Insert data through ajax into mysql database

Question

Newsletter:

<form id="form-search" method="post" action="<?php echo $_SERVER['PHP_SELF'];?>">
    <span><span class="style2">Enter you email here</span>:</span>
    <input name="email" type="email" id="email" required/>

 <input type="submit" value="subscribe" class="submit" onclick="return fun()" />
                        </form>
                        <?php
mysql_connect("localhost","","");
mysql_select_db("");
error_reporting(E_ALL && ~E_NOTICE);

$email=$_POST['email'];
$sql="INSERT INTO newsletter_email(email) VALUES ('$email')";
$result=mysql_query($sql);
if($result){
echo "You have been successfully subscribed.";
}
 if(!$sql)
die(mysql_error());

mysql_close();

?>

But I want to insert my E-Mail to database through Ajax. I don't want my page to get redirected, because every time the page got refreshed, null value got inserted to the database..

I just want my E-mail got inserted to database through Ajax, and after that the email box i.e.

<input name="email" type="email" id="email" required/>  
<input type="submit" value="subscribe" class="submit" onclick="return fun()" />

should get disappeared and there should be the line "you've been successfully subscribed" ..

Any brief code will be very useful.. thank u in advance :)

Answer 1

Try this:

  $(document).on('click','#save',function(e) {
  var data = $("#form-search").serialize();
  $.ajax({
         data: data,
         type: "post",
         url: "insertmail.php",
         success: function(data){
              alert("Data Save: " + data);
         }
});
 });

and in insertmail.php:

<?php 
if(isset($_REQUEST))
{
        mysql_connect("localhost","root","");
mysql_select_db("eciticket_db");
error_reporting(E_ALL && ~E_NOTICE);

$email=$_POST['email'];
$sql="INSERT INTO newsletter_email(email) VALUES ('$email')";
$result=mysql_query($sql);
if($result){
echo "You have been successfully subscribed.";
}
}
?>

Don't use mysql_ it's deprecated.

another method:

Actually if your problem is null value inserted into the database then try this and here no need of ajax.

<?php
if($_POST['email']!="")
{
    mysql_connect("localhost","root","");
    mysql_select_db("eciticket_db");
    error_reporting(E_ALL && ~E_NOTICE);
    $email=$_POST['email'];
    $sql="INSERT INTO newsletter_email(email) VALUES ('$email')";
    $result=mysql_query($sql);
    if($result){
    //echo "You have been successfully subscribed.";
              setcookie("msg","You have been successfully subscribed.",time()+5,"/");
              header("location:yourphppage.php");
    }
     if(!$sql)
    die(mysql_error());
    mysql_close();
}
?>
    <?php if(isset($_COOKIE['msg'])){?>
       <span><?php echo $_COOKIE['msg'];setcookie("msg","",time()-5,"/");?></span> 
    <?php }?>
<form id="form-search" method="post" action="<?php echo $_SERVER['PHP_SELF'];?>">
    <span><span class="style2">Enter you email here</span>:</span>
    <input name="email" type="email" id="email" required/>
    <input type="submit" value="subscribe" class="submit"/>
</form>

Answer 2

The ajax is going to be a javascript snippet that passes information to a small php file that does what you want. So in your page, instead of all that php, you want a little javascript, preferable jquery:

function fun()
{
    $.get('\addEmail.php', {email : $(this).val()}, function(data) {
        //here you would write the "you ve been successfully subscribed" div
    });
}

also you input would have to be:

<input type="button" value="subscribe" class="submit" onclick="fun();" />

last the file addEmail.php should look something like:

mysql_connect("localhost","root","");
mysql_select_db("eciticket_db");

error_reporting(E_ALL && ~E_NOTICE);

$email=mysql_real_escape_string($_GET['email']);
$sql="INSERT INTO newsletter_email(email) VALUES ('$email')";
$result=mysql_query($sql);
if($result){
echo "You have been successfully subscribed.";
}
 if(!$sql)
die(mysql_error());

mysql_close();

Also sergey is right, you should use mysqli. That's not everything, but enough to get you started.