日期:2014-05-18  浏览次数:20466 次

[ajax小问题]关于xmlHttp.open("POST", url, true),详细如下
var   title=vbtrim(document.all.title.value);
var   gpdm=escape(vbtrim(document.all.gpdm.value));
var   url   =   "posttopic1.aspx?title= "+escape(title)+ "&message= "+escape(message)+ "&gpdm= "+gpdm;
xmlHttp.open( "POST ",   url,   true);
xmlHttp.onreadystatechange   =   updatePage;
xmlHttp.send(null);    

如果   message   的内容很少,可以提交

如果message是一篇文章,则报错

*message是input的ID

你们说怎么办

------解决方案--------------------
1。
var url = "posttopic1.aspx?title= "+escape(title)+ "&message= "+escape(message)+ "&gpdm= "+gpdm;
----------
这种是 GET方式,不式POST方式

2。

var title=vbtrim(document.all.title.value);
var gpdm=escape(vbtrim(document.all.gpdm.value));
var url = "posttopic1.aspx?title= "+escape(title)+ "&message= "+escape(message)+ "&gpdm= "+gpdm;
xmlHttp.open( "POST ", url, true);
xmlHttp.onreadystatechange = updatePage;
xmlHttp.send(null);

》》》

var title=vbtrim(document.all.title.value);
var gpdm=escape(vbtrim(document.all.gpdm.value));
var url = "posttopic1.aspx ";
xmlHttp.open( "POST ", url, true);
xmlHttp.onreadystatechange = updatePage;
var args = "title= "+escape(title)+ "&message= "+escape(message)+ "&gpdm= "+gpdm;
xmlHttp.send(args);