Maq

That seems to have done the trick. Thanks! My blog is now working - http://thurfjell.nu/links/ Cool You know, we have a Website Critique section in case you wanted people to check out your site: http://www.phpfreaks.com/forums/website-critique/

This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=323958.0

(moving to MySQL section for now)

This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=323935.0

Sounds like the problem has to do with the cardinality of your query results. Can you show us your table structure?

Where do you store the user data whether they're female or male? Sessions? And what column from your table determines whether the picture is male or female?

(Please use tags next time)

Please do not double post. I'll assume it was an honest mistake because they were only apart by a few seconds.

This topic has been moved to PHP Coding Help. http://www.phpfreaks.com/forums/index.php?topic=323922.0

What's the error?

I tend to like to keep some sort of structure, so I would tend to do something like: echo ""; echo ""; echo "{$row['customers_firstname']} {$row['customers_lastname']}"; echo ""; echo "" . date('M. d, Y', strtotime($row['date'])) . ""; echo "";

You can check by changing this line to: mysql_query("UPDATE LOW_PRIORITY jos_vm_product SET product_in_stock = product_in_stock + $product_quantity WHERE product_sku = $order_item_sku AND STR_TO_DATE($order_item_sku, '%Y-%d-%m') >= CURDATE()") or die(mysql_error()); *Note: This should be temporary. Errors and exceptions should be handled appropriately, read: http://www.phpfreaks.com/blog/or-die-must-die

Show us the troubled code where you are trying to send more than 2 variables.

I'm talking about these lines: mysql_query("UPDATE LOW_PRIORITY jos_vm_product SET product_in_stock = product_in_stock + $product_quantity WHERE product_sku = $order_item_sku AND STR_TO_DATE($order_item_sku, '%Y-%d-%m') $quantity_query = "SELECT * FROM jos_vm_order_item where order_id = $order_id";

Are you getting to your else statement where the INSERT is being performed?

You have 2 fatal errors. You're missing the terminating double quote and the semi colon for your UPDATE query. The negative 1 is explained in the manual:

The same way you pulled 'customers_firstname' and 'customers_lastname'.

Please use tags in the future please.

I don't see anything that relates to creating this URL. You have an example of exactly what the URL should look like, echo out the 'a' tag. Look at the structure of the 'a' element: http://www.w3schools.com/tags/tag_a.asp

What's the error?

You need to either change all your inner double quotes to singles or escape them.

Is there an error? What is happening, or not happening?

If the first query is the only one working, then I would assume the other queries are erroneous but being suppressed by the '@' symbol. Replace that line with: $result = mysql_query($query) or die(mysql_error()); // run the query

Define 'broke'.

First of all, that's not PHP code. Second, what do you mean by: