Question

我有一个问题，当我查询数据库并获得零结果时，我得到一个空数组，但我想抛出'ModelNotFoundException'。这是我的示例代码：

<?php
#App\Http\Controller\MainController
namespace App\Http\Controllers;

use Illuminate\Http\Request;
use Illuminate\Database\Eloquent\ModelNotFoundException;

use App\Http\Requests;
use App\Http\Controllers\Controller;
use App\Demo;
use DB;


class MainController extends Controller
{

  public function getTagRecords($tag, $limit)
  {
    try{
        $results = Demo::getRecordsByTag($tag, $limit);
    }catch (Illuminate\Database\Eloquent\ModelNotFoundException $e){
        return response()->json(['message'=>'hello there error'], 404);
    }
    return $results;
  }

我正在通过AJAX发送请求，并希望返回错误消息和状态。我没有修改过Handler类。

以下是型号代码：

<?php

namespace App;

use Illuminate\Database\Eloquent\Model;
use Illuminate\Database\Eloquent\ModelNotFoundException;  
use DB;

class Demo extends Model
{

    public static function getRecordsByTag($tag, $limit)
    {
        $params = [
            'tag' => $tag, 
            'limit' => $limit,
        ];

        return  DB::select("SELECT d.idDemo, d.idUsuario, u.Referencia, d.Uri_fichero,
            (SELECT Nombre FROM Edad_voz WHERE IdEdad_voz = d.IdEdad_voz) Edad_voz,
            (SELECT Nombre FROM Tono_voz WHERE idTono_voz = d.idTono_voz) Tono_voz,
            (SELECT Idioma_Variante FROM Idioma WHERE idIdioma = d.idIdioma) Idioma,
            (SELECT Nombre FROM Genero WHERE idGenero = d.idGenero) Genero

            FROM Demo AS d
            INNER JOIN logs AS l ON d.idDemo = l.idElemento
            INNER JOIN usuario AS u ON d.idUsuario = u.idUsuario

            WHERE

            l.accion = :tag AND
            l.procesado = '0'

            ORDER BY idDemo

            LIMIT :limit;", $params);
    }
}

Answer 1

我会重写你的getRecordsByTag函数，如果它是这样的话，请小心：

    $result = DB::select("SELECT d.idDemo, d.idUsuario, u.Referencia, d.Uri_fichero,
        (SELECT Nombre FROM Edad_voz WHERE IdEdad_voz = d.IdEdad_voz) Edad_voz,
        (SELECT Nombre FROM Tono_voz WHERE idTono_voz = d.idTono_voz) Tono_voz,
        (SELECT Idioma_Variante FROM Idioma WHERE idIdioma = d.idIdioma) Idioma,
        (SELECT Nombre FROM Genero WHERE idGenero = d.idGenero) Genero

        FROM Demo AS d
        INNER JOIN logs AS l ON d.idDemo = l.idElemento
        INNER JOIN usuario AS u ON d.idUsuario = u.idUsuario

        WHERE

        l.accion = :tag AND
        l.procesado = '0'

        ORDER BY idDemo

        LIMIT :limit;", $params);

    if (is_empty($result)) throw new ModelNotFoundException;

    return $result;

所以基本上：如果你的结果为空则抛出异常。您的控制器不需要对此进行任何修改。

Answer 2

首先，您最好删除命名空间前的注释：#App\Http\Controller\MainController

第二，您可以并且应该检查该类是否存在，然后手动抛出异常

根据您使用的PHP版本，您应该能够执行以下操作：

if (class_exists(Demo::class)) {
    $results = Demo::getRecordsByTag($tag, $limit);
} else {
    // or you can throw an exception from here
    // throw new Illuminate\Database\Eloquent\ModelNotFoundException('hello there error');
    return response()->json(['message'=>'hello there error'], 404);
}

更新：我刚刚看到你提供了getRecordsByTag（）方法调用背后的代码。我要说方法添加文档来解释它可以返回一个空的响应，例如array()并且由调用者决定如何处理特定情况，使模型保持干净并且没有业务逻辑。因此在调用者中检查结果是否为空集合，如果是这种情况则抛出异常，因为它是其运行所需的特定逻辑。

if (count($results = Demo::getRecordsByTag($tag, $limit)) === 0) {
    throw new RecordNotFoundException("Tag: $tag was not found in the Demo table");
}

Laravel返回一个空数组而不是状态为200的异常

2 个答案: