我在这个程序中得到了 403 响应码,但我需要得到 200 才能返回搜索结果,我该怎么办?
String url="http://www.google.com/search?q=";
String charset="UTF-8";
String key="java";
String query = String.format("%s",URLEncoder.encode(key, charset));
URLConnection con = new URL(url+ query).openConnection();
BufferedReader in = new BufferedReader(new InputStreamReader(con.getInputStream()));
String inputLine;
while ((inputLine = in.readLine()) != null)
System.out.println(inputLine);
in.close();
最佳答案
试试JSoup
Document document = Jsoup
.connect("http://www.google.com/search?q=" + query)
.userAgent("Mozilla/5.0 (Windows NT 6.1; WOW64; rv:5.0) Gecko/20100101 Firefox/5.0")
.get();
System.out.println(document.html());
要提取链接,请使用 selector api .
依赖:
<dependency>
<!-- jsoup HTML parser library @ http://jsoup.org/ -->
<groupId>org.jsoup</groupId>
<artifactId>jsoup</artifactId>
<version>1.7.3</version>
</dependency>
关于java - 在java中获取google的搜索结果,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20758824/