laravel 之类的查询不起作用？答案

【问题标题】：laravel like query is not working?laravel 之类的查询不起作用？
【发布时间】：2017-11-08 13:38:39
【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，我不能得到结果，因为喜欢它应该有 % 符号但是，生成没有百分比符号的 laravel 查询

我下面的原始查询运行良好

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，伙计们，

谢谢

【问题讨论】：

在哪里！= 也许在哪里？
你使用的是什么版本的 Laravel？
@Jerodev Laravel 5.6 版

标签： mysql laravel

【解决方案1】：

就个人而言，我使用它并且效果很好......

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

【讨论】：

【解决方案2】：

你的 where 条件也可以是这样的.. 它会工作..

->where(function($query) use ($searchValue)
                       {
                       if(!empty($searchValue)):
                       $query->Where('u.name','like', DB::raw("'%$searchValue%'"));
                       $query->orWhere('u.email','like', DB::raw("'%$searchValue%'"));
                       endif;
                    })

【讨论】：