PHP form : not updating mysql database

Question

I have virtually no programming experience and trying this first project, I am a bit stuck on how to update the database, so I click on edit and the correct record gets loaded into the edit screen update.php

When I click update, I get the message from updated.php saying that the database has been updated, but the database does not get updated, when I display the records they are the same as before the update, thanks in advance for all your help.

the following code:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" 
"http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
    <title>Form Edit Data</title>
</head>

<body>
    <table border=1>
    <tr>
        <td align=center>Form Edit Employees Data</td>
    </tr>
    <tr>
    <td>
        <table>
        <?
        $user_name = "";
        $password = "";
        $database = "";
        $server = "localhost";

        mysql_connect($server, $user_name, $password);
        $db_found = mysql_select_db($database);
        $id = $_GET['id'];
        $order = "SELECT * FROM MY_ID where ID = ' " .$id . " ' ";
        $result = mysql_query($order);
        $row = mysql_fetch_array($result);
        ?>
        <form method="post" action="edit_data.php"?id=<?= $id ?>>
            <input type="text" name="id" value="<? echo "$row[ID]"?>">
            <tr>        
                <td>First Name</td>
                <td>
                    <input type="text" name="FirsName" size="20" value="<? echo "$row[FirstName]"?>">
                </td>
            </tr>
            <tr>
                <td>Sur Name</td>
                <td>
                    <input type="text" name="SurName" size="40" value="<? echo "$row[SurName]"?>">
                </td>
            </tr>
            <tr>
                <td>Address</td>
                <td>
                    <input type="text" name="Address" size="40" value="<? echo "$row[Address]"?>">
                </td>
            </tr>
            <tr>
                <td align="right">
                    <input type="submit" name="submit" value="submit">
                </td>
            </tr>
        </form>
        </table>

    </td>
    </tr>
    </table>
</body>
</html>

and here is the other file

<?php
$user_name = "";
$password = "";
$database = "";
$server = "";

mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database);

$id = $_REQUEST['ID'];
$FirstName = trim(mysql_real_escape_string($_POST["FirstName"]));
$SurName = trim(mysql_real_escape_string($_POST["SurName"]));
$Address = trim(mysql_real_escape_string($_POST["Address"]));

$sql = "UPDATE MY_ID SET FirstName='$FirstName',SurName='$SurName',Address='$Address' WHERE ID='$id'";
$result=mysql_query($sql);


if ($result){
    echo "Successful";
    echo "<BR>";
    echo "<a href='edit.php'>View result</a>";
}
else {
    echo "ERROR";
}

?>

Mansour Fahad · Accepted Answer · 2013-03-28 00:49:45Z

2

Looks like you forget the double quotation mark and the full stop. You should write it as: '".$example."'

$sql = "UPDATE MY_ID SET FirstName='".$FirstName."',SurName='".$SurName."',Address='".$Address.:' WHERE ID='".$id."'";

answered Mar 28, 2013 at 0:49

Mansour Fahad

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Comments

Jean · Accepted Answer · 2013-03-28 00:54:17Z

0

It is because your form method is POST, and you are trying to GET ID. Probably ID returns null. My suggestion is to put a hidden input in your form as with name="ID", then read it in your posted page as $_POST["ID"];

answered Mar 28, 2013 at 0:54

Jean

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Comments

Plipus Tel · Accepted Answer · 2013-03-28 01:16:31Z

0

Yes, the answer is as Mansours said. You should not use single quota to your variable.

So, it's bad practice writing code something like this:

<input type="text" value="<?php echo "$row[name]"; ?>">

it should be

<input type="text" value="<?php echo $row['name']; ?>">

it would be clear, and also, when inserting or updating the record you should write as follow:

$sql = "UPDATE MY_ID SET FirstName='" . $FirstName . "',
                         SurName='" . $SurName . "',
                         Address='" . $Address . "' 
         WHERE ID='" . $id . "'";
mysql_query($sql);

answered Mar 28, 2013 at 1:16

Plipus Tel

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