【发布时间】:2015-04-24 00:48:48
【问题描述】:
我正在使用 cakephp,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;
我得到这样的结果:
{
"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,我希望通过以下方式获得结果:
{
"id": "132",
"title": "Awesome",
"image": "image_25398457.jpg"
"maincategory": "3"
"isfavorite": "1"
}
我怎样才能做到这一点?
谢谢!
【问题讨论】:
标签: php mysql sql json cakephp-2.3