Jsp 如何在浏览器上而不是在java控制台上显示Url链接
我有一个问题，那就是如何在浏览器上显示url的链接 下面是我的servlet文件和jsp文件Jsp 如何在浏览器上而不是在java控制台上显示Url链接,jsp,servlets,Jsp,Servlets,我有一个问题，那就是如何在浏览器上显示url的链接 下面是我的servlet文件和jsp文件 Search.java： import java.io.IOException; import javax.servlet.*; import javax.servlet.http.*; import java.sql.PreparedStatement; import java.sql.ResultSet; import java.sql.SQLException;
Search.java import java.io.IOException;
import javax.servlet.*;
import javax.servlet.http.*;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Statement;
import org.jsoup.Jsoup;
import org.jsoup.nodes.Document;
import org.jsoup.nodes.Element;
import org.jsoup.select.Elements;
import java.io.PrintWriter;
public class Search extends HttpServlet
{
private static final long serialVersionUID = 1L;
public static DB db = new DB();
protected void doGet(HttpServletRequest request,HttpServletResponse
response) throws ServletException, IOException
{
PrintWriter out=response.getWriter();
try {
db.runSql2("TRUNCATE Record;");
}
catch (SQLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
try {
processPage("http://www.yahoo.com");
}
catch (SQLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
public static void processPage(String URL) throws SQLException,
IOException
{
//check if the given URL is already in database
String sql = "select * from Record where URL = '"+URL+"'";
ResultSet rs = db.runSql(sql);
if(rs.next())
{
}
else
{
//store the URL to database to avoid parsing again
sql = "INSERT INTO `tenderMysql`.`Record` " + "(`URL`)
VALUES " + "(?);";
PreparedStatement stmt = db.conn.prepareStatement(sql,
Statement.RETURN_GENERATED_KEYS);
stmt.setString(1, URL);
stmt.execute();
//get useful information
Document doc = Jsoup.connect("http://www.yahoo.com").get();
if(doc.text().contains("crash")){
System.out.println(URL);
}
//get all links and recursively call the processPage method
Elements questions = doc.select("a[href]");
for(Element link: questions)
{
if(link.attr("href").contains("www"))
processPage(link.attr("abs:href"));
}
}
}
}
下面给出的是我的jsp文件，我必须从中显示浏览器上的链接
Search.jsp <%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>
Select website name from DropdownList
</title>
</head>
<body bgcolor="8B4513">
<form action="HelloServlet" >
<center>
<h1> Optimzed Search</h1>
<br>
</center>
</form>
</body>
</html>
从下拉列表中选择网站名称
优化搜索
protected void doGet(...)
{
PrintWriter out=response.getWriter();
processPage("http://www.yahoo.com", out);
}
public static void processPage(String URL, PrintWriter out)
{
....
if(doc.text().contains("crash")){
out.println("<p>" + URL+ "</p>");
}
}
要在JSP中打印哪个url？如果（doc.text（）.包含（“崩溃”）{System.out.println（URL）}这个
。。。它应该显示与浏览器崩溃相关的链接..当我尝试运行.jsp文件时，它只是显示一个内容，即优化搜索。。。nd当我尝试运行servlet文件时，它显示了一个错误..javax.servlet.ServletException:servlet执行引发了一个异常org.apache.tomcat.websocket.server.WsFilter.doFilter（WsFilter.java:52）根因java.lang.ClassNotFoundException:org.jsoup.jsoup org.apache.catalina.loader.WebappClassLoader.loader（WebappClassLoader.java:1720）org.apache.catalina.loader.WebappClassLoader.loadClass（WebappClassLoader.java:1571）Search.processPage（Search.java:61）Search.doGet（Search.java:34）javax.servlet.http.HttpServlet.service（HttpServlet.java:620）javax.servlet.http.HttpServlet.service（HttpServlet.java:727）org.apache.tomcat.websocket.server.WsFilter.doFilter（WsFilter.java:52）根本原因java.lang.NoClassDefFoundError:org/jsoup/jsoup Search.processPage（Search.java:61）Search.doGet（Search.java:34）javax.servlet.http.HttpServlet.service（HttpServlet.java:620）javax.servlet.http.HttpServlet.service（HttpServlet.java:727）org.apache.tomcat.websocket.server.WsFilter.doFilter（WsFilter.java:52）缺少Jsoup jar文件。您需要将其添加到部署资源中。我已经更新了答案。
project -- > properties --> Deployment Assembly -- > Add -
-> Archive from File system ---> Add --> Select your jar (in this case soup)