cakephp 3.0获取两个coulmns中的值作为一个 [英] cakephp 3.0 get values in two coulmns as one

问题描述

这里我试图获取多个两列作为值，但在cakephp 3.0它给出一个错误

Here I am trying to get multiple of two columns as value but in cakephp 3.0 it given a error

错误：SQLSTATE [42000]：语法错误或访问冲突：1064您的SQL语法有错误;请检查与您的MySQL服务器版本对应的手册，以获取在AS（Transactions__amount * PluTransaction FROM 事务 事务 LEF'

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 (Transactions__amount * PluTransaction FROM transactions Transactions LEF'

  $result =  $this->Transaction->find('all', array(
      'conditions' => [
        'Transactions.house_id' => $houseId]
    ))->join([
      [
        'alias' => 'PluTransaction',
        'table' => 'plu_transactions',
        'type' => 'LEFT',
        'conditions' => 'PluTransaction.transaction_id = Transactions.id'
      ]
      ])->select(['Transactions.id',
    '(Transactions.amount * PluTransaction.item_quantity) AS TOTAL',  
  ]);

推荐答案

这不是如何定义计算列，请参考文档

That's not how you define calculated columns, please refer to the docs

• Cookbook>数据库访问& ORM>查询构建器>选择数据

• Cookbook > Database Access & ORM > Query Builder > Selecting Data

Cookbook>数据库访问& ORM>查询生成器>原始表达式

Cookbook > Database Access & ORM > Query Builder > Raw Expressions

c> key => value 格式以单独定义别名和表达式。

You must use the key => value format to define the alias and the expression separately.

$query = $this->Transaction->find('all', [
    'conditions' => [
        'Transactions.house_id' => $houseId
    ]
]);
$query
    ->select([
        'Transactions.id',
        'TOTAL' => $query->newExpr('Transactions.amount * PluTransaction.item_quantity')
    ])
    ->join(/* ... */)
    // ...

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