Question

我正在使用cakephp并且我有一个模型

$db = $this->getDataSource();
$result = $db->fetchAll(
        'SELECT table1.id, 
            table1.title, 
            table1.buy_url, 
            table2.image_file as image, 
            table3.category_id as maincategory, 
            (table4.user_id = "71") AS isfavorite
        FROM table1
        INNER JOIN ...
        LEFT JOIN ...
        LEFT JOIN ...
        where ...);

    return $result;

我得到的结果如下：

{
  "table1": {
    "id": "132",
    "title": "Awesome",
  },
  "table2": {
    "image": "image_25398457.jpg"
  },
  "table3": {
    "maincategory": "3"
  },
  "table4": {
    "isfavorite": "1"
  }
}

但我不想显示表格的名称，我更愿意以下列方式获得结果：

{
    "id": "132",
    "title": "Awesome",
    "image": "image_25398457.jpg"
    "maincategory": "3"
    "isfavorite": "1"
}

我怎么能做到这一点？

谢谢！

Answer 1

从我所看到的，结果按表名分组。

最简单的选择是：

$merged = call_user_func_array('array_merge', $result);

另一种选择是：

$db = $this->getDataSource();
$result = $db->fetchAll(
    'SELECT * FROM (
        SELECT table1.id, 
            table1.title, 
            table1.buy_url, 
            table2.image_file as image, 
            table3.category_id as maincategory, 
            (table4.user_id = "71") AS isfavorite
        FROM table1
        INNER JOIN ...
        LEFT JOIN ...
        LEFT JOIN ...
        where ... '
    ) as final_table
);

return $result;

这就是为什么你会有这样的原因：

{
   "final_table" : {
       "id": "132",
       "title": "Awesome",
       "image": "image_25398457.jpg"
       "maincategory": "3"
       "isfavorite": "1"
   }
}

Cakephp db查询结果没有表名，只是MySQL中的简单结果？

1 个答案: