Question

我想从url收到404错误，但我使用的代码总是会产生错误：

public String getRespond(String request) {

    try{
        URL url=new URL (request);
        URLConnection connection=url.openConnection();
        InputStream in=new BufferedInputStream(connection.getInputStream());

        Scanner scanner =new Scanner(in);
        String response=scanner.nextLine();
        scanner.close();
        return response;

    }
    catch (Exception e){
        return null;
    }
}

错误出现在“in = new ...”中，如果出现404错误，我无法知道。这是错误：

02-23 11:42:42.869  20413-20413/com.mappleapps.tm2ibz E/AndroidRuntime﹕ FATAL EXCEPTION: main
Process: com.mappleapps.tm2ibz, PID: 20413
android.os.NetworkOnMainThreadException
        at android.os.StrictMode$AndroidBlockGuardPolicy.onNetwork(StrictMode.java:1156)
        at java.net.InetAddress.lookupHostByName(InetAddress.java:385)
        at java.net.InetAddress.getAllByNameImpl(InetAddress.java:236)
        at java.net.InetAddress.getAllByName(InetAddress.java:214)
        at com.android.okhttp.internal.Dns$1.getAllByName(Dns.java:28)
        at com.android.okhttp.internal.http.RouteSelector.resetNextInetSocketAddress(RouteSelector.java:216)
        at com.android.okhttp.internal.http.RouteSelector.next(RouteSelector.java:122)
        at com.android.okhttp.internal.http.HttpEngine.connect(HttpEngine.java:292)
        at com.android.okhttp.internal.http.HttpEngine.sendSocketRequest(HttpEngine.java:255)
        at com.android.okhttp.internal.http.HttpEngine.sendRequest(HttpEngine.java:206)
        at com.android.okhttp.internal.http.HttpURLConnectionImpl.execute(HttpURLConnectionImpl.java:345)
        at com.android.okhttp.internal.http.HttpURLConnectionImpl.getResponse(HttpURLConnectionImpl.java:296)
        at com.android.okhttp.internal.http.HttpURLConnectionImpl.getInputStream(HttpURLConnectionImpl.java:179)

感谢!!!

Answer 1

在应用程序的主线程上不再允许HTTP请求。

因此，您应该使用并行Thread，或者更简单地使用AsyncTask

class ResponseTask extends AsyncTask<String, Void, String> {

    public static interface ResponseTaskListener {
        void onResponse(String response);
        void onException(Exception e);
    }

    private ResponseTaskListener listener;
    private Exception e;

    public ResponseTask(ResponseTaskListener listener) {
        this.listener = listener;
    }

    protected String doInBackground(String... urls) {
        try {
            String request = urls[0];

            URL url=new URL (request);
            URLConnection connection=url.openConnection();
            InputStream in=new BufferedInputStream(connection.getInputStream());

            Scanner scanner =new Scanner(in);
            String response=scanner.nextLine();
            scanner.close();
            return response;

        } catch (Exception e){
            this.e = e;
            return null;
        }
    }

    // String response = what is returned by doInBackground
    protected void onPostExecute(String response) {
        if (this.listener != null) {
            if (this.e != null) {
                this.listener.onException(this.e);
            } else {
                this.listener.onResponse(response);
            }
        }
    }
}

然后执行此请求，您可以：

new ResponseTask(new ResponseTaskListener() {
    public void onResponse(String response) {
        // Play with the response
    }
    public void onException(Exception e) {
        // Play with the exception
    }
}).execute(urlToRssFeed);

在Android中从网址收到404错误

1 个答案: