Swe*_*er 17 spring jsp spring-mvc
我创建了一个弹簧形式,并希望从中提交并显示另一个jsp页面.当我运行这个项目时,我得到以下异常.任何人都可以帮助我理解为什么我得到这个错误.
org.apache.jasper.JasperException: /WEB-INF/views/home.jsp (line: [25], column: [1]) Unable to find setter method for attribute: [commandName]
org.apache.jasper.compiler.DefaultErrorHandler.jspError(DefaultErrorHandler.java:42)
org.apache.jasper.compiler.ErrorDispatcher.dispatch(ErrorDispatcher.java:292)
org.apache.jasper.compiler.ErrorDispatcher.jspError(ErrorDispatcher.java:115)
org.apache.jasper.compiler.Generator$GenerateVisitor.evaluateAttribute(Generator.java:2998)
org.apache.jasper.compiler.Generator$GenerateVisitor.generateSetters(Generator.java:3218)
org.apache.jasper.compiler.Generator$GenerateVisitor.generateCustomStart(Generator.java:2404)
org.apache.jasper.compiler.Generator$GenerateVisitor.visit(Generator.java:1894)
org.apache.jasper.compiler.Node$CustomTag.accept(Node.java:1544)
org.apache.jasper.compiler.Node$Nodes.visit(Node.java:2389)
org.apache.jasper.compiler.Node$Visitor.visitBody(Node.java:2441)
org.apache.jasper.compiler.Node$Visitor.visit(Node.java:2447)
org.apache.jasper.compiler.Node$Root.accept(Node.java:470)
org.apache.jasper.compiler.Node$Nodes.visit(Node.java:2389)
org.apache.jasper.compiler.Generator.generate(Generator.java:3657)
org.apache.jasper.compiler.Compiler.generateJava(Compiler.java:256)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:384)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:361)
org.apache.jasper.compiler.Compiler.compile(Compiler.java:345)
org.apache.jasper.JspCompilationContext.compile(JspCompilationContext.java:603)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:369)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:385)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:329)
javax.servlet.http.HttpServlet.service(HttpServlet.java:741)
org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:53)
org.springframework.web.servlet.view.InternalResourceView.renderMergedOutputModel(InternalResourceView.java:170)
org.springframework.web.servlet.view.AbstractView.render(AbstractView.java:312)
org.springframework.web.servlet.DispatcherServlet.render(DispatcherServlet.java:1325)
org.springframework.web.servlet.DispatcherServlet.processDispatchResult(DispatcherServlet.java:1069)
org.springframework.web.servlet.DispatcherServlet.doDispatch(DispatcherServlet.java:1008)
org.springframework.web.servlet.DispatcherServlet.doService(DispatcherServlet.java:925)
org.springframework.web.servlet.FrameworkServlet.processRequest(FrameworkServlet.java:978)
org.springframework.web.servlet.FrameworkServlet.doGet(FrameworkServlet.java:870)
javax.servlet.http.HttpServlet.service(HttpServlet.java:634)
org.springframework.web.servlet.FrameworkServlet.service(FrameworkServlet.java:855)
javax.servlet.http.HttpServlet.service(HttpServlet.java:741)
org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:53)
JSP文件:
<%@ taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c" %>
<%@ page session="false" %>
<%@ taglib prefix="form" uri="http://www.springframework.org/tags/form"%>
<html>
<head>
<title>Home</title>
<style>
.error {
color: #ff0000;
}
.errorblock {
color: #000;
background-color: #ffEEEE;
border: 3px solid #ff0000;
padding: 8px;
margin: 16px;
}
</style>
</head>
<body>
<h2>Spring's form textbox example</h2>
<form:form method="POST" action="/customer" commandName="customer">
<form:errors path="*" cssClass="errorblock" element="div" />
<table>
<tr>
<td>Username :</td>
<td><form:input path="userName" />
</td>
<td><form:errors path="userName" cssClass="error" />
</td>
</tr>
<tr>
<td colspan="3"><input type="submit" />
</td>
</tr>
</table>
</form:form>
</html>
当我使用简单的html表单然后它正常工作,但同样的事情,如果我通过弹簧形式,它给出错误.
控制器类:
@Controller
public class HomeController {
@RequestMapping("/")
public String welcomePage()
{
return "home";
}
@RequestMapping(value="/customer", method=RequestMethod.POST)
public ModelAndView submitForm(@RequestParam("userName") String name)
{
ModelAndView mv = new ModelAndView("success");
mv.addObject("userName", name);
return mv;
}
}
Ogu*_*guz 55
您使用的是哪个版本的Spring MVC？我有同样的问题,maven依赖,即
group id:org.springframework artifact id:spring-webmvc version:5.0.2.RELEASE
在版本5之后,删除了commandName,您应该使用modelAttribute.我在这里找到了它,https://jira.spring.io/browse/SPR-16037
我只是用modelAttribute更改了commandName.
<form:form modelAttribute="goal">
小智 7
1...OLD WAY = 命令名
<form:form action="index" commandName="todo">
2...新方法 = 模型属性
<form:form action="index" modelAttribute="todo">
commandName 是旧的方法，在新的应用程序中你应该使用 modelAttribute
commandName=请求范围或会话范围中包含有关此表单的信息的变量的名称，它应该是been。
参考文献：http://forum.spring.io/forum/spring-projects/web/59966-exact-meaning-of-form-form-commandname
Spring注解与 <form:form commandName="xy" 的关系
在使用表单显示（返回）jsp 文件GET的方法中customer，添加模型属性，如下所示：
@RequestMapping(value="/customer", method=RequestMethod.GET)
public ModelAndView showForm(Model model) {
model.addAttribute("customer",new Customer());
}
然后尝试是否正确显示表单。
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