cakephp 3 使用 table.* 获取所有内容

Question

我正在使用 cakePHP3 查询生成器使用以下查询从两个表中获取记录，其中我想要 table1 中的所有列和 table2 中的选定列：

$this->loadModel('Table1');
$Table1 = $this->Table1->find('all', array('fields' => array('Table1.*'),'conditions'=>$conditions,'order'=>array('Table1.id'=>'DESC')))->contain(['Table2']);

但我收到以下错误

Error: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'AS `Table1__*` FROM Table1 Table1 LEFT JOIN Table2 Table2 ON ' at line 1

我是 CakePHP3 的新手。

Answer 1

为什么不使用查询生成器？

有很多方法可以做到这一点

$query= $this->Table1->find('all') 
    ->where($conditions)
    ->order(['Table1.id'=>'DESC'])
    ->contain(['Table2' => [
            'fields' => ['field1', 'field2']
        ]
    ]);

或

$query= $this->Table1->find('all') 
    ->select($this->Table1) // selects all the fields from Table1
    ->select(['Table2.field1', 'Table2.field2']) // selects some fields from Table2
    ->where($conditions)
    ->order(['Table1.id'=>'DESC'])
    ->contain(['Table2']);

或

$query= $this->Table1->find('all') 
    ->where($conditions)
    ->order(['Table1.id'=>'DESC'])
    ->contain(['Table2' => function($q) {
             return $q->select(['field1', 'field2']);
        }
    ]);

Answer 2

假设表之间的关系正确

示例条件数组

$conditions = ['Table2.id' > 1];

从表中选择行

 $table1 = $this->Table1->find()->select(['Table1.*'])->where($conditions)->contain(['Table2'])->order(['Table1.id'=>'DESC'])->toArray();

或

$table1 => $this->Table1->find('all', array(
    'contain' => ['Table2'],
    'conditions' => $conditions,
    'fields' => ['Table1.*'],
    'order' => ['Table1.id'=>'DESC']
));

这里有更多关于queryBuilder的信息