如何使用 Laravel 的流畅查询构建器选择计数?
这是我使用 fluent 查询构建器的查询.
Here is my query using fluent query builder.
$query = DB::table('category_issue')
->select('issues.*')
->where('category_id', '=', 1)
->join('issues', 'category_issue.issue_id', '=', 'issues.id')
->left_join('issue_subscriptions', 'issues.id', '=', 'issue_subscriptions.issue_id')
->group_by('issues.id')
->order_by(DB::raw('COUNT(issue_subscriptions.issue_id)'), 'desc')
->get();
如您所见,我按连接表中的计数进行排序.这工作正常.但是,我希望此计数与我的选择一起返回.
As you can see, I am ordering by a count from the joined table. This is working fine. However, I want this count returned with my selections.
这是我的原始续集查询,效果很好.
Here is the my raw sequel query that works fine.
Select issues.*, COUNT(issue_subscriptions.issue_id) AS followers
FROM category_issue JOIN Issues ON category_issue.issue_id = issues.id
LEFT JOIN issue_subscriptions ON issues.id = issue_subscriptions.issue_id
WHERE category_issue.category_id = 1
GROUP BY issues.id
ORDER BY followers DESC
我将如何使用 Laravel 的流畅查询构建器进行这个选择?我知道我可以使用原始 sql 查询,但我想尽可能避免这种情况.任何帮助将不胜感激,提前致谢!
How would I go about this select using Laravel's fluent query builder? I am aware I can use a raw sql query but I would like to avoid that if possible. Any help would be appreciated, thanks in advance!
推荐答案
你可以在 select() 中使用一个数组来定义更多的列,你可以在那里使用 DB::raw() 并将其别名给关注者.应该是这样的:
You can use an array in the select() to define more columns and you can use the DB::raw() there with aliasing it to followers. Should look like this:
$query = DB::table('category_issue')
->select(array('issues.*', DB::raw('COUNT(issue_subscriptions.issue_id) as followers')))
->where('category_id', '=', 1)
->join('issues', 'category_issue.issue_id', '=', 'issues.id')
->left_join('issue_subscriptions', 'issues.id', '=', 'issue_subscriptions.issue_id')
->group_by('issues.id')
->order_by('followers', 'desc')
->get();
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