POST of multiple input form fields INTO mySQL db using AJAX

Question

I'm trying to submit Multiple Form Items from HTML using Java-script into mySQL via AJAX request. I can get one field to update but not the other 2. I've commented out the other code I believed would aid in this but was not working.

 HTML part: 
<!DOCTYPE HTML PUBLIC “-//W3C//DTD HTML 4.01 Transitional//EN”
“http://www.w3.org/TR/html4/loose.dtd”>
<head>

<script type="text/javascript">

function insert()  {

      if (window.XMLHttpRequest) {
      xmlhttp = new XMLHttpRequest();
   } else {
       xmlhttp = new ActiveXObject('Microsoft.XMLHTTP');
   }
   xmlhttp.onreadystatechange = function(){
      if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
    document.getElementById('message').innerHTML = xmlhttp.responseText;
  }
  }

 parameters = 'fname='+document.getElementById('fname').value;


   xmlhttp.open("POST", "update.php", true);
   xmlhttp.setRequestHeader('Content-type','application/x-www-form-urlencoded');
   xmlhttp.send(parameters);

} ;

 </script>

 </head>

 <body>

          First Name: <input class="work" type="text" id="fname" ><br>
          Middle Name: <input class="work" id="mname" type="text"><br>
          Last Name: <input class="work" id="lname" type="text"><br>
          <input type="button" value="Submit" onclick="insert();">

          <div id="message"></div>

  </body>
  </html>



 PHP part:
<?php
       //require 'connect.midasproject.php';
   $conn_error = 'could not connect.';

        $dbhost = "localhost";
        $dbname = "mastergolddb";
        $dbuser = "root";
        $dbpass = "";

 if (!@mysql_connect("$dbhost", "$dbuser", "$dbpass")||!@mysql_select_db ("$dbname")) {
   die($conn_error);
 } else {
   echo 'connected.';
   }

$fname = $_POST['fname'];
//$mname = $_POST['mname'] ;
//$lname = $_POST['lname'];

 if (!empty($fname))  {

$query = "INSERT INTO `customers` VALUES ('','$fname','joe','blow')";
    if  ($query_run = mysql_query($query)) {
       echo 'data inserted.' ;
    } else {
  echo 'Query failed.';
    }
   }

?>

You didn't send the other fields, you only sent the fname field. — Musa
– Musa, Commented Sep 20, 2012 at 20:52
And you need to use those fields again in your SQL query, '{$fname}',{$mname}', etc — dbf
– dbf, Commented Sep 20, 2012 at 20:56
I've tried multiple combinations of adding the other mname and lname inputs fields to the 'parameters' but code fails. When I create variables $mname = $_POST['mname'] ; $lname = $_POST['lname']; the also code fails. — user1635052
– user1635052, Commented Sep 20, 2012 at 22:54
I've used them in my SQL query, I put in Joe and blow as constants to show code that works but when I pass in the $mname and $lname it doesnt work. — user1635052
– user1635052, Commented Sep 20, 2012 at 22:56

Teena Thomas · Accepted Answer · 2012-09-20 20:52:21Z

1

you need to pass all 3 values to the ajax function. you seem to be passing just fname here parameters = 'fname='+document.getElementById('fname').value;

answered Sep 20, 2012 at 20:52

Teena Thomas

5,2091 gold badge15 silver badges17 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

user1635052 Over a year ago

How do I pass the other values? BTW- the HTML code of this didnt post.. I'm trying how to edit the question to append with that now

Collectives™ on Stack Overflow

POST of multiple input form fields INTO mySQL db using AJAX

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related