CODEWITHSUNDEEP
phpmysqljquery-mobile
sakin
sakin

Reputation: 11

Unable to insert data into database using PHP and J Query Mobile

I can insert data into the database for the several time before this. I'm using web matrix and XAMPP to develop. I have problem while dealing with ports but then after altering something it become okay, but the database does not receive data anymore. My other system using PHP that was been develop is running good but the problem is with the jQuery mobile.

This is the index.html file 

<!DOCTYPE html>
<html>
<head>
    <meta charset="utf-8" />
    <meta name="viewport" content="width=device-width, initial-scale=1" />
    <meta name="apple-mobile-web-app-capable" content="yes" />
    <meta name="apple-mobile-web-app-status-bar-style" content="black" />
    <title>
    </title>
    <link rel="stylesheet" href="http://code.jquery.com/mobile/1.2.0/jquery.mobile-1.2.0.min.css" />
    <link rel="stylesheet" href="my.css" />
    <script src="http://code.jquery.com/jquery-1.7.2.min.js">
    </script>
    <script src="http://code.jquery.com/mobile/1.2.0/jquery.mobile-1.2.0.min.js">
    </script>
    <script src="my.js">
    </script>
    <!-- User-generated css -->
    <style>
    </style>
    <!-- User-generated js -->
    <script>
        try {

  $(function() {

});

} catch (error) {
  console.error("Your javascript has an error: " + error);
 }
     </script>
 </head>
 <body>
    <!-- Home -->
    <div data-role="page" id="page1">
        <div data-theme="a" data-role="header" data-position="fixed">
            <h3>
               Student Uitm
            </h3>
        </div>
        <div data-role="content">
            <form action="http://localhost/student/save2db.php" method="post">
                <label>Nama</label><input type="text" name="Nama">
                <label>No Kad Pengenalan</label><input type="text" maxlength="12" name="icno">
                <label>Jantina</label><select name="jantina">
                <option value="L">Lelaki</option>
                <option value="P">Perempuan</option>
            </select>
                <input name="sbt" type="submit" value="save" data-inline="true">
            </form>
        </div>
        <div data-theme="a" data-role="footer" data-position="fixed">
            <h3>
                Footer
            </h3>
        </div>
    </div>
</body>
</html>

and this is the save2db.php file:

 <?php
 $con=mysqli_connect("localhost","root"," ","student");
    // Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$sql="INSERT INTO student1 (Nama, icno, jantina)
VALUES
('$_POST[Nama]','$_POST[icno]','$_POST[jantina]')";

 if (!mysqli_query($con,$sql))
 {
  die('Error: ' . mysqli_error($con));
  }
 echo "1 record added";

mysqli_close($con);  
?>

Upvotes: 1

Views: 890

Answers (1)

didierc
didierc

Reputation: 14730

The problems:

  • you check for the db connection, but you let the script run through even when it's failed
  • you are not being cautious enough with the received data
  • you don't log errors
  • internal problem symptoms leak out to users

Here's a possible alternative to your script taking care of these issues:

<?php
$con = mysqli_connect("localhost","root"," ","student");
 // Check connection
if (mysqli_connect_errno()) {
    trigger_error("Failed to connect to MySQL: " . mysqli_connect_error(), 
                   E_USER_ERROR);
    die("unable to complete request");
} else if (!(isset($_POST['Nama']) &&
         isset($_POST['icno']) && isset($_POST['jantina'])) {
    trigger_error( "incomplete POST data", E_USER_ERROR);
    die("unable to complete request");
} else {
    $nama = mysqli_real_escape_string($con, $_POST['Nama']);
    $icno = mysqli_real_escape_string($con $_POST['icno']);
    $jantina = mysqli_real_escape_string($con,$_POST['jantina']);
    $sql = "INSERT INTO student1 (Nama, icno, jantina) VALUES ('$nama','$icno','$jantina')";
    if (!mysqli_query($con,$sql)) {
        trigger_error("query failed :" . mysqli_error($con), E_USER_ERROR);
        die("unable to complete request");
    }
    echo "1 record added";
    mysqli_close($con);  
}

With the above modification done, check the php logs after unsuccessful updates to know the possible reasons of the failure.

Upvotes: 1

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