Harshal Shah- Technically Uncomplicating: Ajax Mysql Php Example

<html>
<head>

<script src=
"http://ajax.googleapis.com/ajax/libs/jquery/1.3/jquery.min.js"
type="text/javascript">
</script>

    <script type="text/javascript">
      function register(){
    //    window.open("/welcome.html",'welcome','width=300,height=200,menubar=yes,status=yes,location=yes,toolbar=yes,scrollbars=yes');

        $.ajax({
            type: "POST",
            url: "/ReturnData.php",
            data:     "username=" + document.getElementById("username").value +
                    "&email=" + document.getElementById("email").value,
            success: function(html){
                $("#response").html(html);
            }
        });
        }
    </script>
</head>

<body>
    <form action="" method="post">
            <p>
                <label for="name">Name:</label><br />
                <input type="text" name="username" id="username" size="25" />
            </p>
            <p>
                <label for="email">Email:</label><br />
                <input type="text" name="email" id="email" size="25" />
            </p>
            <p>
                <input type="button" name="submit" id="submit" value="Subscribe" onclick="register()"/>
            </p>
            <ul id="mylist">
               <li><a rel="3" href="/#dave">Dave's email address</a></li>
            <li><a rel="4" href="/#erik">Erik's email address</a></li>
            </ul>

<p id="info"> </p>
    <div id="response">

    </div>
</form>
</body>
</html>

<?php
                $db_host = 'localhost';
                $db_user = 'root';
                $db_pass = 'root';
                $db_name = 'db';

        $Username = $_POST['username'];
        $Email    = $_POST['email'];

        $connect = mysql_connect( $db_host, $db_user, $db_pass ) or die( mysql_error());
        $connection = $connect;

        mysql_select_db( $db_name, $connect ) or die( mysql_error() );

         echo $Username;
        $qInsertUser = mysql_query(" INSERT INTO test
                                     VALUES ('$Username','$Email')
                                  ");
        $qGetDetails = mysql_query("SELECT * FROM test");


        $num=mysql_numrows($qGetDetails);
        echo $num;
        echo "hi";
        $i=0;
        while ($row=mysql_fetch_array($qGetDetails)) {
        echo "<br>".$row['name'];
        }
    ?>

Harshal Shah- Technically Uncomplicating

Friday, January 28, 2011

Ajax Mysql Php Example

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