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Added new table

php, mysql, columns

   
9:37 pm on Aug 24, 2012 (gmt 0)



Hi there,
First - this isn't my first time to this site, but I was previously signed up with a gmail account, so... here I am looking like a newb!

Okay, I've had to add a new column to the database table that controls our directory so that I can control how certain names appear within their departments.

I did not set this up originally, I took the site over 4 years after it's inception and have never had to change this until now.

I see nothing in the script that will let me control ordering by this new column, though it appears the directory is currently ordered by the auto-incremented id for each entry.

I believe the following is the code controlling what we're currently seeing:

<?php print $row['category']; ?></strong></td>
<td align="center"><?php print $row['main_line']; ?></td>
<td></td>
</tr>

<?php
$resultl = @mysql_query("select first_name, last_name, phone, email, category from directory, categories where department = categories.category_id and department= " . $row['category_id']);

while ($rowl = mysql_fetch_array($resultl)) {

?>
<tr>
<td width="340" valign="top" class="bodytext"><?php print $rowl['first_name']; ?> <?php print $rowl['last_name']; ?></td>
<td width="162" valign="top" class="bodytext" align="center"><?php print $rowl['phone']; ?></td>
<td width="162" valign="top" class="bodytext" align="center"><a href="#" onclick='<?php print despam($rowl['email']); ?>'><?php print $rowl['email']; ?></a></td>
</tr>
<?php
}
}
?>


can anyone please tell me how to add and order by the column "order"?
I'm good at what I do, but a lot of this was in place before I took over so I'm not what you'd call an expert at php - I'm trying to get there though!

Thanks in advance,
Sherry
11:38 pm on Aug 24, 2012 (gmt 0)

5+ Year Member



Assuming the "order" column is part of this same table, decide if you want to list in ascending(asc) or descending(desc) order and simply add 'ORDER BY order ASC'
$resultl = @mysql_query("select first_name, last_name, phone, email, category from directory, categories where department = categories.category_id and department= " . $row['category_id']."ORDER BY order ASC); //or DESC sort


Hope this helps
4:49 am on Aug 26, 2012 (gmt 0)



I appreciate your response, I haven't had time to get to this until today so I apologize for getting to this so late.

I've tried moving the code around and use your suggestions but Dreamweaver keeps producing errors when I change it. I'm not really sure how it's happening; any ideas?

[i49.tinypic.com...] here is the error that DW is producing...

Thanks again, and in advance :)
Sherry
2:36 pm on Aug 26, 2012 (gmt 0)

5+ Year Member



Hi. Looks like you're missing a " after ASC on line 619. I missed it too while quickly typing that code in I gave you 0_o. Oops.

It should look like this ".$row['category_id']." ORDER BY order ASC");

Hope that helps.
9:56 pm on Aug 27, 2012 (gmt 0)



I feel like a nuisance, but that still isn't working - when I put it in (took it out so I didn't get yelled at, lol) it gave header errors before populating some of the rest of the page, sorry, any ideas?
12:06 am on Aug 28, 2012 (gmt 0)

5+ Year Member



Hmmmm. What errors did you get?
12:16 am on Aug 28, 2012 (gmt 0)

5+ Year Member



Again. This should be working code now that I corrected the first error. Tell me what errors you see after this is implemented.
$resultl = @mysql_query("select first_name, last_name, phone, email, category from directory, categories where department = categories.category_id and department= " . $row['category_id'] . " ORDER BY order ASC"); 
7:20 am on Aug 28, 2012 (gmt 0)



Wish I'd seen your response sooner, I was away from the computer trying to pretend I have a life outside of this box ;)

The error that populates above the table that's supposed to be showing the listing is:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/sbloom02/public_html/contact/directory_by_dept.php on line 621

no fewer than 20 times before it populates -- I'd give you the link but I have to keep taking the code out so none of the membership sees the errors. I guess I should have done this on a dev page but really didn't expect it to be this frustrating!

Thanks so much,
Sherry
12:53 pm on Aug 28, 2012 (gmt 0)

5+ Year Member



Yes. I would save a test.php page and run the same script. But echo the $result1 var to see if it's valid. Add echo $result1 just below line 619 and see what it says. It should show the SELECT statement.