PHP - AJAX and MySQL
AJAX can be used to communicate with a database interactively.
AJAX Database Example
The following example demonstrates how a web page can read information from a database via AJAX:
SQL file for the Websites table used in this tutorial:websites.sql。
Example
Select an option, and the website information will be displayed here...
Example Explained - MySQL Database
In the above example, the database table we use is shown below:
mysql> select * from websites; +----+--------------+---------------------------+-------+---------+ | id | name | url | alexa | country | +----+--------------+---------------------------+-------+---------+ | 1 | Google | https://www.google.cm/ | 1 | USA | | 2 | 淘宝 | https://www.taobao.com/ | 13 | CN | | 3 | Example | http://www.example.com/ | 4689 | CN | | 4 | 微博 | http://weibo.com/ | 20 | CN | | 5 | Facebook | https://www.facebook.com/ | 3 | USA | +----+--------------+---------------------------+-------+---------+ 5 rows in set (0.01 sec)
Example Explained - HTML Page
When the user selects a website in the dropdown list above, a function called "showSite()" is executed. The function is triggered by the "onchange" event:
test.html file code:
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>Example (example.com)</title>
<script>
function showSite(str)
{
if (str=="")
{
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{
//IE7+, Firefox, Chrome, Opera, Safari browsers execute code
xmlhttp=new XMLHttpRequest();
}
else
{
//IE6, IE5 browsers execute code
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","getsite_mysql.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>
<form>
<select name="users" onchange="showSite(this.value)">
<option value="">Select a website:</option>
<option value="1">Google</option>
<option value="2">Taobao</option>
<option value="3">Example</option>
<option value="4">Weibo</option>
<option value="5">Facebook</option>
</select>
</form>
<br>
<div id="txtHint"><b>The website information will be displayed here...</b></div>
</body>
</html>
The showSite() function performs the following steps:
- Check if a website is selected
- Create an XMLHttpRequest object
- Create a function to execute when the server response is ready
- Send a request to a file on the server
- Note that a parameter (q) is added to the end of the URL (containing the content of the dropdown list)
PHP File
The server page called via JavaScript above is a PHP file named "getsite_mysql.php".
The source code in "getsite_mysql.php" runs a query against a MySQL database and returns the result in an HTML table:
getsite_mysql.php file code:
$q = isset($_GET["q"]) ? intval($_GET["q"]) : '';
if(empty($q)) {
echo 'Please select a website';
exit;
}
$con = mysqli_connect('localhost','root','123456');
if (!$con)
{
die('Could not connect: ' . mysqli_error($con));
}
//Select database
mysqli_select_db($con,"test");
//Set encoding to prevent garbled Chinese characters
mysqli_set_charset($con, "utf8");
$sql="SELECT * FROM Websites WHERE id = '".$q."'";
$result = mysqli_query($con,$sql);
echo "<table border='1'>
<tr>
<th>ID</th>
<th>Website Name</th>
<th>Website URL</th>
<th>Alexa Ranking</th>
<th>Country</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['url'] . "</td>";
echo "<td>" . $row['alexa'] . "</td>";
echo "<td>" . $row['country'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
Explanation: When the query is sent from JavaScript to the PHP file, the following happens:
- PHP opens a connection to a MySQL database
- Find the selected website
- Creates an HTML table, fills it with data, and sends it back to the "txtHint" placeholder