jQuery无法解析JSP EL

小编典典

jQuery无法解析JSP EL

jsp

我的JQuery process.js文件无法解析JSP
EL表达式。我只是想在JSP中显示一个List对象。如果我只是使用该${students}对象，它将正常工作并显示所有学生，但是如果我尝试使用JQuery显示该对象${students}，它将仅在浏览器中显示该字符串。我已禁用脚本web.xml并使用EL来显示数据。

还有其他方法可以解决吗？

浏览器输出：

Hello World from JQuery!

From JSP:  
[Student{id=1, name=Jack}, Student{id=2, name=Jill}]

From JQuery:
${students}

home.jsp

<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE html>
<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>JSP Page</title>
        <script src="http://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
        <script src="js/process.js"></script>
    </head>
    <body>
        <div id="mydiv"></div>
        <h2>From JSP: <p/> ${students}</h2>
        <h3>From JQuery: <div id="studentDiv"></div></h3>
    </body>
</html>

process.js

$(document).ready(function () {
    //static message
    var msg = "Hello World from JQuery!";
    $("#mydiv").text(msg);

    //dynamic message
    var students = '${students}';
    $("#studentDiv").text(students);
});

HomePageController.java

@WebServlet(name = "HomePageController", urlPatterns = {"/homePageController.do"})
public class HomePageController extends BaseController {

    private static final String VIEW_NAME = "/jsp/home.jsp";

    @Override
    public void processRequest(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        List<Student> students = getAllStudents();
        request.setAttribute("students", students);
        request.getServletContext().getRequestDispatcher(VIEW_NAME).forward(request, response);
    }

    private List<Student> getAllStudents() {
        List<Student> students = new ArrayList<>();
        Student student1 = new Student(1, "Jack");
        Student student2 = new Student(2, "Jill");
        students.add(student1);
        students.add(student2);
        System.out.println("students = " + students);
        return students;
    }
}

BaseController.java

public abstract class BaseController extends HttpServlet {

    @Override
    public void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        processRequest(request, response);
    }

    @Override
    public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        processRequest(request, response);
    }

    public abstract void processRequest(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException;

}

web.xml

<?xml version="1.0" encoding="UTF-8"?>

<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
         version="3.1">

    <!-- Disable scripting -->
    <jsp-config>
        <jsp-property-group>
            <url-pattern>*.jsp</url-pattern>
            <scripting-invalid>true</scripting-invalid>
        </jsp-property-group>
    </jsp-config>

    <session-config>
        <session-timeout>30</session-timeout>
    </session-config>

    <welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
    </welcome-file-list>

</web-app>

阅读 381

2020-06-10

共1个答案

小编典典

EL表达式仅由JspServlet（*.jsp）和FacesServlet（*.xhtml）求值。但是，.js默认情况下，文件不是由那些servlet处理的，而是由容器的内置默认servlet处理的。这不是jQuery的错。这只是你的错

一个快速的技巧是向映射添加*.jsURL模式JspServlet。确定容器的内置名称JspServlet（在Tomcat和克隆中）后jsp，将以下条目添加到您的webapp的web.xml：

<servlet-mapping>
    <servlet-name>jsp</servlet-name>
    <url-pattern>*.js</url-pattern>
</servlet-mapping>

替代方法是：

将其重构为webservice调用并使用$.getJSON()。
将其打印为HTML5 data属性并使用$.data()。
在包含所需脚本之前，将其作为JSP中的内联脚本变量打印。

2020-06-10