ajax request in javascript
fetch('https://api.covid19api.com/summary')
.then(response => response.json())
.then(data => console.log(data))
.catch(err => {
console.log(err)
});ajax request in javascript
fetch('https://api.covid19api.com/summary')
.then(response => response.json())
.then(data => console.log(data))
.catch(err => {
console.log(err)
});ajax syntax in javascript
var id = empid;
$.ajax({
type: "POST",
url: "../Webservices/EmployeeService.asmx/GetEmployeeOrders",
data: "{empid: " + empid + "}",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function(result){
alert(result.d);
console.log(result);
}
});js ajax
// For a plain JS solution, use these functions:
function _GET_REQUEST(url, response) {
var xhttp;
if (window.XMLHttpRequest) {
xhttp = new XMLHttpRequest();
} else {
xhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
response(this.responseText);
}
};
xhttp.open("GET", url, true);
xhttp.send();
}
function _POST_REQUEST(url, params, response) {
var xhttp;
if (window.XMLHttpRequest) {
xhttp = new XMLHttpRequest();
} else {
xhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
response(this.responseText);
}
};
xhttp.open("POST", url, true);
xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhttp.send(params);
}
// and apply like this:
_GET_REQUEST('http://someurl', (response) => {
// do something with variable response
});
_POST_REQUEST('http://someurl', 'paramx=y', (response) => {
// do something with variable response
});Copyright © 2021 Codeinu
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