当我没有将 servlet 附加到 servlet 映射中的 / 时
<servlet-mapping>
<servlet-name>ControllerServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
我的所有 css 和 javascript 框架都正在加载。但是，当我在 web.xml 中添加这部分代码时，没有加载任何 css 和 javascript。如何确保 servlet 知道它是 css/js 文件而不是方法？
web.xml
<context-param>
<param-name>jdbcURL</param-name>
<param-value>jdbc:mysql://localhost:3306/bookstore</param-value>
</context-param>
<context-param>
<param-name>jdbcUsername</param-name>
<param-value>root</param-value>
</context-param>
<context-param>
<param-name>jdbcPassword</param-name>
<param-value>12345</param-value>
</context-param>
<servlet>
<servlet-name>ControllerServlet</servlet-name>
<servlet-class>net.codejava.javaee.bookstore.ControllerServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>ControllerServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<error-page>
<exception-type>java.lang.Exception</exception-type>
<location>/Error.jsp</location>
</error-page>
</web-app>
Controller Servlet
public class ControllerServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
private BookDAO bookDAO;
public void init() {
String jdbcURL = getServletContext().getInitParameter("jdbcURL");
String jdbcUsername = getServletContext().getInitParameter("jdbcUsername");
String jdbcPassword = getServletContext().getInitParameter("jdbcPassword");
bookDAO = new BookDAO(jdbcURL, jdbcUsername, jdbcPassword);
}
protected void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
doGet(request, response);
}
protected void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
String action = request.getServletPath();
try {
switch (action) {
case "/new":
//showNewForm(request, response);
break;
case "/insert":
//insertBook(request, response);
break;
case "/delete":
//deleteBook(request, response);
break;
case "/edit":
//showEditForm(request, response);
break;
case "/update":
//updateBook(request, response);
break;
default:
listBook(request, response);
break;
}
} catch (SQLException ex) {
throw new ServletException(ex);
}
}
private void listBook(HttpServletRequest request, HttpServletResponse response)
throws SQLException, IOException, ServletException {
List<Book> listBook = bookDAO.listAllBooks();
request.setAttribute("listBook", listBook);
RequestDispatcher dispatcher = request.getRequestDispatcher("BookList.jsp");
dispatcher.forward(request, response);
}
}
jsp页面
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<!-- JS -->
<script src='static/js/bootstrap.min.js'></script>
<script src='static/js/vue.min.js'></script>
<!-- CSS -->
<link rel="stylesheet" type="text/css" href="static/css/bootstrap.min.css">
<link rel="stylesheet" type="text/css" href="static/css/common.css">
</head>
<body>
<center>
<h1>Books Management</h1>
<h2>
<a href="new">Add New Book</a>
<a href="list">List All Books</a>
</h2>
</center>
<div align="center">
<table border="1" cellpadding="5">
<caption><h2>List of Books</h2></caption>
<tr>
<th>ID</th>
<th>Title</th>
<th>Author</th>
<th>Price</th>
<th>Actions</th>
</tr>
<c:forEach var="book" items="${listBook}">
<tr>
<td><c:out value="${book.id}" /></td>
<td><c:out value="${book.title}" /></td>
<td><c:out value="${book.author}" /></td>
<td><c:out value="${book.price}" /></td>
<td>
<a href="edit?id=<c:out value='${book.id}' />">Edit</a>
<a href="delete?id=<c:out value='${book.id}' />">Delete</a>
</td>
</tr>
</c:forEach>
</table>
</div>
</body>
</html>
最佳答案
您可以像下面一样添加 CSS 文件
<LINK REL="StyleSheet" HREF="<%=request.getContextPath()%>/CSS/style.css" TYPE="text/css">
所以文件夹结构如下所示
您需要在 WebContent 中添加 css/js 文件
关于javascript - CSS 和 JS 未加载到 JSP servlet 映射上，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/47764341/