ajax response returns entire page
Use condition to verify if $_POST empty in you main page or index.php to not send the whole page in response ajax
and verif your AJAX URL
$.ajax({
type:"POST",
url:"http://mywebsite.com/php/send_email.php",
.......
index.php
------------------------
<?php
if(empty($_POST)){ ?>
//
<html>
the header pages
the body
the include pages
the scripts pages
the footer
<?php } ?>ajax get html response
//your jquery code will be like this:
$.ajax({
type: 'POST',
url: $url,
data: new FormData(this),
dataType: 'json',
contentType: false,
processData:false,//this is a must
success: function(response){
$('your_selector').html(response);
}
});
//php code will be like this
echo '<div>some html code</div>';
die();simple return data jquery
$(document).ready(function(){
$.ajax({
url: 'ajaxfile.php',
type: 'get',
dataType: 'JSON',
success: function(response){
var len = response.length;
for(var i=0; i<len; i++){
var id = response[i].id;
var username = response[i].username;
var name = response[i].name;
var email = response[i].email;
var tr_str = "<tr>" +
"<td align='center'>" + (i+1) + "</td>" +
"<td align='center'>" + username + "</td>" +
"<td align='center'>" + name + "</td>" +
"<td align='center'>" + email + "</td>" +
"</tr>";
$("#userTable tbody").append(tr_str);
}
}
});
});simple return data jquery
<?php
include "config.php";
$return_arr = array();
$query = "SELECT * FROM users ORDER BY NAME";
$result = mysqli_query($con,$query);
while($row = mysqli_fetch_array($result)){
$id = $row['id'];
$username = $row['username'];
$name = $row['name'];
$email = $row['email'];
$return_arr[] = array("id" => $id,
"username" => $username,
"name" => $name,
"email" => $email);
}
// Encoding array in JSON format
echo json_encode($return_arr);Copyright © 2021 Codeinu
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