Reputation: 641
I am trying to write a function that will call getproduct.php?id=xxx
when clicked. I can get the innerHTML
portion to appear, but how do I also call the php
page that actually does the work?
var id = id;
document.getElementById("digital_download").innerHTML =
"Downloading...Please be patient. The process can take a few minutes.";
url = getproduct.php?id=id;
Upvotes: 7
Views: 70014
Reputation: 1044
Hi You can call the below function to perform this it loads the data from server on success you can create fail function as well
function setValue(Id) {
document.getElementById("digital_download").innerHTML =
"Downloading...Please be patient. The process can take a few minutes.";
var data1 = {
message: Id,
};
$.ajax({
data: data1,
type: 'GET',
url: "http://urltoscript/index.php",
cache: false,
dataType: "json",
crossDomain: true,
success: function(data) {
console.log("Response for cancel is: " + data);
document.getElementById("digital_download").innerHTML = data
}
});
}
Upvotes: 1
Reputation: 333
There are many ways by which you can load a page into a division .
The very method is
var xmlhttp;
if (window.XMLHttpRequest) {
xmlhttp = new XMLHttpRequest();
} else {
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById('digital_download').innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET", 'getproduct.php?id=' + id,true);
xmlhttp.send();
}
this is a typical method with no external reference.
If you go with reference then there are 5 ways to make a ajax call with jQuery
Upvotes: 2
Reputation: 1792
You can use get or post request using query
$.ajax({
type: "POST",
url: url,
data: data,
success: success,
dataType: dataType
});
Upvotes: 0
Reputation: 4348
you can call or load php page inside a div using this line as :-
$("#content_div").load("ajax/page_url.php");
the "ajax/page_url.php" its a relative path of php file.
so here you can replace it with external url as well.
please share you knowledge if i am wrong.
Upvotes: 15
Reputation: 16355
Edit: the original question didn't reference jQuery. Leaving this answer here as others may find it useful.
Here's how you would do this using the XHR object for an ajax request without jQuery or Prototype or other JS library.
var xmlhttp;
if (window.XMLHttpRequest) {
xmlhttp = new XMLHttpRequest();
} else {
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById('digital_download').innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET", 'getproduct.php?id=' + id,true);
xmlhttp.send();
}
Upvotes: 1
Reputation: 4353
You can do it with jQuery for example.
var id = 1;
$('#digital_download').html('Downloading...'); // Show "Downloading..."
// Do an ajax request
$.ajax({
url: "getproduct.php?id="+id
}).done(function(data) { // data what is sent back by the php page
$('#digital_download').html(data); // display data
});
Upvotes: 9