使用xmlhttp.open获取2个不同的URL [英] get 2 different url using xmlhttp.open
本文介绍了使用xmlhttp.open获取2个不同的URL的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
需要以下代码的帮助.问题是它只会触发最后一个xmlhttp.open而不是两者.代码如下.
Need help on this following code. the problem is it only trigger the last xmlhttp.open not both. code as below.
<script type="text/javascript">
function loadXMLDoc(obj){
var xmlhttp;
var flag;
var mycode;
mycode = obj;
if (window.XMLHttpRequest){// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}else{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function(){
if (xmlhttp.readyState==4 && xmlhttp.status==200){
}
}
flag = document.getElementById(mycode);
if (flag.value == "Uncall") {
flag.value = "Call Patient";
document.getElementById('myDiv'+mycode).innerHtml="<b><font color='#00FF00'>Called</font></b>";
}else{
flag.value = "Uncall";
document.getElementById('myDiv'+mycode).innerHtml="<b><font color='#00FF00'>Calling..</font></b>";
**xmlhttp.open("GET","http://1.1.1.1/callnumber.cgi?"+mycode,true);
xmlhttp.send();
xmlhttp.open("GET","http://2.2.2.2/displaynumber.php?"+mycode,true);
xmlhttp.send();**
}
}
</script>
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推荐答案
<script type="text/javascript">
function loadXMLDoc(obj){
var xmlhttp;
var flag;
var mycode;
mycode = obj;
flag = document.getElementById(mycode);
if (flag.value != "Uncall"){
loadXMLDocCall(obj,"http://1.1.1.1/callnumber.cgi?");
loadXMLDocCall(obj,"http://2.2.2.2/displaynumber.php?");
flag.value = "Uncall";
document.getElementById('myDiv'+mycode).innerHtml="<b><font color='#00FF00'>Calling..</font></b>";
} else {
flag.value = "Call Patient";
document.getElementById('myDiv'+mycode).innerHtml="<b><font color='#00FF00'>Called</font></b>";
}
}
function loadXMLDocCall(obj,url){
var xmlhttp;
var mycode;
mycode = obj;
if (window.XMLHttpRequest){// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}else{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function(){
if (xmlhttp.readyState==4 && xmlhttp.status==200){
}
}
xmlhttp.open("GET",url+mycode,true);
xmlhttp.send();
}
</script>
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