Java Spring RestTemplate：禁止获取403多部分/表单数据，但使用curl

我正在使用spring rest模板发布一个文件和一个字符串参数的请求

表单数据

，下面是代码：

    HttpHeaders headers = new HttpHeaders();
    headers.set("Authorization", "Bearer ...");
    httpHeaders.setContentType(MediaType.MULTIPART_FORM_DATA);
    File file=new File("src\main\resources\test.json");
    BufferedInputStream in = new BufferedInputStream(new FileInputStream(file));
    byte[] fileBytes = toByteArray(in);

    MultiValueMap<String, String> fileMap = new LinkedMultiValueMap<>();
    ContentDisposition contentDisposition = ContentDisposition
            .builder("form-data")
            .name("file")
            .filename("test.json")
            .build();
    fileMap.add(HttpHeaders.CONTENT_DISPOSITION, contentDisposition.toString());
    fileMap.add(HttpHeaders.CONTENT_TYPE, MediaType.APPLICATION_JSON_VALUE);
    org.springframework.http.HttpEntity<byte[]> fileEntity =
            new org.springframework.http.HttpEntity<>(fileBytes, fileMap);

    MultiValueMap< String, Object > body = new LinkedMultiValueMap<>();
    body.add("stringFormInput", "xyz");
    body.add("file", fileEntity);
    org.springframework.http.HttpEntity< MultiValueMap< String, Object > > entity =
            new org.springframework.http.HttpEntity<>(body, httpHeaders);

    ResponseEntity< JsonNode > responseEntity =
            restTemplate.postForEntity("https://uploadFile", entity, JsonNode.class);

我也试过使用

httpClient

，但仍然是403，没有任何原因消息

HttpPost post = new HttpPost("https://uploadFile");
post.setHeader(HttpHeaders.CONTENT_TYPE, MediaType.MULTIPART_FORM_DATA_VALUE);
post.setHeader(HttpHeaders.AUTHORIZATION, "Bearer ...");

File file=new File("src\main\resources\test.json");
MultipartEntityBuilder builder = MultipartEntityBuilder.create();
builder.setMode(HttpMultipartMode.BROWSER_COMPATIBLE);
builder.addBinaryBody("file", file, ContentType.DEFAULT_BINARY, "test.json");
builder.addTextBody("stringFormInput", "xyz", ContentType.DEFAULT_BINARY);
HttpEntity entity = builder.build();

post.setEntity(entity);
HttpClient httpClient = HttpClientBuilder.create().build();
HttpResponse response = httpClient.execute(post);

编辑

现在，我通过从JAVA执行curl来解决问题

    File file=new File("src\main\resources\test.json");
    String[] command = {"curl", "--location",
            "--request", "POST", "https://uploadFile",
            "--header", "Content-Type: multipart/form-data",
            "--header", "Authorization: Bearer ...",
            "-F", "file=" + "@"+file.getPath(),
            "--form", "stringFormInput=xyz"};

    ProcessBuilder builder = new ProcessBuilder(command);
    builder.redirectErrorStream(true);

    String curlResult = "";
    String line = "";

    try {
        Process process = builder.start();
        BufferedReader r = new BufferedReader(new InputStreamReader(process.getInputStream()));
        while (true) {
            line = r.readLine();
            if (line == null) {
                break;
            }
            curlResult = curlResult + line;
        }
    } catch (Exception e) {
        e.printStackTrace();
    }

---

所以我遇到了一个类似的问题，cURL执行正常，但是Spring代码给出了403。 问题在于接收此请求的目标。像Nginx这样的web服务器很少需要默认情况下可以禁用的“用户代理”这样的强制头

默认情况下，Spring不会传递任何额外的头。如果目标正在使用Ngnix或查找他们正在使用的web服务器，请尝试添加“用户代理”

分享我的代码以供参考

这个代码给出了403

HttpHeaders HttpHeaders=新的HttpHeaders（）； httpHeaders.setContentType（MediaType.MULTIPART\u FORM\u数据）

这个代码运行良好

HttpHeaders HttpHeaders=新的HttpHeaders（）； httpHeaders.setContentType（MediaType.MULTIPART\u FORM\u数据）； 添加（“用户代理”、“PostmanRuntime/7.26.3”）

---

您需要为httpclient请求添加标题。添加了

内容类型

和

授权

标题，仍然是相同的问题此问题可能会有帮助是的，我也尝试了。我尝试了标题

（“用户代理”，“PostmanRuntime/7.26.3”）

仍然给出403，是的，服务器是ngnix。

    File file=new File("src\main\resources\test.json");
    String[] command = {"curl", "--location",
            "--request", "POST", "https://uploadFile",
            "--header", "Content-Type: multipart/form-data",
            "--header", "Authorization: Bearer ...",
            "-F", "file=" + "@"+file.getPath(),
            "--form", "stringFormInput=xyz"};

    ProcessBuilder builder = new ProcessBuilder(command);
    builder.redirectErrorStream(true);

    String curlResult = "";
    String line = "";

    try {
        Process process = builder.start();
        BufferedReader r = new BufferedReader(new InputStreamReader(process.getInputStream()));
        while (true) {
            line = r.readLine();
            if (line == null) {
                break;
            }
            curlResult = curlResult + line;
        }
    } catch (Exception e) {
        e.printStackTrace();
    }