padams

Brilliant, solved it. Thanks, I really should have thought of that myself...

I need to display the results of a query twice on one page and have run into a problem. I have used mysql_data_seek at the end of the first do while loop, setting the internal pointer back to 0, but when I echo out the results a second time I get a blank record at the top. You can see the results on a test page at http://elsombrero.stacnz.com/testing.php, the code is: $courses_sql = "SELECT * FROM courses"; $courses_query = mysql_query($courses_sql) or die(mysql_error()); $rsCourses = mysql_fetch_assoc($courses_query); <ul><?php do { ?> <li><a href="menu.php?courseID=<?php echo $rsCourses['courseID']; ?>"> <?php echo $rsCourses['cName']; ?> </a></li> <?php } while ($rsCourses = mysql_fetch_assoc($courses_query)); mysql_data_seek($courses_query, 0); ?> <?php do { ?> <li><a href="menu.php?courseID=<?php echo $rsCourses['courseID']; ?>"> <?php echo $rsCourses['cName']; ?> </a></li> <?php } while ($rsCourses = mysql_fetch_assoc($courses_query)); mysql_data_seek($courses_query, 0); ?></ul> </html>

Great, the inner join worked a treat. Thanks for the help. SELECT p.`playerID`, p.`playerFirstName`, p.`playerLastName`, IFNULL(g.gamescount,0) as gamescount, IFNULL(t.trycount,0) as trycount FROM players p INNER JOIN (SELECT playerID, COUNT(*) as gamescount FROM games GROUP BY playerID) as g USING (playerID) LEFT JOIN (SELECT playerID, COUNT(*) as trycount FROM tries GROUP BY playerID) as t USING (playerID) ORDER BY gamescount DESC

That shouldn't be a problem, as all we're after are summative statistics for each team in each season/term. I've had a play with that table structure and it all seems to be working fine, and in fact it has helped me tidy up a few other things that were bugging me. Very big help, thanks so much!

Yes. It's not common but has been known to happen.

I'm designing a site for my school sports program and the situation I am struggling with is that students can play for different teams each year (for example they could progress from a B team to an A team), and can play for the same team for several years. I need to be able to generate team lists and season statistics, and I can't work out how to organise my data so that I can select a team from a specific season, and have only those students who actually played for it returned from the query. Has anyone got an idea how to organise the information?

Result was another error: #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '* ) FROM games g WHERE g . playerID = p . playerID ) as gamescount , ( SELECT CO' at line 1

Tried: SELECT p.playerID, p.playerFirstName, p.playerLastName, (SELECT COUNT(*) FROM games g WHERE g.playerID = p.playerID) gamescount, (SELECT COUNT(*) FROM tries t WHERE t.playerID = p.playerID) trycount FROM players p WHERE gamescount > 0 ORDER BY gamescount DESC; Got another error: #1054 - Unknown column 'gamescount' in 'where clause'

Tried that and got the following error: #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '*) FROM games g WHERE g.playerID = p.playerID) gamescount, (SELECT COUNT' at line 2

I'm running a query that combines results from three tables, including counting and grouping data. It has player details, tries scored and games played. Currently it returns the player names even if they have zero games played. Is it possible to modify the query so that if a player has zero games they do not appear in the results? I'm using a left join, but I know that includes results even if there is nothing in one of the columns. Can I get around the problem by using a different join type? I've tried reading up on them and played around with some but can't get the result I want. SELECT p.`playerID`, p.`playerFirstName`, p.`playerLastName`, IFNULL(g.gamescount,0) as gamescount, IFNULL(t.trycount,0) as trycount FROM players p LEFT JOIN (SELECT playerID, COUNT(*) as gamescount FROM games GROUP BY playerID) as g USING (playerID) LEFT JOIN (SELECT playerID, COUNT(*) as trycount FROM tries GROUP BY playerID) as t USING (playerID) ORDER BY gamescount DESC

But what about the style information? The quantity shouldn't increase if the style is different, it should create a new session variable.

I'm just not sure how to best store the information. I've got an array called $_SESSION['cart'][$productID], but whenever a new item is added we need to check if the product has already been purchased so we can increment quantity by 1. That said, the style is a complicating matter as if the product has already been ordered but it is a different style, then it should become a new item in the shopping cart. I just don't know how to best store the information, any ideas?

