Question

我知道这是一个愚蠢的问题，但自从我在Android Studio中编写Java以来，已经有很长一段时间了。所以实际上我遇到的问题是我无法从一个类返回到另一个类。我在这里尝试了不同类型的解决方案，但它们都没有奏效。不知道为什么。我没有在xml中显示TextView。

这是MainActivity.java：

public class MainActivity extends ActionBarActivity {
TextView txt_temp;
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    XMLreference();
    StringRequest();
}

public void XMLreference(){
    txt_temp = (TextView) findViewById(R.id.txt_temp);
}

public void StringRequest(){
    WebServiceRequest webService = new WebServiceRequest();
    String request = webService.main();

    txt_temp.setText(request);
}

}

这是WebServiceRequest.java：

public class WebServiceRequest {
    private static final String username = "*********"; // put your Device Cloud username here
    private static final String password = "*********"; // put your Device Cloud password here
    public String responseContent;

    /**
     * Run the web service request
     */
    public String main() {
        HttpsURLConnection conn = null;

        try {
            // Create url to the Device Cloud server for a given web service request
            URL url = new URL("https://devicecloud.digi.com/ws/DataStream/00000000-00000000-********-********/xbee.serialIn");
            conn = (HttpsURLConnection) url.openConnection();
            conn.setDoOutput(true);
            conn.setDoInput(true);
            conn.setRequestMethod("GET");

            // Build authentication string
            String userpassword = username + ":" + password;

            // can change this to use a different base64 encoder
            String encodedAuthorization = Base64.encodeToString(userpassword.getBytes(), Base64.DEFAULT).trim();

            // set request headers
            conn.setRequestProperty("Authorization", "Basic "
                    + encodedAuthorization);

            InputStream is = conn.getInputStream();

            Scanner isScanner = new Scanner(is);
            StringBuffer buf = new StringBuffer();
            while (isScanner.hasNextLine()) {
                buf.append(isScanner.nextLine() + "\n");
            }
            responseContent = buf.toString();

            // add line returns between tags to make it a bit more readable
            responseContent = responseContent.replaceAll("><", ">\n<");

            // Output response to standard out
            // System.out.println(responseContent);

        } catch (Exception e) {
            // Print any exceptions that occur
            e.printStackTrace();
        } finally {
            if (conn != null)
                conn.disconnect();
        }
        return responseContent;
    }
}

我已经尝试创建

public void getResponseContent() {
    return responseContent;
}

并通过

获取

WebServiceRequest webService = new WebServiceRequest();
String request = webService.getResponsteContent();

但它也没有用。我不确定，但是我犯了一个愚蠢的错误，经过几个小时的编程后，我找不到解决它的线索。

Answer 1

好吧，我不确定你是否想要一个构造函数，但main()永远不会被调用，所以responseContent总是空的。

匹配您的代码剪辑

WebServiceRequest webService = new WebServiceRequest();
String request = webService.getResponsteContent();

将main中的方法WebServiceRequest更改为像这样的构造函数

public WebServiceRequest() { .. } // it was String main() before

一旦实例化WebServiceRequest

，它就会被调用

Answer 2

Web请求必须从第二个线程完成。

Runnable run = new Runnable() {
        @Override
        public void run() {
            try {
                //your main method here
                handler.post(new Runnable() {
                    @Override
                    public void run() {
                            //update your textView here
                    }

                });

            } catch (Exception e) {

            }
        }
    };
    Thread thread = new Thread(run);
    thread.start();

此外，您需要移动响应代码并更新到Web请求方法中的textView。必须将对用户界面的更改发布到处理程序。

另一个选择是让你的类扩展IntentService。如果你想检查一下，请看这里。

https://developer.android.com/training/run-background-service/create-service.html

Android Studio：无法从另一个类获取返回字符串

2 个答案: