【问题标题】:Illegal offset type in isset or empty Laravelisset 或空 Laravel 中的非法偏移类型
【发布时间】:2019-11-28 01:34:00
【问题描述】:

我使用以下查询来获取 Laravel 中特定表的项目 ID 和 ID。

$projectlist = Projects::Where('customer',$request->input('client_name'))->pluck(DB::raw("CONCAT('project_prefix','project_nos') AS projectid"),"id");

我收到以下错误。

message Illegal offset type in isset or empty
exception   ErrorException
file    /var/www/msstone/vendor/laravel/framework/src/Illuminate/Support/Str.php

项目模型

namespace App;

use Illuminate\Database\Eloquent\Model;

class Projects extends Model
{
    protected $guarded = ['id'];

    protected $table = 'projects';

}

【问题讨论】:

    标签: laravel laravel-5 laravel-5.8


    【解决方案1】:

    尝试通过以下方式更改 eloquent 查询:

    $projectlist=Projects::Where('customer',$request->input('client_name'))->pluck(DB::raw("CONCAT('project_prefix','project_nos') AS projectid"),"id");
    

    $projectlist = Projects::select(DB::raw("CONCAT(project_prefix,project_nos) AS projectid"),'id')->where('customer',$request->input('client_name'))->pluck("projectid","id");
    

    【讨论】:

    • 它的返回“project_prefixproject_nos”
    • dd($projectlist) 有什么用?
    • 1 => "project_prefixproject_nos" 2 => "project_prefixproject_nos" 3 => "project_prefixproject_nos"
    • 提供您的Projects 型号代码。还有projects数据库表。
    • projects 表添加database tablemigration 文件。
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