Java URL Read getUrlContent(String url)

Description

If the response code is 200, returns the content at the URL.

License

Apache License

Declaration

public static String getUrlContent(String url) 

Method Source Code

//package com.java2s;
/*//from   w  w  w.j  ava2 s.co  m
 * Copyright (C) 2016 Google Inc.
 *
 * Licensed under the Apache License, Version 2.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 *     http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */

import com.google.common.io.CharStreams;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URL;

public class Main {
    /**
     * If the response code is 200, returns the content at the URL. Otherwise, returns null.
     */
    public static String getUrlContent(String url) {
        try {
            HttpURLConnection urlConnection = (HttpURLConnection) new URL(url).openConnection();
            int responseCode = urlConnection.getResponseCode();
            if (responseCode == 200) {
                return CharStreams.toString(new InputStreamReader(urlConnection.getInputStream()));
            }
        } catch (IOException e) {
        }
        return null;
    }
}

Related

1. getTextFromURL(URL url)
2. getTextFromURL(URL url)
3. getURL(final URL url)
4. getURL(String url)
5. getURL(URL url, String params)
6. getURLContent_old(final String uri, final StringBuffer content)
7. getUrlContentWithRetries(String url, long timeoutMs, long retryDelayMs)
8. getUrlFollowingRedirects(String possibleRedirectionUrl)
9. getUrlInfos(String urlAsString, int timeout)