我想使用下面的脚本将值从AJAX文件传递给PHP,但它失败了。这样做的正确方法是什么?感谢
示例代码如下:
function createNewWindow()
{
var newWindowModel = new DHTMLSuite.windowModel({windowsTheme:true,id:'newWindow1',title:'Response Time to Invitation',xPos:130,yPos:400,minWidth:100,minHeight:100 } );
newWindowModel.addTab({ id:'myTab1',htmlElementId:'myTab1',tabTitle:'TAB',textContent:'Send data', contentUrl:'load.php?loadNo:loadNo' } );
var newWindowWidget = new DHTMLSuite.windowWidget(newWindowModel);
newWindowWidget.init();
}
答案 0 :(得分:0)
传递价值?你的意思是参数?如果是的话:
var http = new XMLHttpRequest();
GET方法:
var url = "load.php";
var params = "loadNo=loadNo¶m=value";
http.open("GET", url+"?"+params, true);
http.onreadystatechange = function() {//Call a function when the state changes.
if(http.readyState == 4 && http.status == 200) {
alert(http.responseText);
}
}
http.send(null);
POST方法:
var url = "laod.php";
var params = "loadNo=loadNo¶m=value";
http.open("POST", url, true);
//Send the proper header information along with the request
http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
http.setRequestHeader("Content-length", params.length);
http.setRequestHeader("Connection", "close");
http.onreadystatechange = function() {//Call a function when the state changes.
if(http.readyState == 4 && http.status == 200) {
alert(http.responseText);
}
}
http.send(params);
答案 1 :(得分:0)
if (window.XMLHttpRequest)
{
xmlhttp=new XMLHttpRequest();
}
else
{
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
var x=xmlhttp.responseText;
alert(x);
}
}
}
xmlhttp.open("GET","load.php?loadNo="+loadNo+"¶m="+value,true);
xmlhttp.send();
答案 2 :(得分:0)
使用Jquery
执行AJAX请求的简单方法var request = $.ajax({
url: "script.php", // script path goes here
type: "GET",
data: {id : param}, // Parameters go here
dataType: "html"
});
request.done(function(msg) {
$("#log").html( msg ); // On success
});
request.fail(function(jqXHR, textStatus) {
alert( "Request failed: " + textStatus ); // On failure
});