问

使用android进行多部分/表单数据构建

跑车世界Y 发布于 2023-01-29 14:37

http

我试图做出HttpPost与multiPart/form-data通过我的Android应用程序.我有一个邮递员测试,正在使用我的api和postman中的请求预览,看起来像这样:

POST /api/0.1/content/upload HTTP/1.1
Host: 54.221.194.167
X-AUTHORIZATION: 166e649911ff424eb14446cf398bd7d6
Cache-Control: no-cache
Postman-Token: 2412eba9-f72d-6f3b-b124-6070b5b26644

----WebKitFormBoundaryE19zNvXGzXaLvS5C
Content-Disposition: form-data; name="file01"

{"mime_type":"image/jpeg","title":"IMG_20140131_111622"}
----WebKitFormBoundaryE19zNvXGzXaLvS5C
Content-Disposition: form-data; name="file01"; filename="addedaslib.jpg"
Content-Type: image/jpeg


----WebKitFormBoundaryE19zNvXGzXaLvS5C

我试图使用我的android HttpPost使用multipart/form-data复制它,但它似乎没有工作.有没有办法"预览"我的请求,看看它实际上是如何发布到api？我究竟做错了什么？我的代码:

public HttpResponse invokeXAUTHPOSTService(String url, String token, File file) {


        client = new DefaultHttpClient();
        HttpPost request = new HttpPost(url);
        HttpResponse response = null; 
        MultipartEntity mpe = new MultipartEntity();
    try {


        Log.v("API", "URL:"+url);
        request.setHeader("Content-Type", "multipart/form-data");
        request.addHeader("X-AUTHORIZATION",token);
        request.addHeader("Cache-Control", "no-cache");

        DRPContentForUpload content = new DRPContentForUpload(file);
        String jsonObject = DRPJSONConverter.toJson(content);

        FormBodyPart part1= new FormBodyPart("file01", new StringBody(jsonObject));
        FormBodyPart part2= new FormBodyPart("file01", new FileBody(file)); 
        mpe.addPart(part1);
        mpe.addPart(part2);

        //

        request.setEntity(mpe);
        Log.v("RAW REQUEST", "request looks like:"+mpe.toString());
        response = client.execute(request);

编辑

我能够和我的API团队交谈,他们说我的帖子实际上是这样的:

--0ieMJK6PPwcrM_K3KQvl6eNDGqooZPzJcvHOm0
Content-Disposition: form-data; name="file01"
Content-Type: text/plain; charset=US-ASCII
Content-Transfer-Encoding: 8bit

{"mime_type":"image/jpeg","title":"IMG_20140131_111622"}
--0ieMJK6PPwcrM_K3KQvl6eNDGqooZPzJcvHOm0
Content-Disposition: form-data; name="file01"; filename="IMG_20140131_111622.jpg"
Content-Type: application/octet-stream
Content-Transfer-Encoding: binary

这说它仍然没有工作,并吐出一个错误,我错过了params

这是我项目中包含的库的屏幕截图:

在此输入图像描述

1 个回答

所以在搜索到高分和低分以及几乎放弃之后,这个链接终于有了帮助

这是最终的工作代码:

 public HttpResponse invokeXAUTHPOSTService(String url, String token, File file) {

    client = new DefaultHttpClient();

    HttpPost request = new HttpPost(url);

    HttpResponse response = null;

    DRPContentForUpload content = new DRPContentForUpload(file);
    String jsonObject = DRPJSONConverter.toJson(content);
    String BOUNDARY= "--eriksboundry--";

    request.setHeader("Content-Type", "multipart/form-data; boundary="+BOUNDARY);
    request.addHeader("X-AUTHORIZATION",token);
    MultipartEntity entity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE,BOUNDARY,Charset.defaultCharset());
    try {


        entity.addPart("file01", new StringBody(jsonObject));

        entity.addPart("file01", new FileBody(file));

        request.addHeader("Accept-Encoding", "gzip, deflate");

    } catch (UnsupportedEncodingException e) {
        Log.v("encoding exception","E::: "+e);
        e.printStackTrace();
    }
    request.setHeader("Accept", "application/json");
    request.setHeader("Content-Type", "multipart/form-data; boundary="+BOUNDARY);
    request.setEntity(entity);

    try {




        response = client.execute(request);



    } catch (ClientProtocolException e) {

        e.printStackTrace();
    } catch (IOException e) {

        e.printStackTrace();
    }


    return response;

}

2023-01-29 14:40 回答

VASTEw

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