Laravel Eloquent模型查询

Question

我想用laravel实现正确的查询，如果有人能提供帮助的话

我有这个问题

SELECT * FROM `users` WHERE ((`is_verified` = 1) AND (`first_name` like '%%' or `middle_name` like '%%' or `last_name` like '%%' or `email` like '%%'))

在我的代码中有这样的代码

$user->where([
                ['is_verified', '=', 1]
            ])
                ->where('first_name', 'like', "%$search%")
                ->orWhere('middle_name', 'like', "%$search%")
                ->orWhere('last_name', 'like', "%$search%")
                ->orWhere('email', 'like', "%$search%");

但它会产生

SELECT * FROM `users` WHERE ((`is_verified` = '1') and `first_name` like '%%' or `middle_name` like '%%' or `last_name` like '%%' or `email` like '%%') 

Answer 1

你必须在"closure-where“子句中使用匿名函数。

$user->where([
                ['is_verified', '=', 1]
            ])->where(function ($query) use ($search) {
                $query->where('first_name', 'like', "%$search%")
                ->orWhere('middle_name', 'like', "%$search%")
                ->orWhere('last_name', 'like', "%$search%")
                ->orWhere('email', 'like', "%$search%");
});

Answer 2

你必须像下面的代码一样使用匿名函数：

User::where(function ($query){
        $query->where('is_verified', 1);
    })->where(function ($query) use ($search) {
        $query->where('first_name', 'like', "%$search%")
        ->orWhere('middle_name', 'like', "%$search%")
        ->orWhere('middle_name', 'like', "%$search%")
        ->orWhere('email', 'like', "%$search%");
    })->toSql();