LostNights

action="infoupdate2.php?id=<?php =$id ?>" gave a parse error action="infoupdate2.php?id=<?php $id ?>" didnt make a difference

When I change [code] $query = "UPDATE corporatelist SET firstname='$_POST[firstname]', lastname='$_POST[lastname]',title='$_POST[title]',department='$_POST[department]',office='$_POST[office]',phone='$_POST[phone]' WHERE sku = '".$id."' "; [/code] to [code] $query = "UPDATE corporatelist SET firstname='$_POST[firstname]', lastname='$_POST[lastname]',title='$_POST[title]',department='$_POST[department]',office='$_POST[office]',phone='$_POST[phone]' WHERE lastname = 'Benko' "; [/code] That changes it for the person whos last name is Benko, so its a problem with the variable.

No error, just cycles to the "Your info has been updated" line, but it doesnt update that line in the database. Could it be the variable from the other page? Its just a link variable. the ID echo's out fine.

It wont update my information in the DB [code] <?php $id = $_GET['id'];     if (isset($_POST['submitbutton'])) { mysql_connect('localhost','user','user') or die ("Cannot connect to server"); mysql_select_db("photodirectory") or die ("Cannot connect to database"); $query = "UPDATE corporatelist SET firstname='$_POST[firstname]', lastname='$_POST[lastname]',title='$_POST[title]',department='$_POST[department]',office='$_POST[office]',phone='$_POST[phone]' WHERE sku = '".$id."' "; $result = mysql_query($query);         if (!$result) {             echo "There was a problem with your entry. <a href='infoupdate2.php'>Back</a>";         } else {             echo "Your information has been changed.";         }     } else { ?> <?php echo $id; ?> <form action="infoupdate2.php" method="post"> <p>First Name: <input type="text" maxlength="30" size="32" name="firstname" /></p> <p>Last Name: <input type="text" maxlength="30" size="32" name="lastname" /></p> <p>Job Title: <input type="text" maxlength="55" size="57" name="title" /></p> <p>Department: <input type="text" maxlength="18" size="20" name="department" /></p> <p>Office: <input type="text" maxlength="4" size="6" name="office" /></p> <p>Phone Number: <input type="text" maxlength="4" size="6" name="phone" /></p> <input type="submit" value="Enter Employee Information" name="submitbutton" /> </form> <?php } ?> [/code]

Im am sending a variable from a link to another page so that I can update a row in the database based on that variable. Im just a little confused exactly how the line of code goes. This is from the page sending the variable sku [code] <?php do { ?>         <div class="employee2">         <p>             <?php echo "Record # " . $row_Recordset1['sku']; ?> |             <?php echo $row_Recordset1['firstname']; ?>             <?php echo $row_Recordset1['lastname']; ?> |             <?php echo $row_Recordset1['title']; ?> |             <?php echo $row_Recordset1['department']; ?> |             <?php echo $row_Recordset1['phone']; ?> |             <?php echo $row_Recordset1['office']; ?> |             <a href="infoupdate.php?id=<?php echo $row_Recordset1['sku']; ?>">Edit Info</a>         </p>         </div>          <?php } while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); ?> [/code] This is the page receiving the variable, do I have to type something special? cause I cant echo out $sku I want this $sku variable to also retrieve the info for that # and put it into form fields that can be used to update. [code] $query = "INSERT INTO directory SET firstname='$_POST[firstname]', lastname='$_POST[lastname]',title='$_POST[title]',department='$_POST[department]',office='$_POST[office]',phone='$_POST[phone]' WHERE sku='".$sku."'"; [/code]

I want to echo out a page <span class="breaker"></span> after every 12 database results. <?php do { ?> <div class="employee"> <p> <img src="images/<?php echo $row_Recordset1['firstname']; ?>_<?php echo $row_Recordset1['lastname']; ?>.jpg" width="140" height="140" /><br /> <?php echo $row_Recordset1['firstname']; ?><br /> <?php echo $row_Recordset1['lastname']; ?><br /> <?php echo $row_Recordset1['office']; ?><br /> <?php echo $row_Recordset1['phone']; ?> </p> </div> <?php } while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); ?>

Im deriving a list of photos with information from my database. I want to be able to print the webpage off without any of the information being broken up between pages. Since its coming from the database I cant just use page breaks cause then I get one image per page. Is there a solution to this?

Im deriving information from a MySQL database to display photos and information. When I go to print or preview the print task, the images will cut off where the page ends and continue on the next printed page. I end up with some heads at the bottom of one page, and their torso on the top of the next page. I was wondering how I could fix it so that the images do not get cut off when printing. www.dgaa.ca/teamroster.php that is the page I am talking about