How do i update mysql database with ajax and php with no page refresh
+2
A:
Here is a good example, it shows a SELECT statement but it should be straight forward and easy to update the script to what you need.
HTML from the example page:
<html>
<head>
<script src="selectuser.js"></script>
</head>
<body>
<form>
Select a User:
<select name="users" onchange="showUser(this.value)">
<option value="1">Peter Griffin</option>
<option value="2">Lois Griffin</option>
<option value="3">Glenn Quagmire</option>
<option value="4">Joseph Swanson</option>
</select>
</form>
<p>
<div id="txtHint"><b>User info will be listed here.</b></div>
</p>
</body>
</html>
JavaScript
var xmlHttp;
function showUser(str)
{
xmlHttp=GetXmlHttpObject();
if (xmlHttp==null)
{
alert ("Browser does not support HTTP Request");
return;
}
var url="getuser.php";
url=url+"?q="+str;
url=url+"&sid="+Math.random();
xmlHttp.onreadystatechange=stateChanged;
xmlHttp.open("GET",url,true);
xmlHttp.send(null);
}
function stateChanged()
{
if (xmlHttp.readyState==4 || xmlHttp.readyState=="complete")
{
document.getElementById("txtHint").innerHTML=xmlHttp.responseText;
}
}
function GetXmlHttpObject()
{
var xmlHttp=null;
try
{
// Firefox, Opera 8.0+, Safari
xmlHttp=new XMLHttpRequest();
}
catch (e)
{
//Internet Explorer
try
{
xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e)
{
xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
}
}
return xmlHttp;
}
PHP called by Ajax
<?php
$q=$_GET["q"];
$con = mysql_connect('localhost', 'peter', 'abc123');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("ajax_demo", $con);
$sql="SELECT * FROM user WHERE id = '".$q."'";
$result = mysql_query($sql);
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Age</th>
<th>Hometown</th>
<th>Job</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "<td>" . $row['Age'] . "</td>";
echo "<td>" . $row['Hometown'] . "</td>";
echo "<td>" . $row['Job'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($con);
?>
Phill Pafford
2009-05-12 14:59:14
I would also add some validation before doing any SQL query on user given values.
Phill Pafford
2009-05-12 15:00:11
+1
A:
http://www.w3schools.com/PHP/php_ajax_database.asp explains it quite nicely.
Jez
2009-05-12 14:59:20
+1
A:
You can use the JQuery library and do something like
$.ajax({
type: "POST",
url: "some.php",
data: "name=John&location=Boston",
success: function(msg){
alert( "Data Saved: " + msg );
}
});
In your some.php file have something like this:
<?php
$name = $_POST['name'];
$location = $_POST['location'];
$query("your insert query");
$result = mysql_query($query);
?>
?>
Rigobert Song
2009-05-12 15:00:27
If you have a follow-up question you should better ask it as a new question, not post it here as an answer. More people would look at it that way. The "Ask Question" button is in the top right...
sth
2009-10-28 23:50:36