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我想将此 HTML 请求帖子“翻译”到Android：

<form name="myForm" action="https://mysite.example" method="POST">
  <input type="hidden" name="Key1" value=1>
  <input type="hidden" name="Key2" value="2">
  <input type="hidden" name="Key3" value="3">
</form>

请注意，第一个值是整数而不是字符串。

所以我尝试使用在stackoverflow上找到的代码：

public String  performPostCall(String requestURL, HashMap<String, String> postDataParams) {
    URL url;
    String response = "";
    try {
        url = new URL(requestURL);
        HttpsURLConnection conn = (HttpsURLConnection) url.openConnection();
        conn.setReadTimeout(15000);
        conn.setConnectTimeout(15000);
        conn.setRequestMethod("POST");
        conn.setDoInput(true);
        conn.setDoOutput(true);

        OutputStream os = conn.getOutputStream();
        BufferedWriter writer = new BufferedWriter(
                new OutputStreamWriter(os, "UTF-8"));
        writer.write(getPostDataString(postDataParams));

        writer.flush();
        writer.close();
        os.close();
        int responseCode=conn.getResponseCode();

        if (responseCode == HttpsURLConnection.HTTP_OK) {
            String line;
            BufferedReader br=new BufferedReader(new InputStreamReader(conn.getInputStream()));
            while ((line=br.readLine()) != null) {
                response+=line;
            }
        }
        else {
            response="";
        }
    } catch (Exception e) {
        e.printStackTrace();
    }
    return response;
}

private String getPostDataString(HashMap<String, String> params) throws UnsupportedEncodingException {
    StringBuilder result = new StringBuilder();
    boolean first = true;
    for(Map.Entry<String, String> entry : params.entrySet()){
        if (first)
            first = false;
        else
            result.append("&");

        result.append(URLEncoder.encode(entry.getKey(), "UTF-8"));
        result.append("=");
        result.append(URLEncoder.encode(entry.getValue(), "UTF-8"));
    }

    return result.toString();
}

在我的 AsyncTask 中，我正在做：

HashMap<String, String> parameters = new HashMap<>();
String url = "https://mysite.example";
String result;

parameters.put("Key1", "1");
parameters.put("Key2", "2");
parameters.put("Key3", "3");
result = performPostCall(url, parameters);

但这不起作用。怎么了？

4

2 回答 2

0

我认为你错过了一条重要的线，它在你的输出流之前建立了实际的连接

conn.connect();

您还应该需要在您的 AndroidManifest.xml 中添加权限

<uses-permission android:name="android.permission.INTERNET"/>
于 2015-06-27T05:04:27.200 回答
0

您在处理之前关闭“作家”：

writer.flush();
writer.close();
os.close();

int responseCode=conn.getResponseCode();

于 2018-02-22T01:11:46.800 回答