Laravel-5.6 'LIKE‘在哪里选择

Question

我使用这个查询并得到一个错误：

 $description = $request->get('description');
        if (!empty($description)){
        $description_query = Transcationhistorique::where(['sender_id' => $user_id, "%$description%", 'LIKE','description'])
            ->orWhere('receiver_id', $user_id)->get();
        }else{
            $description_query  =  "" ;
        }

这就是我得到的错误：

"SQLSTATE42S22:未找到列：'where子句‘1054个未知列'0’(sender_id = 32和0 = %salaire%和1 = LIKE和2 = description)或receiver_id =32)

我真正想要的是：

select * from `transcation_historique` where (`sender_id` = 32 and `description` = %salaire%) or `receiver_id` = 32)

Answer 1

尝尝这个,

$description_query = Transcationhistorique::where(
                         function($query) use ($user_id, $description){
                             return $query->where('sender_id', $user_id)
                                          ->where('description', 'like', '%' . $description .'%');

                         }
                     )
                     ->orWhere('receiver_id', $user_id)
                     ->get();

Answer 2

您的where查询结构似乎不正确。如果要使用"=“Source以外的运算符，则应使用以下结构

$query->where([
    ['column_1', '=', 'value_1'],
    ['column_2', '<>', 'value_2'],
    [COLUMN, OPERATOR, VALUE],
    ...
])

我的建议是：

 $description = $request->get('description');
 if (!empty($description)){
     $description_query = Transcationhistorique::where([
            ['sender_id', '=', $user_id],
            ['description', 'LIKE', "%{$description}%"]
     ])
     ->orWhere('receiver_id', $user_id)->get();
 }else{
     $description_query  =  "" ;
 }

Answer 3

试试这个：

Transcationhistorique::where('receiver_id', '=', $user_id)
        ->orWhere(function ($query) use ($user_id, $description) {
            $query->where('sender_id', '=', $user_id)->where('description', 'LIKE', "%".$description."%");
        })->get();