我正在使用cakePHP3查询构建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']);
但是我收到以下错误
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。
答案 0 :(得分:4)
为什么不使用查询构建器?
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']);
}
]);
答案 1 :(得分:-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();
OR
$table1 => $this->Table1->find('all', array(
'contain' => ['Table2'],
'conditions' => $conditions,
'fields' => ['Table1.*'],
'order' => ['Table1.id'=>'DESC']
));
在这里,您有关于queryBuilder
的更多信息