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[英]How to write Laravel eloquent where clause as normal mysql query?
[英]How to write a where not in query in laravel?
我想在 laravel 中写一个 Where not in query。我在 sql 中写了它工作正常请帮我将查询转换为 laravel。
这是查询...
SELECT *
FROM `apiaccessauth`
WHERE `site_name` NOT IN (
SELECT `site_name`
FROM `API_site_access_log`
WHERE `created_at` LIKE '%2021-10-15%'
);
如果你想使用Eloquent,它会像下面这样
Eloquent model
<?php
namespace App\Modules\Vehicle\Models;
use Illuminate\Database\Eloquent\Model;
class ApiAccessAuth extends Model
{
protected $table = 'apiaccessauth';
}
您的查询
ApiAccessAuth::whereNotIn('site_name', function ($query) {
$query->from('API_site_access_log')
->select('site_name')
->whereDate('created_at', '2021-10-15');
})->get();
如果你想使用查询构建器,它会像下面这样
use Illuminate\Support\Facades\DB;
DB::table('apiaccessauth')
->whereNotIn('site_name', function ($query) {
$query->from('API_site_access_log')
->select('site_name')
->whereDate('created_at', '2021-10-15');
})->get();
也许您想将查询重构为
SELECT a.* FROM `apiaccessauth` a, `API_site_access_log` b WHERE a.`site_name` = b.`site_name` AND b.`site_name` NOT LIKE '%2021-10-15%'
所以使用查询生成器
$result = DB::table('apiaccessauth') ->join('API_site_access_log', 'API_site_access_log.site_name', '=', 'apiaccessauth.site_name') ->where('API_site_access_log.site_name', 'not like', "%2021-10-15%") ->get();
也许你需要一个 left 或 outer Join,这取决于你想要什么
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