Question

我已经开始测试http客户端apache API了。我需要它，因为我想发送请求并接收对virustotal API的响应。病毒总API需要发布请求中的参数：

api键值（每个用户的唯一值）
我从他们的网站上了解到的文件本身。

例如：

>>> url = "https://www.virustotal.com/vtapi/v2/url/scan"
>>> parameters = {"url": "http://www.virustotal.com",
...               "apikey": "-- YOUR API KEY --"}
>>> data = urllib.urlencode(parameters)
>>> req = urllib2.Request(url, data)

目前，我正在尝试用Java而不是Python做同样的事情。以下是我的源代码的一部分，用于指导整个步骤：

  CloseableHttpClient httpClient = HttpClientBuilder.create().build();
  //create post request
  HttpPost request = new HttpPost("https://www.virustotal.com/vtapi/v2/file/scan");
  //http json header
  request.addHeader("content-type", "application/json");

  String str  = gson.toJson(param);

  String fileName = UUID.randomUUID().toString() + ".txt";

  try {
    //API key
    StringEntity entity = new StringEntity(str);
    Writer writer = new BufferedWriter(new FileWriter(fileName));
    writer.write(VirusDefinitionTest.malware());
    request.setEntity(entity);
  } catch (IOException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
  }
  MultipartEntityBuilder builder = MultipartEntityBuilder.create();
  FileBody fileBody = new FileBody(new File(fileName)); 

  builder.addTextBody("my_file", fileName);
  HttpEntity entity = builder.build();
  request.setEntity(entity);

  HttpResponse response;
  try {
    response = httpClient.execute(request);

...

不幸的是，我收到了HTTP / 1.1 403 Forbidden。显然，错误是在实体的某个地方，但我找不到如何做到这一点。任何帮助都会受到热烈欢迎。

Answer 1

这适用于Apache 4.5.2 HttpClient：

CloseableHttpClient httpclient = HttpClients.createDefault();
try {
    HttpPost httppost = new HttpPost("https://www.virustotal.com/vtapi/v2/file/scan");

    FileBody bin = new FileBody(new File("... the file here ..."));
    // the API key here
    StringBody comment = new StringBody("5ec8de.....", ContentType.TEXT_PLAIN);

    HttpEntity reqEntity = MultipartEntityBuilder.create()
            .addPart("apikey", comment)
            .addPart("file", bin)
            .build();

    httppost.setEntity(reqEntity);

    System.out.println("executing request " + httppost.getRequestLine());
    CloseableHttpResponse response = httpclient.execute(httppost);
    try {
        System.out.println("----------------------------------------");
        System.out.println(response.getStatusLine());
        HttpEntity resEntity = response.getEntity();
        if (resEntity != null) {
            System.out.println("ToString:" + EntityUtils.toString(resEntity));
        }
        EntityUtils.consume(resEntity);
    } finally {
        response.close();
    }
} finally {
    httpclient.close();
}

重要的部分是reqEntity，必须有两个专门命名的字段"apikey"和"file"。使用有效的API密钥运行此代码可以获得API的预期响应。

Answer 2

问题似乎是首先添加明确的＆＃34;内容类型＆＃34;标题是＆＃34; application / json＆＃34;最后你发送了Muiltipart实体。您需要将所有参数和文件添加到Muiltipart实体。现在参数不会发送，因为它们被Muiltipart实体覆盖：

CloseableHttpClient httpClient = HttpClientBuilder.create().build();
//create post request
HttpPost request = new HttpPost("https://www.virustotal.com/vtapi/v2/file/scan");
//http json header
request.addHeader("content-type", "application/json");

String str  = gson.toJson(param);

String fileName = UUID.randomUUID().toString() + ".txt";

try {
//API key
StringEntity entity = new StringEntity(str);
Writer writer = new BufferedWriter(new FileWriter(fileName));
writer.write(VirusDefinitionTest.malware());
// --> You set parameters here !!!
request.setEntity(entity);
} catch (IOException e) {
  // TODO Auto-generated catch block
  e.printStackTrace();
}
MultipartEntityBuilder builder = MultipartEntityBuilder.create();
FileBody fileBody = new FileBody(new File(fileName)); 

builder.addTextBody("my_file", fileName);
HttpEntity entity = builder.build();
// --> You overwrite the parameters here !!!
request.setEntity(entity);

HttpResponse response;
try {
 response = httpClient.execute(request);

如何使用apache httpClient API将file + api密钥和其他参数作为单个请求发送？

2 个答案: