如何使用 Laravel Query Builder 从子查询中进行选择? [英] How to select from subquery using Laravel Query Builder?

问题描述

我想通过以下 SQL 使用 Eloquent ORM 获取价值.

I'd like to get value by the following SQL using Eloquent ORM.

- SQL

 SELECT COUNT(*) FROM 
 (SELECT * FROM abc GROUP BY col1) AS a;

然后我考虑了以下内容.

Then I considered the following.

- 代码

 $sql = Abc::from('abc AS a')->groupBy('col1')->toSql();
 $num = Abc::from(DB::raw($sql))->count();
 print $num;

我正在寻找更好的解决方案.

I'm looking for a better solution.

请告诉我最简单的解决方案.

Please tell me simplest solution.

推荐答案

除了@delmadord 的回答和您的评论:

In addition to @delmadord's answer and your comments:

目前没有在FROM子句中创建子查询的方法，因此需要手动使用raw语句，然后，如果需要，您将合并所有绑定:

Currently there is no method to create subquery in FROM clause, so you need to manually use raw statement, then, if necessary, you will merge all the bindings:

$sub = Abc::where(..)->groupBy(..); // Eloquent Builder instance

$count = DB::table( DB::raw("({$sub->toSql()}) as sub") )
    ->mergeBindings($sub->getQuery()) // you need to get underlying Query Builder
    ->count();

请注意，您需要以正确的顺序合并绑定.如果您有其他绑定子句，则必须将它们放在 mergeBindings 之后:

Mind that you need to merge bindings in correct order. If you have other bound clauses, you must put them after mergeBindings:

$count = DB::table( DB::raw("({$sub->toSql()}) as sub") )

    // ->where(..) wrong

    ->mergeBindings($sub->getQuery()) // you need to get underlying Query Builder

    // ->where(..) correct

    ->count();

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