使用改造上传图像到 Slim 框架 API

Question

所以我想将使用retrofit 的图像上传到我的Slim framework API。现在我可以在我的另一个项目中使用单独的.php 文件成功上传图像，该文件通过$_FILES 对象接收图像，就是这样

<?php
ini_set('display_errors',1);
ini_set('upload_max_filesize', '20M');
ini_set('post_max_size', '20M');
ini_set('max_input_time', 30000);
ini_set('max_execution_time', 30000);

require_once '../include/Functions.php';
$db = new Functions();


$result['error'] = true;
$result['message'] = 'Error occurred, try again1';
$userfk = $_POST['userid'];
$fileNames = array();



for($i = 0; $i < count ( $_FILES ['file'] ['name'] ); $i ++) {
try {

    $extension  = pathinfo($_FILES ['file'] ["name"][$i], PATHINFO_EXTENSION);
    array_push($fileNames,microtime_float().'_'.rand(1, 99999999).'.'.$extension);

    if (move_uploaded_file( $_FILES ['file'] ["tmp_name"][$i], "../images/iduploads/".$fileNames[$i])) {

            $result['error'] = false;
            $result['message'] = 'Upload success';

    } else {

            $result['error'] = true;
            $result['message'] = 'Something went wrong, try again';

        throw new Exception('Could not move file');
    }
} catch (Exception $e) {

            $result['error'] = true;
            $result['message'] = 'Error, photos did not upload, try again';
}

}


if(count ( $_FILES ['file'] ['name']) > 0 && !empty($fileNames))
    $db->uploadIdentificationDocs($userfk, $fileNames[0], $fileNames[1],    $fileNames[2]);

  echo json_encode($result);




function microtime_float(){
 list($usec, $sec) = explode(" ", microtime());
 return ((float)$usec + (float)$sec);
}


?>

从我的 Android 端我使用

@POST("idUpload.php")
Call<Result> uploadMultiFile(@Body RequestBody file);

和

    APIService service = retrofit.create(APIService.class);
    MultipartBody.Builder builder = new MultipartBody.Builder();
    builder.setType(MultipartBody.FORM);
    User user = SharedPreferencesUtils.getUserPreferences(this);
    builder.addFormDataPart("userid", user.getUserId());
    builder.addFormDataPart("file[]", profilePicture.getName(), RequestBody.create(MediaType.parse("multipart/form-data"), profilePicture));
    MultipartBody requestBody = builder.build();
    ProgressRequestBody progressRequestBody = new ProgressRequestBody(requestBody, this);
Call<Result> call = service.uploadMultiFilee(progressRequestBody);

这可行，但我不喜欢离开Slim 上传图片的想法，我选择它的原因是因为它的简单和干净的代码。这感觉不对，所以我尝试使用slims'提供的函数上传

$app->post('/uploadprofilepicture/{userid}', function (Request $request, Response $response) {
    $db = new Functions();
    $userFk = $request->getAttribute('userid');
    $result['error'] = true;
    $result['message'] = 'Received here';
    try {
        $uploadedFiles = $request->getUploadedFiles();
        if ($uploadedFile->getError() === UPLOAD_ERR_OK) {
            $filename = moveUploadedFile($directory, $uploadedFile);
            $result['error'] = true;
            $result['message'] = 'Upload successs image '.$filename;
        }
    }catch (Exception $e) {

         $result['error'] = true;
         $result['message'] = 'Error occured: '.$e;
    }

    $response->getBody()->write((json_encode($result)));
});

但它失败了，它只是返回error 500，在运行postman Call to a member function getError() on null 意味着$request->getUploadedFiles() 为空

Answer 1

@Zamrony 是对的，至少他帮我弄清楚了我的问题，最后这里是工作代码

use Slim\Http\UploadedFile;
$profilePhotoContainer = $app->getContainer();
$profilePhotoContainer['upload_directory'] = '../images/iduploads/';


 $app->post('/uploadprofilepicture', function (Request $request, Response $response) {
 $db = new Functions();
 $directory = $this->get('upload_directory');
 $result['error'] = true;
 $result['message'] = 'Received here';
 try {
    $uploadedFiles = $request->getUploadedFiles();
    $uploadedFile = $uploadedFiles['profilephoto'];
    if ($uploadedFile->getError() === UPLOAD_ERR_OK) {
        $filename = moveUploadedFile($directory, $uploadedFile);
        $result['error'] = true;
        $result['message'] = 'Upload successs image '.$filename;
     }
  }catch (Exception $e) {

     $result['error'] = true;
     $result['message'] = 'Error occured: '.$e;
 }

$response->getBody()->write((json_encode($result)));
});



function moveUploadedFile($directory, UploadedFile $uploadedFile){
 $extension = pathinfo($uploadedFile->getClientFilename(), PATHINFO_EXTENSION);
 $basename = bin2hex(random_bytes(8));
 $filename = sprintf('%s.%0.8s', $basename, $extension);
 $uploadedFile->moveTo($directory . DIRECTORY_SEPARATOR . $filename);
 return $filename;
}

Answer 2

应该是 $uploadedFiles 而不是 $uploadedFile 。这就是你得到 null 的原因。

编辑：$uploadedFile 变量未初始化。这就是你得到空值的原因。例如如果html定义如下

<form method="post" enctype="multipart/form-data">
    <input type="file" name="myUploadedFile">
</form>

那么在处理文件上传的代码中，应该是

$uploadedFiles = $request->getUploadedFiles();
$uploadedFile = $uploadedFiles['myUploadedFile'];
if ($uploadedFile->getError() === UPLOAD_ERR_OK) {
     //TODO: move file
}

阅读Uploading files using POST forms了解更多信息。