action里面这样写:
[align=left] List msgList = new LinkedList();
msgList.add("msg1");
msgList.add("msg2");
msgList.add("msg3");
result_map.put("messages", msgList);
return new ModelAndView(this.getSuccess_view(), result_map);[/align]jsp里面:
<p>Your current messages:</p>
<c:forEach items="${messages}" var="item" begin="0" end="9" step="1"
varStatus="var">
<c:if test="${var.index % 2 == 0}">
*
</c:if>
${item}<br>
</c:forEach>输出结果是:
Your current messages:
${item}不明白为什么没有结果输出.
ModelAndView(this.getSuccess_view(), result_map) ,这里把result_map设置到哪里去了?request,session?
[align=left] List msgList = new LinkedList();
msgList.add("msg1");
msgList.add("msg2");
msgList.add("msg3");
result_map.put("messages", msgList);
return new ModelAndView(this.getSuccess_view(), result_map);[/align]jsp里面:
<p>Your current messages:</p>
<c:forEach items="${messages}" var="item" begin="0" end="9" step="1"
varStatus="var">
<c:if test="${var.index % 2 == 0}">
*
</c:if>
${item}<br>
</c:forEach>输出结果是:
Your current messages:
${item}不明白为什么没有结果输出.
ModelAndView(this.getSuccess_view(), result_map) ,这里把result_map设置到哪里去了?request,session?
result_map这个是HashMap吗?
return new ModelAndView(this.getSuccess_view(), result_map);
你的这个类ModelAndView实现的是什么?
Action返回的不都是ActionForward对象吗,你们是不是经过了封装的?
onSubmit确实是返回ModelAndView,ModelAndView是标准的spring的类,不需要自己实现什么
<p>Your current messages: </p>
<c:forEach items="${messages}" var="item" begin="0" end="9" step="1"
varStatus="var">
<c:if test="${var.index % 2 == 0}">
*
</c:if>
${item} <br>
</c:forEach> 现在搞定了,开始问题出在jsp没有支持EL,修改web.xml后就可以了.
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
version="2.4" xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
....