Javascript responseText - XMLHttpRequest
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responseText - XMLHttpRequest
提问by dotnetrocks
in my code responseText is not working. It is supposed to display, text entered in the text box +" :Your request has been seen by syam"
在我的代码 responseText 中不起作用。它应该显示，在文本框中输入的文本+“：您的请求已被syam看到”
<html>
<head id="Head1" runat="server">
<title></title>
<script type="text/javascript">
var xmlHttpRequest;
function sSignature(str) {
xmlHttpRequest = new XMLHttpRequest();
xmlHttpRequest.onreadystatechange = function() {
if (xmlHttpRequest.readyState == 4 && xmlHttpRequest.status == 200) {
document.getElementById("target").innerHTML = xmlHttpRequest.responseText;
}
}
xmlHttpRequest.open("GET", "AjaxResponse.aspx?q=" + str, true);
xmlHttpRequest.send();
}
</script>
</head>
<body>
<form id="form1" runat="server">
<div>
enter a string :<input type="text" id="textbox" onkeyup="sSignature(this.value)"/>
<span id="target">text should change here</span>
</div>
</form>
</body>
</html>
In the code-behind page, in page_load()
在代码隐藏页面中，在 page_load()
string sRequest = Request.QueryString["q"];
var sResponse = sRequest+ " :Your request has been seen by syam";
Response.Write(sResponse);
回答by nikc.org
I believe the error is in your onreadystatechangedhandler. It will receive an eventparam, in which the targetproperty points to the XHR-instance.
我相信错误出在您的onreadystatechangedhandler. 它将接收一个event参数，其中target属性指向 XHR 实例。
Try swapping it out with this:
尝试用这个换掉它：
xmlHttpRequest.onreadystatechange = function (event) {
var xhr = event.target;
if (xhr.readyState === 4 && xhr.status === 200) {
document.getElementById("target").innerHTML = xhr.responseText
}
};
回答by run
send your request first
首先发送您的请求
function sSignature(str) {
xmlHttpRequest = new XMLHttpRequest();
xmlHttpRequest.open("GET", "AjaxResponse.jsp?q=" + str, true);
xmlHttpRequest.send();
xmlHttpRequest.onreadystatechange = function() {
if (xmlHttpRequest.readyState == 4 && xmlHttpRequest.status == 200) {
document.getElementById("target").innerHTML = xmlHttpRequest.responseText;
}
}
}
回答by Acn
Mind that your code will not work in Microsoft Internet Explorer.
请注意，您的代码将无法在 Microsoft Internet Explorer 中运行。
Secondly, modify one line of code to make it look better -
其次，修改一行代码，让它看起来更好——
xhr.send()by xhr.send(null);
xhr.send()经过 xhr.send(null);
