用javascript+ajax做的一个仿digg程序,多个form,同一个提交动作,苦于无法用javascript确定提交form的id,所以只能同一个动作复制了N遍,换了换动作的ID,很啰嗦。请指导一下,谢谢。动作:function OnCheckAvailability()
{
if(window.XMLHttpRequest)
{
oRequest = new XMLHttpRequest();
}
else if(window.ActiveXObject)
{
oRequest = new ActiveXObject("Microsoft.XMLHTTP");
}oRequest.open("POST", "/action/digg.asp", true);
oRequest.onreadystatechange = UpdateCheckAvailability;oRequest.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
oRequest.send("strUsername=" + document.form1.keycode.value);
}function UpdateCheckAvailability()
{
if(oRequest.readyState == 4)
{
if(oRequest.status == 200)
{
document.getElementById("Available1").innerHTML = oRequest.responseText;
document.getElementById("Upbut1").innerHTML = ("已完成");
}
else
{
document.getElementById("Available1").innerHTML = "A-Error";
document.getElementById("Upbut1").innerHTML = ("有错误");
}
}
}在页面里有11个digg项目,所以这个动作也复制了11遍... 我只有10分了,希望别嫌少啊...
{
if(window.XMLHttpRequest)
{
oRequest = new XMLHttpRequest();
}
else if(window.ActiveXObject)
{
oRequest = new ActiveXObject("Microsoft.XMLHTTP");
}oRequest.open("POST", "/action/digg.asp", true);
oRequest.onreadystatechange = UpdateCheckAvailability;oRequest.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
oRequest.send("strUsername=" + document.form1.keycode.value);
}function UpdateCheckAvailability()
{
if(oRequest.readyState == 4)
{
if(oRequest.status == 200)
{
document.getElementById("Available1").innerHTML = oRequest.responseText;
document.getElementById("Upbut1").innerHTML = ("已完成");
}
else
{
document.getElementById("Available1").innerHTML = "A-Error";
document.getElementById("Upbut1").innerHTML = ("有错误");
}
}
}在页面里有11个digg项目,所以这个动作也复制了11遍... 我只有10分了,希望别嫌少啊...
function XXXX(objForm)
{
代码块;
}调用:
<form>
XXXX(this.form)
</form>或者:
<form id="ABC"></form>
XXXX(ABC) 这里ABC不用双引号扩起来,直接传DOM对象的ID,JavaScript直接操作对象。
<div id="l-con-left"><div id="digg-it"><span id="Available<%=i%>"><%=rs("ups")%></span></div>
<div id="l-con-bot"><span id="Upbut<%=i%>"><a onClick="OnCheckAvailability<%=i%>();" style="CURSOR:hand">投一票</a></span></form>js中的动作:其中的【i】与上面的1到10个数字对应,以此类推:十个js相应的接收十个form发来的信息,然后更新function OnCheckAvailability【i】()
{
if(window.XMLHttpRequest)
{
oRequest = new XMLHttpRequest();
}
else if(window.ActiveXObject)
{
oRequest = new ActiveXObject("Microsoft.XMLHTTP");
}oRequest.open("POST", "up-it.asp", true);
oRequest.onreadystatechange = UpdateCheckAvailability【i】;oRequest.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
oRequest.send("strUsername=" + document.form【i】.keycode.value);
}function UpdateCheckAvailability0()
{
if(oRequest.readyState == 4)
{
if(oRequest.status == 200)
{
document.getElementById("Available【i】").innerHTML = oRequest.responseText;
document.getElementById("Upbut【i】").innerHTML = ("已完成");
}
else
{
document.getElementById("Available【i】").innerHTML = "A-Error";
document.getElementById("Upbut【i】").innerHTML = ("有错误");
}
}
}我对js不熟悉,不知道更为科学简洁的方式如何写。