Question

javascript 跳转 action

<script language="JavaScript">
function del(argnews_id){
document.dAOForm.news_id.value=argnews_id;
document.dAOForm.action="delete.do";
document.dAOForm.submit();

}
function updata(argnews_title,argnews_editor,argnews_content){
alert(argnews_title+argnews_editor+argnews_content);
//document.dAOForm.elements["news_title"].value=argnews_title;
// document.dAOForm.elements["news_editor"].value=argnews_editor;
// document.dAOForm.elements["news_content"].value=argenews_content;

document.dAOForm.action="updataNews.do?title=argnews_title&editor=argnews_editor&content=argnews_content";
document.dAOForm.submit();
}
<button onclick="updata('<bean:write property='news_title' name='DAO'/>','<bean:write property='news_editor' name='DAO'/>','<bean:write property='news_content' name='DAO'/>');" >修改</button>
这是页面的的一个按钮！调用updata时他弹出你传递给的参数！但跳转到action里用request.getParameter("title");拿出的值为什么变了空的？

Answer 1

document.dAOForm.action="updataNews.do?title=argnews_title&editor=argnews_editor&content=argnews_content";

修改如下：

document.dAOForm.action="updataNews.do?title=“+argnews_title+”&editor=“+argnews_editor+"&content="+argnews_content;

但是原来那样写，request.getParameter("title")也不可能是空啊 ，应该得到字符串“argnews_title” 检查页面和代码是不是有错哦

Answer 2

页面表单中有没有title这项啊，提交的时候你提交的是表单的内容，设个隐藏域也行啊