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mysql - laravel 之类的查询不起作用?

转载 作者:行者123 更新时间:2023-11-29 03:59:42 26 4
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这是我的代码:

 $lists = DB::table('connection_request as cr')
->leftJoin('users as u', function($join)
{
$join->on('u.id', '=', 'cr.sender_id')->orOn('u.id','=', 'cr.receiver_id');
})
->select('cr.id as connection_id','cr.sender_id as sen_id','cr.receiver_id as rec_id','cr.approve_status','u.id','u.user_type','u.user_type_id',DB::raw("IF(u.avatar = '', 'uploads/avatar/default.jpg', u.avatar) as avatar"),'u.name','u.email')
->where(function($query) use ($user_id)
{
if(!empty($user_id)):
$query->Where('cr.receiver_id','=', $user_id);
endif;
if(!empty($user_id)):
$query->orWhere('cr.sender_id','=', $user_id);
endif;
})
->where(function($query) use ($searchValue)
{
if(!empty($searchValue)):
$query->Where('u.name','like', '%' . $searchValue . '%');
$query->orWhere('u.email','like', '%' . $searchValue . '%');
endif;
})
->where('cr.approve_status','=',1)
->where('u.id','!=',$user_id)
->get();

它提供了这样的mysql查询

select `cr`.`id` as `connection_id`, `cr`.`sender_id` as `sen_id`, `cr`.`receiver_id` as `rec_id`, `cr`.`approve_status`, `u`.`id`, `u`.`user_type`, `u`.`user_type_id`, IF(u.avatar = '', 'uploads/avatar/default.jpg', u.avatar) as avatar, `u`.`name`, `u`.`email` from `connection_request` as `cr` left join `users` as `u` on `u`.`id` = `cr`.`sender_id` or `u`.`id` = `cr`.`receiver_id` where (`cr`.`receiver_id` = 10 or `cr`.`sender_id` = 10) and (`u`.`name` LIKE 'pri' or `u`.`email` LIKE 'pri') and `cr`.`approve_status` = 1 and `u`.`id` != 10

所以,我无法得到结果,因为在它应该有 % 符号之后,生成的 laravel 查询没有百分比符号

我下面的原始查询运行完美

select `cr`.`id` as `connection_id`, `cr`.`sender_id` as `sen_id`, `cr`.`receiver_id` as `rec_id`, `cr`.`approve_status`, `u`.`id`, `u`.`user_type`, `u`.`user_type_id`, IF(u.avatar = '', 'uploads/avatar/default.jpg', u.avatar) as avatar, `u`.`name`, `u`.`email` from `connection_request` as `cr` left join `users` as `u` on `u`.`id` = `cr`.`sender_id` or `u`.`id` = `cr`.`receiver_id` where (`cr`.`receiver_id` = 10 or `cr`.`sender_id` = 10) and (`u`.`name` LIKE '%pri%' or `u`.`email` LIKE '%pri%') and `cr`.`approve_status` = 1 and `u`.`id` != 10

请检查并帮助我,

谢谢

最佳答案

就我个人而言,我使用它并且效果相当好......

$query->where('u.name','LIKE', "%{$searchValue}%")->get();

关于mysql - laravel 之类的查询不起作用?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44408610/

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