function ajax() {
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("demo").innerHTML = xhttp.responseText;
alert(this.responseText);
}
};
}
xhttp.open("POST", "abcdef.xyz/abc/logincheck.php?mail=abc@gmail.com&password=abc", true);
xhttp.send();
<p id="demo">
The content of the body element is displayed in your browser.
</p>
<button onclick="ajax()">
Click on me!
</button>
I have a code as seen above and the ajax operations do result in failure. What is wrong with my code? The text in p never changes.The php file starts like this:
<?php
$mail = $_POST["mail"];
$password = $_POST["password"];
</div>
xhttp.open()
and xhttp.send()
need to be inside the ajax()
function.
function ajax() {
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("demo").innerHTML = xhttp.responseText;
alert(this.responseText);
}
};
xhttp.open("POST", "abcdef.xyz/abc/logincheck.php?mail=abc@gmail.com&password=abc", true);
xhttp.send();
}
due to xhttp.open and send are define out side of function block.
xhttp.open("POST", "abcdef.xyz/abc/logincheck.php?mail=abc@gmail.com&password=abc", true);
xhttp.send();