0

我只是想问一下如何按照这个 SQL 语法在 CakePHP 中进行查询连接

SELECT a.id, SUM( r.jumlah_realisasi) AS jumlah_realisasi, SUM(b.jumlah_budget) AS jumlah_budget, SUM(c.jumlah_contrapos)
FROM accountposts a
LEFT JOIN
(
SELECT accountpost_id, agency_id, SUM(budget_after_changes) AS jumlah_budget
FROM budgets
GROUP BY accountpost_id
) b
ON a.id = b.accountpost_id
LEFT JOIN
(
SELECT accountpost_id, agency_id, SUM(realisation_value) AS jumlah_realisasi
FROM realisations
GROUP BY accountpost_id
) r
ON a.id = r.accountpost_id
LEFT JOIN
(
SELECT accountpost_id, agency_id, SUM(contrapos_value) AS jumlah_contrapos
FROM contraposts
GROUP BY accountpost_id
) c
ON a.id = c.accountpost_id
GROUP BY
a.id

我尝试了这种语法（我使用的是 CakePHP 2.x）：

$joins = array(
               array(
                  'table' => 'budgets',
                  'alias' => 'Budget',
                  'type' => 'LEFT',
                  'conditions' => array('Accountpost.id = Budget.accountpost_id')
                ),
                array(
                  'table' => 'realisations',
                  'alias' => 'Realisation',
                  'type' => 'LEFT',
                  'conditions' => array('Accountpost.id = Realisation.accountpost_id')
                ),
                array(
                   'table' => 'contraposts',
                   'alias' => 'Contrapost',
                   'type' => 'LEFT',
                   'conditions' => array('Accountpost.id = Contrapost.accountpost_id')
                ),
         );

        $this->paginate = array(
            'limit' => 60,
            'joins' => $joins,
            'fields' => array('Accountpost.id','Accountpost.explanation','Accountpost.account_code',
                              'SUM(Budget.budget_after_changes) AS jumlah_budget','SUM(Realisation.realisation_value) AS jumlah_realisasi','SUM(Contrapost.contrapos_value) AS jumlah_contrapos'),
            'group' => array('Accountpost.id'),
            'order' => array('Accountpost.id' => 'ASC'),
        );

这是来自 CakePHP 的 SQL 转储：

SELECT `Accountpost`.`id`, `Accountpost`.`explanation`, `Accountpost`.`account_code`, SUM(`Budget`.`budget_after_changes`), `Budget`.`budget_after_changes`, `Realisation`.`realisation_value`, `Contrapost`.`contrapos_value` FROM `realisasi_anggaran`.`accountposts` AS `Accountpost` LEFT JOIN `realisasi_anggaran`.`budgets` AS `Budget` ON (`Accountpost`.`id` = `Budget`.`accountpost_id`) LEFT JOIN `realisasi_anggaran`.`realisations` AS `Realisation` ON (`Accountpost`.`id` = `Realisation`.`accountpost_id`) LEFT JOIN `realisasi_anggaran`.`contraposts` AS `Contrapost` ON (`Accountpost`.`id` = `Contrapost`.`accountpost_id`) WHERE 1 = 1 GROUP BY `Accountpost`.`id` ORDER BY `Accountpost`.`id` ASC LIMIT 60

但是 SQL 语法版本和 CakePHP 版本的结果不同，SQL 语法在检查 SUM 时没有重复值，但在 CakePHP 版本中，检查 SUM 时有重复值。如何以正确的方式实现我的 SQL 语法到 cakePHP？

4

1 回答 1

0

您的$options数组构造不正确。这样，数组就有了两个嵌套的命名键joins，Cake 不会构造正确的 SQL 连接。

数组应如下所示：

$joins = array(
  array(
    'table' => 'budgets',
    'alias' => 'Budget',
    'type' => 'LEFT',
    'conditions' => array('Accountpost.id = Budget.accountpost_id')
  )
);

$this->paginate = array(
  'limit' => 60,
  'joins' => $joins,
  'fields' => array('Accountpost.id','Accountpost.explanation','Accountpost.account_code',
                                  'SUM(Budget.budget_after_changes) AS jumlah_budget'),
  'group' => array('Accountpost.id'),
  'order' => array('Accountpost.id' => 'ASC'),
);

或者构造如下：

$options = array(
  'limit' => 60,
  'fields' => array('Accountpost.id','Accountpost.explanation','Accountpost.account_code',
                                  'SUM(Budget.budget_after_changes) AS jumlah_budget'),
  'group' => array('Accountpost.id'),
  'order' => array('Accountpost.id' => 'ASC'),
);

$options['joins'] = array(
  array(
    'table' => 'budgets',
    'alias' => 'Budget',
    'type' => 'LEFT',
    'conditions' => array('Accountpost.id = Budget.accountpost_id')
  )
);

$this->paginate = $options;
于 2012-07-29T13:26:30.057 回答