发布请求错误404 REST API、JAVA、Springboot[关闭]

Question

编辑问题以包括所需的行为、特定问题或错误，以及再现问题所需的最短代码。这将帮助其他人回答这个问题。

我试着用postman向我的Spring Boot应用程序发送一个简单的post请求。不幸的是，我得到这个404错误。我看过一些教程和文档，但现在仍然知道如何解决这个问题。

控制器

@RestController
public class FileUploadController {
 @PostMapping("/uploadFile")
  public ResponseEntity<FileUploadResponse> uploadFile(@RequestParam("file") MultipartFile multipartFile) throws IOException {
    FileUploadResponse response = new FileUploadResponse();
    //FileUploadUtil is the class we programme for saving file to data server.
    FileUploadUtil fileUploadUtil = new FileUploadUtil();
    //StringUtils.cleanPath:remove invalid characters from the path
    String fileName = StringUtils.cleanPath(Objects.requireNonNull(multipartFile.getOriginalFilename()));
    //
    response.setFileName(fileName);
    response.setDownloadUri("/downloadFile/" + fileUploadUtil.saveFile(fileName, multipartFile));

    return new ResponseEntity<>(response, HttpStatus.OK);
  }
}

Util

  String saveFile(String fileName, MultipartFile multipartFile) throws IOException {
    //Files-Upload is name of Folder that we save the file
    String uploadFileLocation = "Files-Upload";
    Path uploadPath = Paths.get(uploadFileLocation);

    if (!Files.exists(uploadPath.resolve(fileName))) {
      Files.createDirectories(uploadPath);
    }
    /*
    -try catch for caching possible Error.
    -multipartFile:A representation of an uploaded file received in a multipart request.
    -getInputStream:Return an InputStream to read the contents of the file from.
    -Input Stream:you can read data from a Java InputStream as an ordered sequence of bytes. This is useful when reading data
     from a file, or received over the network
     */
    try (InputStream inputStream = multipartFile.getInputStream()) {
      /*.resolve:It returns a string with absolute path, An absolute path always contains the root element and the complete
      directory list required to locate the file*/
      //path is for addressing a folder locally, so this line tell us to get the address of the file in java that we want to save
      // our uploaded file.
      //we search for "file name" in our upload directory
      //(InputStream in, Path target, CopyOption... options)
      Files.copy(inputStream, uploadPath.resolve(fileName), StandardCopyOption.REPLACE_EXISTING);
    } catch (IOException e) {
      throw new RuntimeException("Error Uploading file", e);
    }
    return fileName + RandomStringUtils.randomAlphanumeric(8);
  }
} 

发布人：在此处输入图像描述

Answer 1

如果没有任何错误代码，很难找到问题。请添加它。

Answer 2

解决方案1：通过在属性中设置上下文路径来修复404错误。在应用程序中设置属性。属性/yml文件：

server.servlet.context-path=/easymark

解决方案2：将URL映射添加到@RestController中。

@RestController("/easymark")

解决方案2：更改邮递员URL：

http://localhost:8080/easymark/uploadFile

为此：

http://localhost:8080/uploadFile