Insert MySQL record with PHP and AJAX

Question

trying to insert a record in my DB using AJAX for the very first time. I have the following...

Form

 <form>
     <input type="text" id="salary" name="salary">
     <input type="button" onclick="insertSalary()">
 </form>

AJAX

<script type="text/javascript">

function insertSalary()
{
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    document.getElementById('current-salary').innerHTML=xmlhttp.responseText;
    }
  };
xmlhttp.open("POST","insert_salary.php",true);
xmlhttp.send("salary=" + document.getElementById("salary").value);
}

</script>

PHP

$uid = $_SESSION['oauth_id'];      
$monthly_income = mysql_real_escape_string($_POST['salary']);

#Insert a new Record
$query = mysql_query("INSERT INTO `income` (user_id, monthly_income) VALUES ($uid, '$monthly_income')") or die(mysql_error());
$result = mysql_fetch_array($query);
return $result;

Nowmy data is being inserted into the table BESIDES the 'salary' which is being inputted as '0'

Once inserted I also have a div 'current-salary' that should then be populated with there inputted value only it isnt, Can anybody help me to understand where im going wrong?

Can you debug it in firebug or something and see if salary is being properly passed to your php? — rwilliams
– rwilliams, Commented Apr 12, 2012 at 19:16

James · Accepted Answer · 2012-04-12 19:21:37Z

4

If you want to save your self a lot of time, effort, and heartache, use the jquery library for your ajax requests. You can download it at http://jquery.com/

After adding a reference(Script tag) to the jquery script your javascript for the ajax request would become:

function insertSalary()
{
    var salary = $("#salary").val();
    $.post('insert_salary.php', {salary: salary}, function(data) 
    {
        $("#current-salary").html(data); 
    });
}

Also keep in mind that using "insert_salary.php" as the url means it is a relative path and must be in the folder of the current running script.

Your php script needs to echo whatever you would like to be injected into your current-salary tag also.

answered Apr 12, 2012 at 19:21

James

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6 Comments

Liam Over a year ago

wow, that easy! It works fine, my only problem is it doesnt show a response in the div...

James Over a year ago

Eh, you caught me before I could edit it. :) You need php to echo whatever you would like in your div tag.

Liam Over a year ago

So if I was to add 'echo $monthly_income;' beneath return $result that should work? It's not however for some strange reason... @james

James Over a year ago

Remove the return $result line altogether and try echoing a hardcoded value just to see if it will work, then replace your hardcoded value with a variable that you want to use.

Liam Over a year ago

ahh my bad @james! Im running an if statement at the top of the page to say if the user has no row in table do this, else show the form, on return, I need to page to refresh, or run the IF query again somehow?

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