Laravel 5.使用USING运算符 [英] Laravel 5. Using the USING operator

问题描述

我试图找到它很长时间，而且我不敢相信Laravel没有此功能.

I tried to find it for a long time, and I can't believe that Laravel doesn't have this functionality.

所以，我可以写:

select * from a join b where a.id = b.id

或更漂亮:

select * from a join b using(id)

对于Laravel，第一种情况很简单:

First case is simple for Laravel:

$query->leftJoin('b', 'a.id', '=', 'b.id')

但是如何写第二种情况?我希望它应该简单而简短，例如:

But how to write second case? I expect that it should be simple and short, like:

$query->joinUsing('b', 'id')

但是没有这样的方法，我找不到它.

But thereis no such method and I can't find it.

PS:答案很可能很简单，因为很难找到使用"一词，所以很难找到答案，因为它无处不在.

PS: it's possible that the answer is very simple, it's just hard to find by word "using", because it's everywhere.

更新

我将更深入地探究源代码，尝试建立作用域或通过一个函数来联接，但是即使在此函数内部，使用此$ query也无法执行任何操作.示例:

I'm going deeper to source, trying to make scope or pass a function to join, but even inside of this function I can't to anything with this $query. Example:

public function scopeJoinUsing($query, $table, $field) {
    sql($query->join(\DB::raw("USING(`{$field}`)")));
    // return 
    // inner join `b` on USING(`id`)  ``
    // yes, with "on" and with empty quotes
    sql($query->addSelect(\DB::raw("USING(`{$field}`)")));
    // return 
    // inner join `b` USING(`id`) 
    // but in fields place, before "FROM", which is very logic :)
}

所以即使忘记范围，我也无法在DB :: raw()中做到这一点，所以这是不可能的...第一次我看到Laravel中有些不可能的事情.

So even if forget about scope , I can't do this in DB::raw() , so it's impossible... First time I see that something impossible in Laravel.

推荐答案

因此，答案是-这是不可能的.

So, the answer is - it's impossible.

Laravel不支持此功能，这实在令人遗憾.

Laravel doesn't support this functionality, which is really sad.

我已在Laravel源代码中对其进行了修复，请在此处进行我的请求- https://github.com /laravel/framework/pull/12773

I fix it in Laravel source code, my PULL REQUEST here - https://github.com/laravel/framework/pull/12773

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