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所以我想使用 TestFairy API（网址：https ://app.testfairy.com/api/upload ）。此 API 调用需要 3 个 post 参数：

  • api_key（细绳）

  • apk_file（.apk 文件）

  • testers_groups（细绳）

到目前为止，我想出了这个：

HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);

List <NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("testers_groups", testers));
params.add(new BasicNameValuePair("api_key", key));

httppost.setEntity(new UrlEncodedFormEntity(params, HTTP.UTF_8));

MultipartEntity reqEntity = new MultipartEntity();
reqEntity.addPart("apk_file", new FileBody(file));

httppost.setEntity(reqEntity);

System.out.println("executing request " + httppost.getRequestLine());
HttpResponse response = httpclient.execute(httppost);
HttpEntity resEntity = response.getEntity();

System.out.println(response.getStatusLine());
if (resEntity != null) {
    System.out.println(EntityUtils.toString(resEntity));
}
if (resEntity != null) {
    resEntity.consumeContent();
}

httpclient.getConnectionManager().shutdown();

但它不起作用。

谁能指出我正确的方向？

4

1 回答 1

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尝试这样的事情：

MultipartEntity entity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
entity.addPart("api_key", new StringBody(key));
entity.addPart("apk_file", new FileBody(file));
entity.addPart("testers_groups", new StringBody(testers));
HttpPost httpPost = new HttpPost(url);
httpPost.setEntity(entity);

HttpClient httpClient = new DefaultHttpClient();
HttpResponse response = httpClient.execute(httpPost);

StringBody可以选择在and构造函数中添加 mime 类型FileBody（但可能不是必需的）。

于 2013-11-28T08:47:21.737 回答