SQLSTATE [23502]：非空违规：7 错误 - laravel

Posted 2023-03-14

【中文标题】SQLSTATE [23502]：非空违规：7 错误 - laravel

【英文标题】：SQLSTATE[23502]: Not null violation: 7 ERROR - laravel

【发布时间】：2021-09-18 01:57:30

【问题描述】：

以下是上传文件的 API。 它显示错误 -

SQLSTATE[23502]: Not null violation: 7 ERROR: null value in column \"Userid\" of relation \"files\" violates not-null constraint\nDETAIL: Failing row contains (7, null, Distance Learning Promo.mp4, null, null, null, 2021-07-07 13:38:07, 2021-07-07 13:38:07). (SQL: insert into \"files\" (\"FileName\", \"updated_at\", \"created_at\") values (Distance Learning Promo.mp4, 2021-07-07 13:38:07, 2021-07-07 13:38:07) returning \"id\").

Postman 显示状态500 internal status error。

我不认为我的 Userid 列是 null 。请帮我提些建议

控制器

<?php
namespace App\Http\Controllers;
use Illuminate\Http\Request;
use App\Models\Uploads;
use Illuminate\Support\Facades\Auth;

class UploadController extends Controller

    public function __construct()
    
        $this->middleware('auth:api');
    
    public function upload(Request $request)
    
        $file=$request->file('file_name');
        $fileName=$file->getClientOriginalName();

        $folder = uniqid().'-'.now()->timestamp;
        $file->storeAs('public/other-document/'.$folder, $fileName);
       
       $filePath = $folder.'/'.$fileName;
       $dataToInsert = array();
       $dataToInsert['FileName'] = $fileName;
       $dataToInsert['FilePath'] = $filePath;
       if($request->has('uploadType'))
       
           $dataToInsert['FileType'] = $request->uploadType;
       
       $dataToInsert['uploadedBy'] = Auth::user()->id;
       $dataToInsert['Userid'] = $request->bearerToken();
       Uploads::create($dataToInsert); 
    

迁移文件

Schema::create('files', function (Blueprint $table) 
            $table->id();
            $table->string('Userid');
            $table->string('FileName');
            $table->string('FilePath');
            $table->string('FileType');
            $table->integer('uploadedBy');
            $table->timestamps();
        );

【问题讨论】：

请检查$dataToInsert['Userid'] = $request->bearerToken();是否有任何值

@JEJ 它具有价值。即使Userid 列也插入了这样的虚拟值-$dataToInsert['Userid'] =1，它也会显示相同的错误。

【参考方案1】：

 $dataToInsert['Userid'] = $request->bearerToken(); // error is here

if(!$dataToInsert['Userid'])
 return "Null value in userID";

检查 $dataToInsert['Userid'] 是否持有任何值。

还要确保您在标头中传递 BearerToken

【讨论】：

它具有价值。即使 Userid 列插入了一个像这样的虚拟值 - $dataToInsert['Userid'] =1 ，它也会显示相同的错误。

@MG 检查模型中的可填充项