The other approach I considered was changing the permissions when I created the folder. So I used mkdir("flashgallery/$galleryName", 0777); However, the permissions reverted back to 755 as soon as the folder was created. Is there any way around this?

I've got multiple session variables called $_SESSION['cart'][$productID], with data in each called 'stylename', 'quantity' and 'price'. For example, there will be $_SESSION['cart'][1], $_SESSION['cart'][2], etc. Is there any way to combine data from each into one variable? So I could have "productID1, stylename1, quantity1 and product2, stylename2, quantity2", etc.

Not sure about the ob_flush stuff. I needed it in an old site and have been reusing a bit of code. Once removed it worked fine. Thanks.

Tried this: $gallery="gallery".$_POST['galleryID']; $target_path = "$gallery/"; chmod("/$gallery", 777); Where $gallery will be something like gallery1 or gallery2, etc. Got this error though: Warning: chmod() [function.chmod]: No such file or directory in... The gallery definitely exists, so I think the problem is having a variable in the middle of the chmod parameters. How would I fix this?

Is there any way to temporarily change the permissions on a newly-created folder on a server? I've created a series of folders for photo galleries, but when I want to upload the files it tells me I can't. Warning: move_uploaded_file(gallery1/telusawallpaper.jpg) [function.move-uploaded-file]: failed to open stream: No such file or directory in ... I know I could go onto the server and change the permissions manually, but ideally I would like the user to be able to upload photos themselves. Current code is: $gallery="gallery".$_POST['galleryID']; $target_path = "$gallery/"; $photo = $_FILES['uploadedfile']['name']; $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) { $result = "Photo has been successfully uploaded into the $gallery folder"; } else { $result = "There has been an error. Try again later"; }

Having a small problem with the headers command. I thought this would work okay, but I get the error: Warning: Cannot modify header information - headers already sent by (output started at /home/leotajs/public_html/createopponent.php:6). The dbconnect includes file just has the database connection stuff. ob_start(); include("includes/dbconnect.php"); session_start(); ob_flush(); if(isset($_SESSION['admin'])) { $teams_sql = "SELECT * FROM teams"; $teams_query = mysql_query($teams_sql) or die(mysql_error()); $rsTeams = mysql_fetch_assoc($teams_query); $galleries = "SELECT * FROM gallery"; $galleries_query = mysql_query($galleries) or die(mysql_error()); $rsGalleries = mysql_fetch_assoc($galleries_query); } else { header("Location: admin.php");}

I need to upload two files, a main image and a thumbnail, and would like to use the same form. I'm using the code below, but for some reason when it runs I get a strange issue. The first file uploads fine, but the second one ends up with the first file name appended on the start of its filename. For example, I uploaded two files, mainpic.jpg and thumbnail.jpg, and the resulting filenames on the server are mainpic.jpg and mainpic.jpgthumbnail.jpg. How do I stop this happening? $target_path = "Wallpapers/"; $name = $_POST['name']; $wallpaper = $_FILES['uploadedfile']['name']; /* Add the original filename to our target path. Result is "uploads/filename.extension" */ $target_path = $target_path . basename( $_FILES['uploadedfile']['name']); if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) { $result = "Wallpaper has been successfully uploaded"; } else { $result = "There has been an error. Try again later"; } $thumbnail = $_FILES['uploadedthumbnail']['name']; $target_path = $target_path . basename( $_FILES['uploadedthumbnail']['name']); if(move_uploaded_file($_FILES['uploadedthumbnail']['tmp_name'], $target_path)) { $thumbnailresult = "Thumbnail has been successfully uploaded"; } else { $thumbnailresult = "There has been an error. Try again later"; }

