Question

我正试图在我的Android应用程序中从Apache HTTP转移到HttpUrlConnection。我被卡住了，我试着到处寻找，但我无法通过它。这是我正在尝试的。

以下是我的HTTP代码：

HttpParams timeoutParams = new BasicHttpParams();
HttpConnectionParams.setConnectionTimeout(timeoutParams, 60000);
HttpConnectionParams.setSoTimeout(timeoutParams, 60000);

DefaultHttpClient httpClient = null;
httpClient = new DefaultHttpClient(timeoutParams);
Cookie podCookie = getPodCookie();
    if (podCookie != null) {
        httpClient.getCookieStore().addCookie(podCookie);
    }
HttpPost postMethod = null;
postMethod.addHeader("Authorization", "<auth-header>");
try {
    List<NameValuePair> nvps = new ArrayList<NameValuePair>();
    for (Entry<String, String> entry : parameters.entrySet()) {
        nvps.add(new BasicNameValuePair(entry.getKey(), entry.getValue()));
    }
    String queryString = URLEncodedUtils.format(nvps, HTTP.UTF_8);
    String modUrl = url + "?" + queryString;
    postMethod = new HttpPost(modUrl);

StringEntity entity = new StringEntity(<String JSON to send>, HTTP.UTF_8);
postMethod.setEntity(entity);
postMethod.setHeader("Content-type", "application/json");
HttpResponse reply = httpClient.execute(postMethod);

这是上述代码的HttpUrlConnection等价物：

List<NameValuePair> nvps = new ArrayList<NameValuePair>();
    for (Entry<String, String> entry: parameters.entrySet()) {
        nvps.add(new BasicNameValuePair(entry.getKey(), entry.getValue()));
     }

String queryString = URLEncodedUtils.format(nvps, HTTP.UTF_8);
String modUrl = baseUrl + "?" + queryString;
HttpURLConnection urlConnection = null;
try {
    URL url1 = new URL(modUrl);
    urlConnection = (HttpURLConnection) url1.openConnection();
    CookieManager cookieManager = new CookieManager();
    CookieHandler.setDefault(cookieManager);
    cookieManager.getCookieStore().add(new URI(url), podCookie);
    urlConnection.setConnectTimeout(60000);
    urlConnection.setReadTimeout(60000);
    urlConnection.setDoOutput(true);
    urlConnection.setRequestMethod("POST");

    urlConnection.setRequestProperty("Authorization", <auth-header>);
    urlConnection.setRequestProperty("Content-Type", "application/json");

    OutputStream os = urlConnection.getOutputStream();
    BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(os, "UTF-8"));
    writer.write(<String JSON to send>);
    writer.flush()
    writer.close();
    os.close();

    InputStream is = new BufferedInputStream(urlConnection.getInputStream());
    String responseString = WebService.convertInputStreamToString(is);

当我尝试上述操作时，我得到401 Unauthorized error。我使用Charles，标题相同。

当我尝试在BufferedWriter而不是网址中添加查询参数时，我将网址更改为基本网址，如下所示：

URL url1 = new URL(baseUrl);

并将以下行添加到writer中，如下所示：

writer.write(modUrl)

当我这样做时，我得到500 Internal Server Error。

在这两种情况下，我都会在IOException行上获得FileNotFoundException InputStream。

关于如何解决这个问题的任何想法？

Answer 1

您应该从成功的httpclient调用中转储标头，这样您就可以准确地知道正在使用OK请求发送哪些标头。

目前尚不清楚如何设置“授权”标题

尚不清楚将什么Json值设置到'StringEntity'中。

在尝试使用HttpUrlConnection之前，您应该确切地知道在一个好的调用中发送了什么（httpClient调用或Curl CLI调用）。然后在那里设置相同的头，将相同的JSON写入Connections的outputStream，你应该得到相同的结果。

两个标题的卷曲表达式以及curl.-d中的JSON实体切换值....

curl -v -X POST \
  -H "Content-type: application/json" \
  -H "Authorization: aValueforAuth" \
  -d '{"score":1337,"playerName":"Sean Plott",...}' \
  http://domain/path?parm1=$urlEncodedVal-1&parm2=$urlEncodedVal-2

Android上的HTTPPost vs HttpUrlConnection POST

1 个答案: