C#后台写脚本代码
string s = @" <script type='text/javascript'>
function doAJAX() {
http_request = null;
if (window.XMLHttpRequest) {
http_request = new XMLHttpRequest();
}
else (window.ActiveXObject)
{
http_request = new ActiveXObject('Microsoft.XMLHTTP');
}
var S = document.getElementById('txtMovName').value;
var url = 'AJAX.ashx?keyword=' + escape(S);
http_request.open('get', url, true);
http_request.onreadystatechange = function () {
if (http_request.readyState == 4) {
if (http_request.status == 200) {
var a = http_request.responseText.split(',');
if (a != "") {
document.getElementById('sl11').style.display = 'block';
//清空原来的OPTIONS
var nL = document.getElementById('sl11').options.length;
while (nL > 0) {
document.getElementById('sl11').remove(document.getElementById('sl11').options.length - 1);
nL = document.getElementById('sl11').options.length;
}
for (i = 0; i < a.length; i++) {
var opt = new Option();
opt.value = a[i];
opt.text = a[i];
var sel1Object = document.getElementById('sl11');
sel1Object.add(opt);
}
if (a.length > 15) {
document.getElementById('sl11').size = 15;
}
}
else {
document.getElementById('sl11').style.display = 'none';
}
}
}
}
http_request.send(null);
}
</script> ";
string conSignStr =<input type=\"text\" id=\"__conTxt\" onkeyup=\"'"+s+"'\" value=\"\" /><br/>
为什么运行的时候,把javascript代码显示出来了,我要的是执行onkeyup事件啊