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')->select('really_long_table_name.id')->get();
我正在寻找与此 SQL 等效的方法:
really_long_table_name AS short_name
当我必须输入大量选择和位置时，这将特别有用(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).没有任何表格别名，我的打字量会增加很多，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?
 解决方案 
Laravel 支持 AS 表和列的别名.试试
$users = DB::table('really_long_table_name AS t')->select('t.id AS uid')->get();
让我们用一个很棒的 tinker 工具来看看它的实际效果
<上一页>$ php工匠修补匠[1] > Schema::create('really_long_table_name', function($table) {$table->increments('id');});//空值[2] > DB::table('really_long_table_name')->insert(['id' => null]);//真的[3] > DB::table('really_long_table_name AS t')->select('t.id AS uid')->get();//大批(//0 => 对象(stdClass)(//'uid' => '1'//)//)

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'
//   )
// )

这篇关于如何在 Laravel Eloquent 查询(或使用查询生成器)中为表起别名?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！