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11 December, 2018

AJAX - Database Operations

 Programing Coderfunda     December 11, 2018     Ajax, AJAX - Database Operations     No comments   

AJAX - Database Operations

To clearly illustrate how easy it is to access information from a database using AJAX, we are going to build MySQL queries on the fly and display the results on "ajax.html". But before we proceed, let us do the ground work. Create a table using the following command.
NOTE − We are assuming you have sufficient privilege to perform the following MySQL operations.
CREATE TABLE 'ajax_example' (
'name' varchar(50) NOT NULL,
'age' int(11) NOT NULL,
'sex' varchar(1) NOT NULL,
'wpm' int(11) NOT NULL,
PRIMARY KEY ('name')
)
Now dump the following data into this table using the following SQL statements −
INSERT INTO 'ajax_example' VALUES ('Jerry', 120, 'm', 20);
INSERT INTO 'ajax_example' VALUES ('Regis', 75, 'm', 44);
INSERT INTO 'ajax_example' VALUES ('Frank', 45, 'm', 87);
INSERT INTO 'ajax_example' VALUES ('Jill', 22, 'f', 72);
INSERT INTO 'ajax_example' VALUES ('Tracy', 27, 'f', 0);
INSERT INTO 'ajax_example' VALUES ('Julie', 35, 'f', 90);

Client Side HTML File

Now let us have our client side HTML file, which is ajax.html, and it will have the following code −
<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
.innerHTML = ajaxRequest.responseText;
}
}

// Now get the value from user and pass it to
// server script.
var age = document.getElementById('age').value;
var wpm = document.getElementById('wpm').value;
var sex = document.getElementById('sex').value;
var queryString = "?age = " + age ;

queryString
+= "&wpm = " + wpm + "&sex = " + sex;
ajaxRequest
.open("GET", "ajax-example.php" + queryString, true);
ajaxRequest
.send(null);
}
//-->
</script>

<form name = 'myForm'>
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>
NOTE − The way of passing variables in the Query is according to HTTP standard and have formA.
URL?variable1 = value1;&variable2 = value2;
The above code will give you a screen as given below −
Max Age: 
Max WPM:
Sex:
Your result will display here in this section after you have made your entry.
NOTE − This is a dummy screen.

Server Side PHP File

Your client-side script is ready. Now, we have to write our server-side script, which will fetch age, wpm, and sex from the database and will send it back to the client. Put the following code into the file "ajax-example.php".
<?php
$dbhost
= "localhost";
$dbuser
= "dbusername";
$dbpass
= "dbpassword";
$dbname
= "dbname";

//Connect to MySQL Server
mysql_connect
($dbhost, $dbuser, $dbpass);

//Select Database
mysql_select_db
($dbname) 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
$query
= "SELECT * FROM ajax_example WHERE sex = '$sex'";

if(is_numeric($age))
$query
.= " AND age <= $age";

if(is_numeric($wpm))
$query
.= " AND wpm <= $wpm";

//Execute query
$qry_result
= mysql_query($query) or die(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
while($row = mysql_fetch_array($qry_result)) {
$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>";
}

echo
"Query: " . $query . "<br />";
$display_string
.= "</table>";

echo $display_string
;
?>
Now try by entering a valid value (e.g., 120) in Max Age or any other box and then click Query MySQL button.
Max Age: 
Max WPM:
Sex:
Your result will display here in this section after you have made your entry.
If you have successfully completed this lesson, then you know how to use MySQL, PHP, HTML, and Javascript in tandem to write AJAX applications.

 

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