Laravel加入查询AS [英] Laravel join queries AS

问题描述

为查询定义 AS 的任何方式?

Any way of defining an AS for a query??

我尝试了以下方法:

$data = News::order_by('news.id', 'desc')
    ->join('categories', 'news.category_id', '=', 'categories.id')
    ->left_join('users', 'news.user_id', '=', 'users.id') // ['created_by']
    ->left_join('users', 'news.modified_by', '=', 'users.id') // ['modified_by']
    ->paginate(30, array('news.title', 'categories.name as categories', 'users.name as username'));

问题在于，类别中的 ['name'] 将替换为 users 中的用户.可以用不同的名称来命名它们吗?

The problem is that ['name'] from categories will be replaces with the one from users. Any way of having them with different names?

具有上面的别名...如何创建两个连接都返回 users.name 的别名?

Having the aliases above... how can I create an alias where both joins return users.name ?

推荐答案

paginate()方法的第二个参数接受表列的 array 以便在查询中进行选择.所以这部分:

paginate() method's second parameter accepts array of table columns to select in the query. So this part:

paginate(30, array('news.title, category.name'));

必须是这样的:

paginate(30, array('news.title', 'category.name'));

更新 (更改问题后)

尝试一下:

->paginate(30, array('news.title', 'categories.name as category_name', 'users.name as user_name'));

更新2 (再次更改问题后)

您也可以在表上使用别名:

You can use alias on tables, too:

$data = News::order_by('news.id', 'desc')
    ->join('categories', 'news.category_id', '=', 'categories.id')
    ->join('users as u1', 'news.user_id', '=', 'u1.id') // ['created_by']
    ->join('users as u2', 'news.modified_by', '=', 'u2.id') // ['modified_by']
    ->paginate(30, array('news.title', 'categories.name as categories', 'u1.name as creater_username', 'u2.name as modifier_username'));

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