Question

//Below is the entire class from which i'm trying to start an activity from 


public class SimpleHttpClient extends Activity {

  public static final int HTTP_TIMEOUT = 30 * 1000;

private static HttpClient mHttpClient;


private static HttpClient getHttpClient() {
    if (mHttpClient == null) {
        mHttpClient = new DefaultHttpClient();
        final HttpParams params = mHttpClient.getParams();
        HttpConnectionParams.setConnectionTimeout(params, HTTP_TIMEOUT);
        HttpConnectionParams.setSoTimeout(params, HTTP_TIMEOUT);
        ConnManagerParams.setTimeout(params, HTTP_TIMEOUT);
    }
    return mHttpClient;
}

public static String executeHttpPost(String url, ArrayList<NameValuePair> postParameters) throws Exception {
    BufferedReader in = null;
    try {
        HttpClient client = getHttpClient();
        HttpPost request = new HttpPost(url);
        UrlEncodedFormEntity formEntity = new UrlEncodedFormEntity(postParameters);
        request.setEntity(formEntity);
        HttpResponse response = client.execute(request);
        in = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));

        StringBuffer sb = new StringBuffer("");
        String line = "";
        String NL = System.getProperty("line.separator");
        while ((line = in.readLine()) != null) {
            sb.append(line + NL);
        }
        in.close();

        String result = sb.toString();
        String disabled = "Disabled";
         if (result.length() == 2)
        {
          return "Disabled";
        }
        else if (result.length() == 3)
      {


         // return "Logged In";
          Intent i = new Intent(this, MainActivity.class);
          startActivity(i);
         //it tells me i cannot reference from a static context
      }
      else if (result.length() == 5)
      {
          return "Unknown";
      }
      else if (result.length() == 4)
      {
          return "Incorrect Login Details";
      }
        else {
          return result+"Unknown"+result.length();
      }
    }

    finally {
        if (in != null) {
            try {
                in.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }
}



public static String executeHttpGet(String url) throws Exception {
    BufferedReader in = null;
    try {
        HttpClient client = getHttpClient();
        HttpGet request = new HttpGet();
        request.setURI(new URI(url));
        HttpResponse response = client.execute(request);
        in = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));

        StringBuffer sb = new StringBuffer("");
        String line = "";
        String NL = System.getProperty("line.separator");
        while ((line = in.readLine()) != null) {
            sb.append(line + NL);
        }
        in.close();

        String result = sb.toString();
        return result;
    }
    finally {
        if (in != null) {
            try {
                in.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }
}
}

Answer 1

startActivity是Activity类中的非静态方法，您尝试从静态方法调用它，这就是编译器不满意的原因。您可以参考api签名here

因此要么从getHttpClient，executeHttpPost方法和mHttpClient字段中删除static关键字，它应该向前移动。

或者您可以使用上下文来调用startActivity，如：

context.startActivity(i);

在创建意图时传递上下文而不是此引用。

Answer 2

您应该将上下文传递给静态方法：

public static String executeHttpPost(Context context, String url, ArrayList<NameValuePair> postParameters) throws Exception {
    ...
    // return "Logged In";
    Intent i = new Intent(context, MainActivity.class);
    context.startActivity(i);
    ...
}

Answer 3

您需要提供Context来启动Intent。在你的情况下你应该改变：

 Intent i = new Intent(this, MainActivity.class);
 startActivity(i);

为：

 Context context;
 // init context here or provide it as a static method argument
 Intent i = new Intent(this, MainActivity.class);
 context.startActivity(i);

因此该方法应如下所示：

public static String executeHttpPost(Context context, String url, ArrayList<NameValuePair> postParameters) throws Exception {
       // ...
      else if (result.length() == 3)
      {
          Intent i = new Intent(this, MainActivity.class);
          context.startActivity(i);
      }
      // ...
}

从静态内容开始活动

3 个答案: