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I have a very simple form, containing a textbox and a submit button. When the user enters something into the form, then clicks submit, I would like to use PHP and Ajax (with jQuery) to insert the result of the form into a MySQL database. this result should be displayed on the same page in the form of a table which is updated after every insert.
Can anyone please help?
The code I have used that isn’t working:
ajax.html:
<html>
<body>
<script language="javascript" type="text/javascript">
<!--
//Browser Support Code
function ajaxFunction(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
}catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
}catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
// Create a function that will receive data
// sent from the server and will update
// div section in the same page.
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
var ajaxDisplay = document.getElementById('ajaxDiv');
ajaxDisplay.value = ajaxRequest.responseText;
}
}
// Now get the value from user and pass it to
// server script.
var name = document.getElementById('name').value;
var age = document.getElementById('age').value;
var wpm = document.getElementById('wpm').value;
var sex = document.getElementById('sex').value;
var queryString = "&name=" +name+ "&age=" + age ;
queryString += "&wpm=" + wpm + "&sex=" + sex;
ajaxRequest.open("GET", "ajax-example.php" +
queryString, true);
ajaxRequest.send(null);
}
//-->
</script>
<form name='myForm'>
Name: <input type='text' id='name' /><br/>
Max Age: <input type='text' id='age' /> <br />
Max WPM: <input type='text' id='wpm' />
<br />
Sex: <select id='sex'>
<option value="m">m</option>
<option value="f">f</option>
</select>
<input type='button' onclick='ajaxFunction()'
value='Query MySQL'/>
</form>
<div id='ajaxDiv'>Your result will display here</div>
</body>
</html>
ajax-example.php:
<?php
$dbhost = "localhost";
$dbuser = "demo";
$dbpass = "demo";
$dbname = "test_db";
//Connect to MySQL Server
mysql_connect($dbhost, $dbuser, $dbpass);
//Select Database
mysql_select_db($test_db) or die(mysql_error());
// Retrieve data from Query String
$age = $_GET['age'];
$sex = $_GET['sex'];
$wpm = $_GET['wpm'];
// Escape User Input to help prevent SQL Injection
$age = mysql_real_escape_string($age);
$sex = mysql_real_escape_string($sex);
$wpm = mysql_real_escape_string($wpm);
//build query
$query = "INSERT INTO form2 (name,age,sex,wpm) VALUES ('$name','$age','$sex','$wpm')";;
mysql_select_db('test_db');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not enter data: ' . mysql_error());
}
//Build Result String
$display_string = "<table>";
$display_string .= "<tr>";
$display_string .= "<th>Name</th>";
$display_string .= "<th>Age</th>";
$display_string .= "<th>Sex</th>";
$display_string .= "<th>WPM</th>";
$display_string .= "</tr>";
// Insert a new row in the table for each person returned
$result1=mysql_query("SELECT * FROM form2 WHERE name='$name'");
while($row = mysql_fetch_array($result1))
{
$display_string .= "<tr>";
$display_string .= "<td>$row[name]</td>";
$display_string .= "<td>$row[age]</td>";
$display_string .= "<td>$row[sex]</td>";
$display_string .= "<td>$row[wpm]</td>";
$display_string .= "</tr>";
}
$display_string .= "</table>";
echo $display_string;
?>
When you receive the data in the php script make query to the database and get your result
hope this would help
There are many tutorials available on internet for ajax with PHP and Jquery. You can go through any of these and get the desired result.
See an example here http://www.tutorialspoint.com/ajax/ajax_database.htm