I am trying to get my shopping cart system working for a website that sells linen. The product pages have two dropdown boxes, - the client can select a product and a style that the product comes in. At the moment, the product page is pointing to an addcart page, which works fine - it is bringing the productID, productname, styleID and the price through from my database. But when you click "checkout" and continue through to 'viewcart.php' the price and product name work fine, but the style is not continued through. Here is one of the product pages: www.humphreyb.stacnz.com/pages/bedlinen.php Here are the queries from the addcart.php page: if (isset ($_GET['productID'])) { // Check for a print ID. $styleID = (int) $_GET['styleID']; $stylequery = "SELECT style FROM styles WHERE styleID=$styleID"; $stylename=mysql_query($stylequery); $style=mysql_fetch_assoc($stylename); $productID = (int) $_GET['productID']; $productquery = "SELECT name FROM products WHERE productID = $productID"; $productname = mysql_query ($productquery); if ($productname) { while ($row = mysql_fetch_array ($productname, MYSQL_ASSOC)) { $name = $row['name']." (".$style['style'].")"; }} And here is the relevant code from the viewcart.php page: //Get info for the contents session variable $_SESSION['names']{$row['productID']} = $row['name']; $_SESSION['styles']{$row['styleID']} = $row['stylename']; // Calculate the total and sub-totals. $subtotal = $_SESSION['cart'][$row['productID']]['quantity'] * $_SESSION['cart'][$row['productID']]['price']; $total += $subtotal; // Print the row. echo "<tr> <td align=\"left\" class=\"text\">{$row['name']} ({$row['stylename']})</td> <td align=\"right\" class=\"text\">\$" . number_format ($_SESSION['cart'][$row['productID']]['price'], 2) . "</td> <td align=\"center\" class=\"text\"><input type=\"text\" size=\"3\" name=\"qty[{$row['productID']}]\" value=\"{$_SESSION['cart'][$row['productID']]['quantity']}\" /></td> <td align=\"right\" class=\"text\">$" . number_format ($subtotal, 2) . "</td> </tr>\n"; } // End of the WHILE loop. The assignment is due tomorrow and everything has to be 100% working! Appreciate all the help.

The mysql_data_seek() was lazy coding - we had copied and pasted something from earlier. Gone now! There are four dropdown menus, each populated by different queries, hitting the same table (styles) but querying using different criteria. The first one seems to work in Dreamweaver Live, but the three after do not. To complicate matters, there is this empty query error when we run it live on the web server.

Keep getting the error "Query was empty". It sounds obvious what the problem is, but when I run any of the queries on the page by themselves in phpMyAdmin I get the results back as I would expect. There is no guidance as to which one of the many queries could be causing the problem. Also, when I run it in Dreamweaver Live it appears to work up to a point, but leaves several dropdown menus blank (which are populated by the results of a query). Here is a sample of the code I have been running: $silkrobestyle = "SELECT * FROM styles WHERE categoryID=13 ORDER BY styleID"; $silkrobestyle_query = mysql_query($silkrobestyle) or die(mysql_error()); $rsSilkrobestyle = mysql_fetch_assoc($silkrobestyle_query); The dropdown menu code is: <select name="style" class="dropdownboxstyles" id="style"> <?php while ($rsSilkrobestyle = mysql_fetch_assoc($silkrobestyle_query)) { ?> <option value="<?php echo $rsSilkrobestyle['styleID']; ?>"><?php echo $rsSilkrobestyle['style']; ?></option> <?php } ?> <?php mysql_data_seek($silkrobestyle_query, 0); ?> </select>

Would that put all three bits of data into the one field? I'm not sure how to then query the database to get the style information as well as the product info. The query I posted seems to get the product information I need, but how would I include something that grabs the style that the shopper chose also? The styles table has two fields: styleID and style (which is the name of the style).

Is it possible to carry three bits of information in one session array? I'm working on a shopping cart for a store that sells linen. Each product has a variety of styles, so when I take the data to the checkout page I need the session to contain the product ID, style ID, and quantity ordered. Is this possible? Once in the checkout page I'd need to iterate through the session array and query the database, returning the product name, style name, quantity, and also the price of the product. Fortunately the style does not affect the price! Session is called $_SESSION['cart'][$productID][$styleID]['quantity'] I'm currently using this code to get the product name from the database: $query = "SELECT productID, name FROM products WHERE productID IN ("; foreach ($_SESSION['cart'] as $pid => $value) { $query .= $pid . ','; } $query = substr ($query, 0, -1) . ') ORDER BY name ASC'; $result = mysql_query ($query); Is there any way I can add in a bit of code that also picks up the style name from the styles table, based on the $styleID in the session?

Thanks, that works fine now. I don't quite understand why there would be a difference. I know that do while loops always execute at least once, but I thought given the syntax it would still work the same as a simple while loop. What's the reason for the difference?