This article mainly introduces the method of implementing simple ajax Loading loading function in PHP. It analyzes the principle, operation skills and related precautions of ajax loading in the form of examples. Friends in need can refer to it
The details are as follows :
var xmlHttp; function createXmlHttpReq() { if(window.ActiveXObject) { xmlHttp = new ActiveXObject('Microsoft.XMLHTTP'); } else if(window.XMLHttpRequest) { xmlHttp = new XMLHttpRequest(); } } function funMy(url) { createXmlHttpReq(); try { xmlHttp.onreadystatechange = cb;//一定要在open()前,下边会有说明。在此处犯错了 xmlHttp.open("GET","for.php?id="+url,true); xmlHttp.send(null); } catch(e) { alert("您访问的资源不存在"); } } //回调函数 function cb() { if(xmlHttp.readyState==1) { alert("1-------------->"); //在Google Chrome 浏览器里不显示loading图片,三秒后显示内容,问题已解决,下边有说明 document.getElementById('ajax').innerHTML = "<img src=loading2.gif>"; //document.getElementById('ajax').innerHTML = "Loading......"; } if(xmlHttp.readyState == 4 && xmlHttp.status == 200) { var data = xmlHttp.responseText; document.getElementById('ajax').innerHTML = data; } }
When I was testing, I got stuck on Chrome. Please see the explanation below:
If you write it this way, you will not receive a response of .readyState==1
Because 1 means that .open() has been called and completed
But .open() is called before the .onreadystatechange event, so it should be impossible for you to receive a response of .readyState==1
Therefore, if you want to receive .readyState==1 =>.onreadystatechange, you must Before .open()
So why do you sometimes receive it?
Because you use the same global variable...During continuous operations, it may happen because an xhr request is still waiting for php. To initialize it again
You should first decide on the data processing method onreadystatechange, and then send the data to be processed open()
The above is the entire content of this article, I hope it will be helpful to Everyone’s learning helps.
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