jvk jvk - 4 months ago 9
JSON Question

how to call name onkey up using ajax?

i am new to ajax. I want to get name from the database according to onkey up but i am getting this error in console "Invalid left-hand side in assignment". and when i type its showing "showname function is not defined".?
This is my code.

Search:




This is myscript
function showname(d) {
var xhttp;
if (window.XMLHttpRequest) {
xhttp= new XMLHttpRequest();
}else{
xhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xhttp.onreadystatechange = function() {
if (xhttp.readyState == 4 && xhttp.status == 200) {
document.getElementById("demo").innerHTML = xhttp.responseText;
}
};
xhttp.open("POST", "ajax2.php?name"=+d, true);
xhttp.send();
}
this ajax2.php code

$name = $_GET["name"];
$db = new mysqli("localhost", "root", "", "customername");
$conn = $db->query("select customerName from customers like '".$name."%'");
$fetch = $conn->fetch_array();

?>
<?php foreach ($fetch as $key => $value) :?>
<p><?php echo $value ?></p>
<?php endforeach; ?>


Thanks you in advance.

Answer

xhttp.open("POST", "ajax2.php?name"=+d, true);

I think you got a typo there, put the = into the quote marks.

xhttp.open("POST", "ajax2.php?name=" + d, true);