在 Laravel Eloquent 中计算另一个表作为名称答案

【问题标题】：count another table as name in Laravel Eloquent在 Laravel Eloquent 中计算另一个表作为名称
【发布时间】：2019-02-07 03:41:39
【问题描述】：

这个 sql 查询有效

SELECT 
    users.*,
    (SELECT COUNT(*) FROM villaaz.elanlar
        WHERE elanlar.user_id = villaaz.users.id) AS elan_sayi
FROM villaaz.users 
INNER JOIN villaaz.elanlar on users.id = elanlar.user_id
WHERE elanlar.country_id=2 AND deleted_at IS NULL
GROUP BY users.id 
ORDER BY users.id ASC;

我想用 Laravel 编写它，注意：这是我的 Laravel 代码的重要部分，问题出在此处，其余部分有效，粗线是问题。

$users = User::join('elanlar', 'elanlar.user_id', 'users.id')
    ->select([
       'users.*', 
        DB::select(DB::raw('COUNT(*) as elan_sayi FROM elanlar WHERE elanlar.user_id = users.id '))
    ])->where('elanlar.country_id', 19)->groupBy('users.id);

错误是：DB::select(DB::raw('COUNT(*) as elan_sayi FROM elanlar WHERE elanlar.user_id = users.id '))

【问题讨论】：

浏览器显示哪个错误？
你忘了select 里面的DB::raw。不应该是这样吗DB::select(DB::raw('select COUNT(*) as elan_sayi FROM elanlar WHERE elanlar.user_id = users.id '))
sql语法错误

标签： php sql laravel count eloquent

【解决方案1】：

DB::select() 执行查询。您必须删除它并将SELECT 添加到原始 SQL：

$users = User::join('elanlar', 'elanlar.user_id', 'users.id')
    ->select([
       'users.*', 
        DB::raw('(SELECT COUNT(*) FROM elanlar WHERE elanlar.user_id = users.id) as elan_sayi')
    ])->where('elanlar.country_id', 19)->groupBy('users.id');

【讨论】：