在 Laravel 查询构建器中运行“存在”查询
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Running "exists" queries in Laravel query builder
提问by Anthony Vipond
I'm using MySQL and have a table of 9 million rows and would like to quickly check if a record (id) exists or not.
我正在使用 MySQL 并且有一个包含 900 万行的表，并且想快速检查记录 (id) 是否存在。
Based on some research it seems the fastest way is the following sql:
根据一些研究，似乎最快的方法是以下 sql：
SELECT EXISTS(SELECT 1 FROM table1 WHERE id = 100)
SELECT EXISTS(SELECT 1 FROM table1 WHERE id = 100)
Source: Best way to test if a row exists in a MySQL table
How can I write this using Laravel's query builder?
我如何使用 Laravel 的查询构建器来编写这个？
回答by Jarek Tkaczyk
Use selectOnemethod of the Connectionclass:
类的使用selectOne方法Connection：
$resultObj = DB::selectOne('select exists(select 1 from your_table where id=some_id) as `exists`');
$resultObj->exists; // 0 / 1;
回答by Mochamad Gufron Efendi
see here http://laravel.com/docs/4.2/queries
见这里http://laravel.com/docs/4.2/queries
Scroll down to Exists Statements, you will get what you need
向下滚动到 Exists Statements，你会得到你需要的
DB::table('users')
->whereExists(function($query)
{
$query->select(DB::raw(1))
->from('table1')
->whereRaw("id = '100'");
})
->get();
回答by AbstractVoid
This is an old question that was already answered, but I'll post my opinion - maybe it'll help someone down the road.
这是一个已经回答的老问题，但我会发表我的意见 - 也许它会帮助某人。
As mysql documentationsuggests, EXISTS will still execute provided subquery. Using EXISTS is helpful when you need to have it as a part of a bigger query. But if you just want to check from your Laravel app if record exists, Eloquent provides simpler way to do this:
正如mysql 文档所建议的， EXISTS 仍将执行提供的子查询。当您需要将 EXISTS 作为更大查询的一部分时，使用 EXISTS 会很有帮助。但是如果你只是想从你的 Laravel 应用程序中检查记录是否存在，Eloquent 提供了更简单的方法来做到这一点：
DB::table('table_name')->where('field_name', 'value')->exists();
this will execute query like
这将执行查询
select count(*) as aggregate from `table_name` where `field_name` = 'value' limit 1
// this is kinda the same as your subquery for EXISTS
and will evaluate the result and return a true/false depending if record exists.
并将评估结果并根据记录是否存在返回真/假。
For me this way is also cleaner then the accepted answer, because it's not using raw queries.
对我来说，这种方式也比接受的答案更清晰，因为它没有使用原始查询。
Update
更新
In laravel 5 the same statement will now execute
在 laravel 5 中，现在将执行相同的语句
select exists(select * from `table_name` where `field_name` = 'value')
Which is exactly, what was asked for.
这正是所要求的。
