当我访问“http://localhost:8080/project/Song”时,不会出现任何错误,它会在页面中列出一些歌曲。当我点击一首歌时,我会转到一个网址“http://localhost:8080/SongPage?name=A+Tribe&id=201”,但此处显示404错误,可能是因为该网址有参数,访问此网址时出现错误:
HTTP状态404 - 未找到
描述源服务器没有找到目标资源的当前表示,或者不愿意透露该目标资源是否存在。
你知道为什么吗?也许是因为下面这个urlPattern:
@WebServlet(name =“SongPage”,urlPatterns = {“/ SongPage”})
我在下面有这个servlet:
@WebServlet(name ="Song", urlPatterns = {"/Song"})
public class Song extends HttpServlet {
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
QueryManager qm = new QueryManager();
ArrayList<ArrayList<String>> result = qm.Songs();
qm.closeConnections();
request.setAttribute("result", result.get(0));
request.setAttribute("id", result.get(1));
RequestDispatcher view=request.getRequestDispatcher("songList.jsp");
view.forward(request,response);
}
}
在songList.jsp中我有这个:
<c:forEach items="${result}" var="item" varStatus="status">
<a href="/SongPage?name=${result[status.index].replace(" ","+")}&id=${id[status.index]}"> ${result[status.index]} </a> <br />
</c:forEach>
SongPage.java:
@WebServlet(name ="SongPage", urlPatterns = {"/SongPage"})
public class SongPage extends HttpServlet {
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String name = request.getParameter("name").replace("+"," ");
...
}
答案 0 :(得分:1)
您原来的网址&#34; http://localhost:8080/project/Song&#34;有&#34;项目&#34;作为基本路径,但另一个网址&#34; http://localhost:8080/SongPage?name=A+Tribe&id=201&#34;没有那条路。您很可能需要更新JSP以将其包含在&#34; href&#34;标记为快速修复:
<a href="project/SongPage?name=${result[status.index].replace(" ","+")}&id=${id[status.index]}"> ${result[status.index]} </a> <br />
或者很可能,如果它实际上是一个上下文路径:
<a href="${pageContext.request.contextPath}/SongPage?name=${result[status.index].replace(" ","+")}&id=${id[status.index]}"> ${result[status.index]} </a> <br />