1

我有一个用于 android 的登录应用程序，它使用 JSON 将数据从数据库解析到应用程序。接受 http 请求的 php api 由标签识别，即“登录”或“注册”，如下所示：

if (isset($_POST['tag']) && $_POST['tag'] != '') {
    "Do stuf
    } else {
       echo "access denied";

该应用程序一直运行良好，但现在我只收到“访问被拒绝”的响应。

public JSONObject loginUser(String email, String password){
    // Building Parameters
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("tag", "login"));
    params.add(new BasicNameValuePair("email", email));
    params.add(new BasicNameValuePair("password", password));
    JSONObject json = jsonParser.getJSONFromUrl(loginURL, params);
    // return json
    // Log.e("JSON", json.toString());
    return json;
}

这是发送请求的 JSONobject，我怀疑它没有正确发送标签。有没有人知道发生了什么？

更新：添加 JSONparser.class

public class JSONParser {

static InputStream is = null;
static JSONObject jObj = null;
static String json = "";

// constructor
public JSONParser() {

}

public JSONObject getJSONFromUrl(String url, List<NameValuePair> params) {

    // Making HTTP request
    try {
        // defaultHttpClient
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(url);
        httpPost.setEntity(new UrlEncodedFormEntity(params));

        HttpResponse httpResponse = httpClient.execute(httpPost);
        HttpEntity httpEntity = httpResponse.getEntity();
        is = httpEntity.getContent();

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

    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                is, "UTF-8"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "n");
        }
        is.close();
        json = sb.toString();
        Log.e("JSON", json);
    } catch (Exception e) {
        Log.e("Buffer Error", "Error converting result " + e.toString());
    }

    // try parse the string to a JSON object
    try {
        jObj = new JSONObject(json);
    } catch (JSONException e) {
        Log.e("JSON Parser", "Error parsing data " + e.toString());
    }

    // return JSON String
    return jObj;

}

}

4

1 回答 1

0

实际上，看起来您似乎在使用错误的 JSON 查询，除非您没有向我们展示更多代码。你错过了 jsonParser 类的创建

public class JSONParser {

static InputStream is = null;
static JSONObject jObj = null;
static String json = "";

// constructor
public JSONParser() {

}

public JSONObject getJSONFromUrl(String url) {

    // Making HTTP request
    try {
        // defaultHttpClient
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(url);

        HttpResponse httpResponse = httpClient.execute(httpPost);
        HttpEntity httpEntity = httpResponse.getEntity();
        is = httpEntity.getContent();          

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

    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                is, "iso-8859-1"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
        is.close();
        json = sb.toString();
    } catch (Exception e) {
        Log.e("Buffer Error", "Error converting result " + e.toString());
    }

    // try parse the string to a JSON object
    try {
        jObj = new JSONObject(json);
    } catch (JSONException e) {
        Log.e("JSON Parser", "Error parsing data " + e.toString());
    }

    // return JSON String
    return jObj;

} }

有关更多信息，请参阅此站点http://www.androidhive.info/2012/01/android-json-parsing-tutorial/

于 2013-02-28T14:35:49.160 回答