php - Laravel:让 Eloquent 创建嵌套 SELECT 的正确方法

Question

我试图 Eloquent 地生成的查询是

SELECT *, (SELECT COUNT(comment_id) FROM comment AS c WHERE c.approved=true AND c.blog_fk=b.blog_id) AS comment_count FROM blog AS b

这是结果

blog_id |  title            | author       | blog           | image            | tags    | created             | updated             | comment_count
--------|-------------------|--------------|----------------|------------------|---------|---------------------|---------------------|--------------
     21 | A day..           | dsyph3r      | Lorem ipsum... | beach.jpg        | symf... | 2014-12-22 19:14:34 | 2014-12-22 19:14:34 | 2
     22 | The pool ..       | Zero Cool    | Vestibulum ... | pool_leak.jpg    | pool,.. | 2011-07-23 06:12:33 | 2011-07-23 06:12:33 | 10
     23 | Misdirection...   | Gabriel      | Lorem ipsum... | misdirection.jpg | misd... | 2011-07-16 16:14:06 | 2011-07-16 16:14:06 | 2
     24 | The grid ...      | Kevin Flynn  | Lorem commo... | the_grid.jpg     | grid... | 2011-06-02 18:54:12 | 2011-06-02 18:54:12 | 0
     25 | You're either ... | Gary Winston | Lorem ipsum... | one_or_zero.jpg  | bina... | 2011-04-25 15:34:18 | 2011-04-25 15:34:18 | 2

我目前使用 DB::select( DB::raw()) 来运行它，这可能不是执行此操作的正确方法。

问题是什么是使生成这些结果的查询变得 Eloquent 正确方法？

Answer 1

改用这个:http://softonsofa.com/tweaking-eloquent-relations-how-to-get-hasmany-relation-count-efficiently

对于嵌套的select/join语句，你需要这样:

$sub = Comment::selectRaw('count(comment_id) as count')
       ->where('approved', '?')
       ->where('comment.blog_fk', '?')
       ->toSql();

Blog::selectRaw(DB::raw("blog.*, ({$sub}) as comment_count"))
       ->setBindings([true, DB::raw('blog.blog_id')], 'select')
       ->get();

或者简单地将所有内容放入selectRaw。