Question

我有一个Android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被标记识别，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，但现在我只得到响应“拒绝访问”。

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记。有没有人知道最新情况？

更新：添加了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;

}

}

Answer 1

实际上看起来它看起来你正在使用JSON查询错误无论如何unles有更多的代码你没有向我们展示。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;

} }

有关详情，请参阅此网站http://www.androidhive.info/2012/01/android-json-parsing-tutorial/

解析JSONobject时出错

1 个答案: