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Php Laravel 5.4显示属于用户的结果_Php_Laravel_Laravel 5.4 - Fatal编程技术网

Php Laravel 5.4显示属于用户的结果

Php Laravel 5.4显示属于用户的结果,php,laravel,laravel-5.4,Php,Laravel,Laravel 5.4,下午好 我试图显示分配给用户的结果,逻辑是,在管理中,用户只看到分配的内容 我的模型采用以下方式,其中在scopeName函数中,我尝试将admins和Propositions表的数据传递给控制器: <?php namespace App; use Illuminate\Database\Eloquent\Model; class Proposal extends Model { /** * The attributes that are mass assignab

下午好

我试图显示分配给用户的结果,逻辑是,在管理中,用户只看到分配的内容

我的模型采用以下方式,其中在scopeName函数中,我尝试将admins和Propositions表的数据传递给控制器:

<?php

namespace App;

use Illuminate\Database\Eloquent\Model;

class Proposal extends Model
{
    /**
     * The attributes that are mass assignable.
     *
     * @var array
     */
    protected $table = 'proposals';

    protected $fillable = [
        'user_id', 'admin_id', 'prefix', 'order', 'title', 'province_id', 'municipalities', 'address', 'description', 'status', 'comments', 'closing_comments', 'institution_id', 'table_id'
    ];

    public function users()
    {
        return $this->belongsTo('App\User');
    }

    public function userid()
    {
        return $this->belongsTo(User::class, 'user_id');
    }

    public function admins()
    {
        return $this->belongsTo('App\Admin');
    }

    public function adminId()
    {
        return $this->belongsTo(Admin::class, 'admin_id');
    }

    public function scopeName($query, $name)
    {
        $query->join('admins', 'proposals.admin_id', '=', 'admins.id')
              ->select(
                'proposals.*',
                'admins.name as admin_name',
                'admins.last_name as admin_last_name',
                'admins.area as admin_area',
                'admins.job_title as admin_job_title',
                'admins.email as admin_email',
                'admins.phone as admin_phone',
                'admins.movil as admin_movil')
              ->where('proposals.order','like',"%$name%")
              ->orWhere('proposals.title','like',"%$name%");
    }
}
/**
 * Show the application dashboard.
 *
 * @return \Illuminate\Http\Response
 */
public function index(Request $request)
{
  $data = Proposal::name($request->get('criteria'))->where('admin_id','=',Auth::id())->orderBy('id','asc')->paginate(5);

  return view('administracion.propuestas.inicio', compact('data'))->with('i', ($request->input('page', 1) - 1) * 5);
}
它抛出的视图中的错误如下:

(2/2) QueryException

SQLSTATE[42S22]: Column not found: 1054 Unknown column 'admins.last_name' in 'field list' (SQL: select `proposals`.*, `admins`.`name` as `admin_name`, `admins`.`last_name` as `admin_last_name`, `admins`.`area` as `admin_area`, `admins`.`job_title` as `admin_job_title`, `admins`.`email` as `admin_email`, `admins`.`phone` as `admin_phone`, `admins`.`movil` as `admin_movil` from `proposals` inner join `admins` on `proposals`.`admin_id` = `admins`.`id` where (`proposals`.`order` like %% or `proposals`.`title` like %%) and `admin_id` = 1 order by `id` asc limit 5 offset 0)

检查admins表中是否有姓氏字段。若否,则

public function scopeName($query, $name)
    {
        $query->join('admins', 'proposals.admin_id', '=', 'admins.id')
              ->select(
                'proposals.*',
                'admins.name as admin_name',
                'admins.area as admin_area',
                'admins.job_title as admin_job_title',
                'admins.email as admin_email',
                'admins.phone as admin_phone',
                'admins.movil as admin_movil')
              ->where('proposals.order','like',"%$name%")
              ->orWhere('proposals.title','like',"%$name%");
    }

在您的方法scopeName中,您忘记返回$queryYes,这是正确的,我没有注意到简单的细节,即低水平咖啡因的影响。