如何在 Laravel Eloquent 查询(或使用查询生成器)中为

问题描述

假设我们正在使用 Laravel 的查询构建器:

$users = DB::table('really_long_table_name')-> 选择('really_long_table_name.id')->get();

我正在寻找与此 SQL 等效的语句:

really_long_table_name AS short_name

当我必须输入大量选择和 wheres(或者通常我在选择的列别名中也包含别名，并且它在结果数组中使用)时，这将特别有用.如果没有任何表别名，我就需要打更多的字，而且一切都变得不那么可读了.在 laravel 文档中找不到答案，有什么想法吗?

Laravel 支持使用 AS 为表和列设置别名.试试

$users = DB::table('really_long_table_name AS t')->select('t.id AS uid')->get();

让我们用一个很棒的 tinker 工具来看看它的作用

Lets say we are using Laravel's query builder:

$users = DB::table('really_long_table_name')
           ->select('really_long_table_name.id')
           ->get();

I'm looking for an equivalent to this SQL:

really_long_table_name AS short_name

This would be especially helpful when I have to type a lot of selects and wheres (or typically I include the alias in the column alias of the select as well, and it get's used in the result array). Without any table aliases there is a lot more typing for me and everything becomes a lot less readable. Can't find the answer in the laravel docs, any ideas?

Laravel supports aliases on tables and columns with AS. Try

$users = DB::table('really_long_table_name AS t')
           ->select('t.id AS uid')
           ->get();

Let's see it in action with an awesome tinker tool

$ php artisan tinker
[1] > Schema::create('really_long_table_name', function($table) {$table->increments('id');});
// NULL
[2] > DB::table('really_long_table_name')->insert(['id' => null]);
// true
[3] > DB::table('really_long_table_name AS t')->select('t.id AS uid')->get();
// array(
//   0 => object(stdClass)(
//     'uid' => '1'
//   )
// )

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