No errors and database not updating

davidjones1990 · September 7, 2011

Here is my code to update a blog entry. When submitted there is no error but the database is not updated with the new info. I have a similar script on a different page and it works fine so I am confused to why this is happening. I am also a PHP beginner.

Here is my code:

<?php
include_once ("scripts/checkuserlog.php"); 
include_once ("scripts/connectToMysql.php");

if (!$_SESSION['idx']) { 
    $msgToUser = '<br /><br /><font color="#FF0000">Only site members can do that</font><p><a href="register.php">Join Here</a></p>';
    include_once 'msgToUser.php'; 
    exit(); 
} else if ($logOptions_id != $_SESSION['id']) {
$msgToUser = '<br /><br /><font color="#FF0000">Only site members can do that</font><p><a href="register.php">Join Here</a></p>';
    include_once 'msgToUser.php'; 
    exit(); 
}

$date = date("m.d.y");
$workoutName = '';
$workoutDescription = '';

$id = $_GET['id'];

if(isset($_POST['workoutName'])){
$workoutName = $_POST['workoutName'];
$workoutDescription = $_POST['workoutDescription'];

$blogUpdate = mysql_query("UPDATE blog SET workoutName='$workoutName', workoutDescription='$workoutDescription' WHERE id='$id' LIMIT 1") or die (mysql_error());

     if ($blogUpdate){
            $successMsg = 'Blog entry updated successfully';
     } else {
	    $errorMsg = 'Problems arose during the information exchange, please try again later.';
     }
}

$sql = mysql_query("SELECT * FROM blog WHERE id='$id' LIMIT 1");

while($row = mysql_fetch_array($sql)){
$workoutName = $row['workoutName'];
$workoutDescription = $row['workoutDescription'];
$dateOfEntry = $row['datetime'];
}

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>S.F.I - Edit Blog Entry</title>
<link href="style/main.css" rel="stylesheet" type="text/css" />
<link href="style/layout.css" rel="stylesheet" type="text/css" />
</head>

<body>
<div id="container">
<?php include_once ("bannerFiles/bannerTemplate.php"); ?>
<?php include_once ("bannerFiles/bannerMenu.php"); ?>
    <br /><br />
<div id="content">
    <table bgcolor="#DBE4FD" width="950px">
    <form action="editBlogPost.php" method="post" enctype="multipart/form-data">
    <tr>
    <td width="200px"><span class="blackText">Workout Name:</span></td><td width="650px"><input name="workoutName" type="text" id="workoutName" value="<?php echo ("$workoutName");?>"/>    <span class="blackText">Date of entry: <?php echo ("$dateOfEntry"); ?></span></td></tr>
    <tr>
    <td><span class="blackText">Workout Description:</span></td><td><textarea name="workoutDescription" cols="75" rows="10" id="workoutDescription" /><?php echo ("$workoutDescription"); ?></textarea></td></tr>
    <tr><td><input name="submit" id="submit" type="submit" value="Update" /></td><td><span class="errorMsg"><?php echo ("$errorMsg"); ?><?php echo ("$successMsg"); ?></span></td>
    </tr>
    </form>
    </table>
</div>
    <br /><br />
    <?php include_once ("footerFiles/footerTemplate.php"); ?>
</div>
</body>
</html>

Thanks in advance for any help

AyKay47 · September 7, 2011

have you outputted $id to make sure it is what you expect before using it in your update query?

gristoi · September 7, 2011

heres your problem:

$id = $_GET['id'];

your POSTING data from the form. Not sending the variable through $_GET[]

davidjones1990 · September 7, 2011

Yep, the ID corresponds to the correct ID in the database

davidjones1990 · September 7, 2011

Im getting the blog entry ID from the URL so I can use it to update that blog entry with the new posted information

Is that right? or do I need to go back to basics

gristoi · September 7, 2011

close, you are getting the blog entry via the url , but you are then posting the form back to itself which wipes out the id. add this into your form

<input type="hidden" name ='id' value="".$_GET['id'].""/>

that will in effect auto forward your get variable along with the form. then:

if(isset($_POST['workoutName'])){
        $id = $_POST['id'];
$workoutName = $_POST['workoutName'];
$workoutDescription = $_POST['workoutDescription'];

$blogUpdate = mysql_query("UPDATE blog SET workoutName='$workoutName', workoutDescription='$workoutDescription' WHERE id='$id' LIMIT 1") or die (mysql_error());

davidjones1990 · September 7, 2011

Works now, although I put this in the form instead of what you said, I know your way is better but i just want to get it working for now

<input type="hidden" name ='id' value="<?php echo ("$id"); ?>"/>

Thanks for your help

Sign In

No errors and database not updating

Recommended Posts

davidjones1990

Link to comment

Share on other sites

AyKay47

Link to comment

Share on other sites

gristoi

Link to comment

Share on other sites

davidjones1990

Link to comment

Share on other sites

davidjones1990

Link to comment

Share on other sites

gristoi

Link to comment

Share on other sites

davidjones1990

Link to comment

Share on other sites

Join the conversation

Browse

Activity

Important Information