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Home/ Questions/Q 7941353
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Editorial Team
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Editorial Team
Asked: June 3, 20262026-06-03T23:38:45+00:00 2026-06-03T23:38:45+00:00

I’ve read about the jQuery.ajax method and believe this should be what I need

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I’ve read about the jQuery.ajax method and believe this should be what I need — but so far can’t get it to work.

I created a test mysql database with a table called “users”, added rows to that table for “name” and “location”, and then made sure I could save data to it using the command line, which I could. Then I made a test page with a button on it and added this copy to my scripts file (the $.ajax part comes straight from the jQuery api examples):

$('#button').click(function(){
    saveData();
});

function saveData(){  
$.ajax({
       type: "POST",
   url: "process.php",
   data: { name: "John", location: "Boston" }
    }).done(function( msg ) {
 alert( "Data was saved: " + msg );
    });
}

I do indeed get an alert message, “Data was saved”, but nothing has actually been saved to my database. I must be doing something wrong with process.php, but not sure what. Here’s the code in process.php (I set variables for database, db_host, user, etc that I don’t display here):

// 1. Create a connection to the server. 
$connection = mysql_connect($db_host, $db_user,$db_pwd);

// make sure a connection has been made
if (!$connection){
die("Database connection failed: " . mysql.error());
}

// 2. Select the database on the server
$db_select = mysql_select_db($database, $connection);
if (!$db_select){
die("Database selection failed: " . mysql.error());
}

// START FORM PROCESSING
if (isset($_POST['submit'])) { // Form has been submitted.
    $name = trim(mysql_prep($_POST['name']));
$location = trim(mysql_prep($_POST['location']));

// INSERT THE DATA 
$query = "INSERT INTO user ( name, location )
              VALUES ( '{$name}','{$location}' )";
        // Confirm if the query is successful.
        $result = mysql_query($query, $connection);
}
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  1. Editorial Team
    Editorial Team
    2026-06-03T23:38:48+00:00Added an answer on June 3, 2026 at 11:38 pm

    It’s has already been told by others, about using ‘success’ and ‘error’ which would be indeed a better method to capture the success/error callback. But that’s really not the problem why nothing is inserted. Your Ajax call looks good besides that.

    Another problem that i see is that you’re only passing the following params:

    data: { name: "John", location: "Boston" }
    

    But in your PHP code you do check if a submit button was set:

    if (isset($_POST['submit']))
    

    This is obviously not the case because the only things that are sent are the things you pass in the ‘data’ attribute of the Ajax call. Instead use:

    if ( $_SERVER['REQUEST_METHOD'] == 'POST' )
    
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