Answers for "how to call ajax function in php"

0

ajax jquery php

<!doctype html>
<html>
<head>
<script src="https://code.jquery.com/jquery-3.3.1.js" integrity="sha256-2Kok7MbOyxpgUVvAk/HJ2jigOSYS2auK4Pfzbm7uH60=" crossorigin="anonymous"></script>
</head>
<body>
<form id="loginform" method="post">
    <div>
        Username:
        <input type="text" name="username" id="username" />
        Password:
        <input type="password" name="password" id="password" />    
        <input type="submit" name="loginBtn" id="loginBtn" value="Login" />
    </div>
</form>
<script type="text/javascript">
$(document).ready(function() {
    $('#loginform').submit(function(e) {
        e.preventDefault();
        $.ajax({
            type: "POST",
            url: 'login.php',
            data: $(this).serialize(),
            success: function(response)
            {
                var jsonData = JSON.parse(response);
 
                // user is logged in successfully in the back-end
                // let's redirect
                if (jsonData.success == "1")
                {
                    location.href = 'my_profile.php';
                }
                else
                {
                    alert('Invalid Credentials!');
                }
           }
       });
     });
});
</script>
</body>
</html>
Posted by: Guest on May-30-2020
15

ajax jquery php

$.ajax({

   url     : "file.php",
   method  : "POST",

       data: { 
         //key                      :   value 
         action                   	:   action , 
         key_1   					:   value_key_1,
         key_2   					:   value_key_2
       }
   })

   .fail(function() { return false; })
	// Appel OK
   .done(function(data) {

   console.log(data);

 });
Posted by: Guest on May-05-2020
-2

ajax call php

$.ajax({ url: '/my/site',
         data: {action: 'test'},
         type: 'post',
         success: function(output) {
                      alert(output);
                  }
});
Posted by: Guest on December-09-2020

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