ajax data post call in javascript
$.ajax({
url: 'ajaxfile.php',
type: 'post',
data: {name:'yogesh',salary: 35000,email: '[email protected]'},
success: function(response){
}
});ajax data post call in javascript
$.ajax({
url: 'ajaxfile.php',
type: 'post',
data: {name:'yogesh',salary: 35000,email: '[email protected]'},
success: function(response){
}
});javascript send post data with ajax
function makeRequest (method, url, data) {
return new Promise(function (resolve, reject) {
var xhr = new XMLHttpRequest();
xhr.open(method, url);
xhr.onload = function () {
if (this.status >= 200 && this.status < 300) {
resolve(xhr.response);
} else {
reject({
status: this.status,
statusText: xhr.statusText
});
}
};
xhr.onerror = function () {
reject({
status: this.status,
statusText: xhr.statusText
});
};
if(method=="POST" && data){
xhr.send(data);
}else{
xhr.send();
}
});
}
//POST example
var data={"person":"john","balance":1.23};
makeRequest('POST', "https://www.codegrepper.com/endpoint.php?param1=yoyoma",data).then(function(data){
var results=JSON.parse(data);
});ajax open a request
<script>
function loadDoc() {
var xhttp = new XMLHttpRequest();
//looking for a change or state , like a request or get.
xhttp.onreadystatechange = function() {
//if equal to 4 means that its ready.
// if equal to 200 indicates that the request has succeeded.
if (this.readyState == 4 && this.status == 200) {
document.getElementById("demo").innerHTML = this.responseText;
}
};
//GET method for gettin the data // the file you are requesting
xhttp.open("GET", "TheFileYouWant.html", true);
//sending the request
xhttp.send();
}Copyright © 2021 Codeinu
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