用form.action进行页面跳转问题
跳转不起作用,点击编辑字段按钮,还是提交到本页面。请帮忙看代码如下:
<%@ page contentType= "text/html; charset=utf-8 " language= "java "%>
<%@ page import= "java.sql.* "%>
<html>
<head>
<script type= "text/javascript " src= "/html/ytx/standard.js " > </script>
<link href= "/html/ytx/wstyle.css " rel= "stylesheet " type= "text/css ">
<script>
function columnclick(form){
form.aciton = "zYTXChartColumn.jsp ";
form.submit();
}
</script>
</head>
<body>
<br>
<form name= "form1 " method= "post " action= " ">
<table align= "center " cellspacing= "1 " width= "100% " class= "normaltable " border= "1 ">
<%
out.println( " <td class= 'ytxtd ' width= '8% ' align=\ "center\ " > <input type= 'button ' name= 'button ' value= '编辑字段 ' onclick=\ "columnclick(form1)\ "> </td> ");
%>
</table>
</form>
</body>
</html>
------解决方案--------------------form.aciton = "zYTXChartColumn.jsp ";
这一句出错了,是action而不是aciton......
------解决方案--------------------hehe. need more careful
------解决方案--------------------呵呵 仔细点 我刚学那会也老出这种问题