如何在 laravel 8 中创建具有一对多关系的 API？答案

【问题标题】：如何在 laravel 8 中创建具有一对多关系的 API？
【发布时间】：2022-01-22 16:00:25
【问题描述】：

我想为报告存储创建一个 API，并且可以选择添加多个违规者。

我有一个问题，当我添加一个违规者时，在 POSTMAN 中它显示错误“调用成员函数 getClientOriginalName() on null”。

如何解决？或者如何正确地创建一个具有一对多关系的附加数据的 API？

以下，我自己做的代码；

Controller.php

    public function store(ReportRequest $request)
    {
        try {
            
            $report = Report::create([
                'category_id'       => $request->category_id,
                'user_id'           => Auth::user()->id,
                'title'             => $request->title,
                'description'       => $request->description,
                'incident_date'     => $request->incident_date,
                'injured_victims'   => $request->injured_victims,
                'survivors'         => $request->survivors,
                'dead_victims'      => $request->dead_victims,
                'location'          => $request->location,
                'latitude'          => $request->latitude,
                'longitude'         => $request->longitude
            ]);

            $data[] = array(
                    'violator_photo' => $request->file('violator_photo'),
                    'violator_name'    => $request->violator_name,
                    'violator_age'   => $request->violator_age,            
                    'violator_phone' => $request->violator_phone 
                    );
            
            foreach ($data as $key ) {   

                $photo = $key['violator_photo'];
                $fileName = time().'_'.$photo->getClientOriginalName();
                $filePath = $photo->storeAs('images/violators', $fileName, 'public');

                Violator::create([
                    'report_id' => $report->id,
                    'name'      => $key['violator_name'],
                    'photo'     => $filePath,
                    'age'       => $key['violator_age'],
                    'phone'     => $key['violator_phone'],
                ]);       
            }    
            
            return response()->json([
                'status' => '200',
                'message' => 'success',
                'data' => $report,
            ]);
        } catch (\Exception$err) {
            return $this->respondInternalError($err->getMessage());
        }
    }

【问题讨论】：

标签： php laravel api eloquent laravel-8

【解决方案1】：

用这段代码试试

public function store(ReportRequest $request)
{
    try {

        $report = Report::create([
            'category_id'       => $request->category_id,
            'user_id'           => Auth::user()->id,
            'title'             => $request->title,
            'description'       => $request->description,
            'incident_date'     => $request->incident_date,
            'injured_victims'   => $request->injured_victims,
            'survivors'         => $request->survivors,
            'dead_victims'      => $request->dead_victims,
            'location'          => $request->location,
            'latitude'          => $request->latitude,
            'longitude'         => $request->longitude
        ]);
        if($report){
            if($request->hasFile('violator_photo')) {
                $violator_photo=$request->file('violator_photo');
                $fileName = time().'_'.$violator_photo->getClientOriginalName();
                $filePath = $violator_photo->storeAs('images/violators', $fileName, 'public');
            }

            Violator::create([
                'report_id' => $report->id,
                'name'      => $request->violator_name,
                'photo'     => $filePath?:'',
                'age'       => $request->violator_age,
                'phone'     => $request->violator_phone,
            ]); 
        }
        
        return response()->json([
            'status' => '200',
            'message' => 'success',
            'data' => $report,
        ]);
    } catch (\Exception$err) {
        return $this->respondInternalError($err->getMessage());
    }
}

【讨论】：

【解决方案2】：

错误getClientOriginalName()，我认为是因为“violator_photo”不是文件，而是字符串/文本

所以，如果你想通过 json 请求发送文件，简单的方法是将文件转换为 base64

在控制器上，您可以对其进行解码，并保存文件

这里是如何解码它的例子，还有一件事你需要检查base64。一些 base64 编码，添加有关文件类型的额外信息，例如，如果文件是图像 png，则数据如下所示

data:image/png;base64,iVBORw0KGg..rest of base64

所以如果数据看起来像这样，您需要删除信息并获取 base64 只需您使用此代码即可完成

$str ="base64 code";
$converted = substr($str, strpos($str,',')+1);
$image = base64_decode($base64_str);
$filename = "the file name.png"; // or you can get the extension name by the information the include on the first of base64
$path = "images/violators/";
Storage::disk('your disk')->put($path.$filename, $image, 'public');

【讨论】：