laravel之类的查询无法正常工作? [英] laravel like query is not working?
本文介绍了laravel之类的查询无法正常工作?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
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
and it provide mysql query like this
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
因此,我无法获得结果,因为类似它应该具有%
符号,但是生成没有百分比符号的laravel查询
So, I can't able to get result because after like it should have %
symbol but, generate laravel query not having percentage symbol
我下面的原始查询运行正常
my below raw query is working perfect
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
推荐答案
我个人使用它,并且效果很好...
Personally, I use that and it works rather well ...
$query->where('u.name','LIKE', "%{$searchValue}%")->get();
这篇关于laravel之类的查询无法正常工作?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文