如何在laravel8中創建具有一對多關系的API？

我想為報告存盤創建 API，并且可以選擇添加多個違規者。

我有一個問題，當我添加一個違規者時，在 POSTMAN 中它顯示錯誤“呼叫成員函式 getClientOriginalName() on null”。

如何解決？或者如何正確創建一個帶有一對多關系附加資料的 API？

以下是我制作的代碼；

控制器.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());
        }
    }

uj5u.com熱心網友回復：

用這個代碼試試

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());
    }
}

uj5u.com熱心網友回復：

錯誤 getClientOriginalName()，我認為這是因為“violator_photo”不是檔案，而是字串/文本

所以，如果你想在 json 請求上發送檔案，簡單的方法是將檔案轉換為 base64

在控制器上，您可以對其進行解碼并保存檔案

這是如何解碼它的示例，您需要檢查base64。一些base64編碼，添加關于檔案型別的額外資訊，例如如果檔案是影像png，資料看起來像這樣

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

因此，如果資料看起來像這樣，您需要洗掉資訊并僅獲取 base64，您可以使用此代碼進行操作

$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');

標籤：php 拉拉维尔 接口 雄辩 laravel-8

上一篇：如何從json中的api創建url

下一篇：調整公開可用的旅行時間腳本以允許航點