Load data from MySQL database to HTML textboxes on button click

Question

I have a small form for updating existing records.

enter image description here

I'm loading the Service IDs to the dropdown box using PHP. And when the user clicks the Load button, it is supposed to display the other details related to that ID in the textboxes below. Here is the code I have so far.

<html>
<head>
</head>
<body>

<?php
//Database initialization
require_once("db_handler.php");

$conn = iniCon();
$db = selectDB($conn);

$query = "SELECT * FROM taxi_services ORDER BY SID";
$result2 = mysql_query($query, $conn);

?>

<div id="upserv">
<b id="caption2">Update location</b>
<br/><br/>
    <form name="upServForm" action="<?php echo $PHP_SELF; ?>" method="post" >
        <?php
        $dropdown = "<select name='codes'>";
        while($row = mysql_fetch_assoc($result2)) 
        {
            $dropdown .= "\r\n<option value='{$row['SID']}'>{$row['SID']}</option>";
        }
        $dropdown .= "\r\n</select>";
    ?>
     Service ID  <?php echo $dropdown; ?> <input type="submit" value="Load" name="loadbtn">
        <table width="300" border="0">
          <tr>
            <td>Name</td>
            <td><input type="text" name="upName" style="text-align:right" value="<? echo $savedName; ?>" /></td>
          </tr>
          <tr>
            <td>Cost</td>
            <td><input type="text" name="upCost" style="text-align:right" value="<? echo $savedCost; ?>" /></td>
          </tr>
          <tr>
            <td>Active</td>
            <td><input type="checkbox" name="upActive" value="<? echo $savedActive; ?>" /></td>
          </tr>
        </table>
</div>
<br/>
<div id="buttons">
    <input type="reset" value="Clear" /> <input type="submit" value="Save" name="updatebtn" />
</div>
    </form>

<?php

if(isset($_POST["loadbtn"]))
{
    $id = $_POST["codes"];

    $query = "SELECT Name, Cost, Active FROM taxi_services WHERE SID = '$id' ";
    $result = mysql_query($query, $conn);
    $details = mysql_fetch_array($result);

    $savedName = $details["Name"];
    $savedCost = $details["Cost"];
    $savedActive = $details["Active"];
}

?>

</body>
</html>

The query gets executed just fine but the data doesn't get displayed in the textboxes. Can anyone please tell me what I am missing here?

Thank you.

I THINK your server doesn't support short tags. In other words, replace <? with <?php — Adi
– Adi, Commented Jun 30, 2012 at 10:08
@diEcho here it is. @Adnan I tried with <?php ?> too but to no avail. — Isuru
– Isuru, Commented Jun 30, 2012 at 10:13

Thomas · Accepted Answer · 2012-06-30 11:04:37Z

3

Your query has to be before the output:

Also note the typecast (integer) of the id to secure against sql injections.

Also note the security issues with $PHP_SELF http://php.about.com/od/learnphp/qt/_SERVER_PHP.htm I changed the code to $_SERVER['SCRIPT_NAME']

ALso note to not use register_globals and disable it in the configuration if you can (use $_SERVER['SCRIPT_NAME'] instead of$SCRIPT_NAME`) : http://www.php.net/manual/en/security.globals.php

If you learn php from a book and this is based on sourcecode out of this book you should throw it away immediately.

<?php

//Database initialization
require_once("db_handler.php");

$conn = iniCon();
$db = selectDB($conn);

$query = "SELECT * FROM taxi_services ORDER BY SID";
$result2 = mysql_query($query, $conn);

if(isset($_POST["loadbtn"]))
{
    $id = (integer) $_POST["codes"];

    $query = "SELECT Name, Cost, Active FROM taxi_services WHERE SID = '$id' ";
    $result = mysql_query($query, $conn);
    $details = mysql_fetch_array($result);

    $savedName = $details["Name"];
    $savedCost = $details["Cost"];
    $savedActive = $details["Active"];
}

?>

<html>
<head>
</head>
<body>

<div id="upserv">
<b id="caption2">Update location</b>
<br/><br/>
    <form name="upServForm" action="<?php echo $_SERVER['SCRIPT_NAME']; ?>" method="post" >
        <?php
        $dropdown = "<select name='codes'>";
        while($row = mysql_fetch_assoc($result2)) 
        {
            $dropdown .= "\r\n<option value='{$row['SID']}'>{$row['SID']}</option>";
        }
        $dropdown .= "\r\n</select>";
    ?>
     Service ID  <?php echo $dropdown; ?> <input type="submit" value="Load" name="loadbtn">
        <table width="300" border="0">
          <tr>
            <td>Name</td>
            <td><input type="text" name="upName" style="text-align:right" value="<? echo $savedName; ?>" /></td>
          </tr>
          <tr>
            <td>Cost</td>
            <td><input type="text" name="upCost" style="text-align:right" value="<? echo $savedCost; ?>" /></td>
          </tr>
          <tr>
            <td>Active</td>
            <td><input type="checkbox" name="upActive" value="<? echo $savedActive; ?>" /></td>
          </tr>
        </table>
</div>
<br/>
<div id="buttons">
    <input type="reset" value="Clear" /> <input type="submit" value="Save" name="updatebtn" />
</div>
    </form>

</body>
</html>

edited Jun 30, 2012 at 11:04

answered Jun 30, 2012 at 10:12

Thomas

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5 Comments

Isuru Over a year ago

You're right! That was the case. And thank you for the tip on SQLi.

Thomas Over a year ago

@nK0de SQLi is not the right term it refers to the "new" MySQL php Extension: de2.php.net/mysqli please don't forget to mark the question answered :-)

Isuru Over a year ago

Right. Got it. I always do, just had to wait till the 15 minute time limit is up. :)

Thomas Over a year ago

@nK0de $PHP_SELF is a bad idea. There are security issues with it: php.about.com/od/learnphp/qt/_SERVER_PHP.htm

Isuru Over a year ago

@Thomas yes, I usually use external PHP files. This page is just a small front-end to a database that only I will be using. It will not be open to the public. So I'm not developing with security in mind too much. :)

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