ON borrower.id = loan.borrower_id
    AND loan.field = '123'
WHERE borrower.email = 'abc'

您的结果将在一个数组中

{Borrower: {field:abc} LOAN: {field: 123} }

您将在本文档中找到更多信息.