phrozenflame

Yeah I aggree with emehrky it needs a more better design and you need to make it look more appealing to other viewers mabe a bit more colour and images.

My little application is a simple file upload system. There is only ever going to be one user who will upload, and one user who will download the file. There is also an archive of current uploaded files for the download person, and a separate archive page for the uploader which is the same, but includes a delete function to delete the db record and also to delete the actual file. When a file is uploaded, an email is sent to the person who will download the file when he is ready, by visiting the archive page and choosing which file he wishes to dowload and clicking the 'download' link. So far so good. What I want to do now is to have a 'Yes'/'No' next to the archive of each file which will let the user know whether the file has been downloaded or not. The question is.....how do I detect if the file has been downloaded or not? Thanks for any suggestions

Oh yes thanks

Thanks it helps

I have been a long time user with using Dynamic SSI Iframes, but I am looking for an easier way to make a layout work. How can I make a layout with ease without having todo php includes? I want it to function like an iframe, but not actually be one

Thanks for reply

Yes thanks that works

I tried this by going onto my profile and clicking show last post's from this user and it hasent got all of mine only some.

I know this is easy question but i haven't tried this before... I want to add addtional features to my website counter... Now, i want to put a condition to get the I.P. address of the visitor... Just a simple snippet codes will do... no problem with the counter stuffs. i just want to... record the I.P. address of the visitor... Thank you very much in advance...

Hi, I have an excel file with macro and was wondering if there is a way to open this excel file from PHP and run the macro. The short  cut key to run macro is CTRL+SHIFT+A and is it als possible to send shortcut key in PHP? Any suggestion/input is much appreciated. thanks in advance,

Its not even working

You might need to CHmod your folder permissions

Sessions

i am working on designing a webpage. I give a list of options for the user to choose from and then once the user has choosen what they want the products avaliable on the option are displayed. The code i have used is: foreach($food_categories  as $key => $subarray)        {     echo "<h3>$key</h3>";     echo "<ul>";     foreach($subarray as $type)     {       echo "<input type='radio' name='interest'                     value='$type'>$type \n";     }     echo "</ul>"; } Instead of having the list like this i want the user to be able to choose from a set of pictures. I have stored pictures in my database under the heading "pic". Now how do i change my code to enable me to display pictures which the user can select rather than a list??? Thanks

Thanks

Um no

I am currently doing a download page on my site. In the admin control area when a new file is submitted the script automatically should obtain the size of the file. now before we say use filesize() it wont work .. this is because the file may or maynot be on the same server as the php script. filesize needs a realpath ( c:\dir\file.zip ) as it doesn't work with a url ( [a href=\"http://www.site.com/file.zip\" target=\"_blank\"]http://www.site.com/file.zip[/a] ). How else can i obtain the file size?

Im tring to create the second of two pages. The first page has a table that displays a row number, a count, and an error code. I want the next page to create a table by clicking on the the error code or row number. The table is suppost to diaplay the error code and all case numbers for that code. Here is an array with dummy data. In this array the counts are 3,6,10,12, the error codes are 4,5,6,7, and the case numbers are 31,41,51 for the top row, 32,42,52 for the second row and ect. The code below creates the table on the first page. Please help me create the second page. Thanks $errcount=array(array(3,array(array(31,4),array(41,4),array(51,4))),                 array(6,array(array(32,5),array(42,5),array(52,5))),   array(10,array(array(33,6),array(43,6),array(53,6))),                 array(13,array(array(34,7),array(44,7),array(54,7)))) <!Set up table headings>      <html> <table border="0" cellpadding="3"> <tr bgcolor=#CCCCCC> <th> </th> <th> # Of Events </th> <th> Error Code </th> </tr> <!setup table> <? $i=0;?> <? foreach($errcount as $C){ ?>         <tr bgcolor= <?if($i%2 == 0) {   echo "#EEEEEE";       }   else {       echo "#CCCCCC";       }?>>         <td>         <?echo $i++; ?>         </td>         <td>         <? echo $C[0]; ?>         </td>         <td>         <? echo $C[1][0][1]; ?>         </td>         </tr> <? } ?> </table> </html>

Thanks again

Thank you

Here's the error I'm getting: 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 'Group = 'KK'' at line 1 The variable $group is being stored in the url as a link from a previous page: [a href=\"http://ecolab.7cpco.com/pricegroups.php?group=KK\" target=\"_blank\"]http://ecolab.7cpco.com/pricegroups.php?group=KK[/a] Here's the php code that's causing the error: $query = "SELECT * FROM item_list_price_groups where Group = '$group'"; I have other pages that work just the same as this is intended to (result based on url-stored variable). I pretty much copied the code verbatim from another application I wrote that works fine. I don't understand what could be wrong with this query. Thanks for your help

Ok thanks for the links

this is my first time to use cron jobs, so i have 3 questions: How do i set a cron job using php? How do i update a cron job using php? How do i delete a cron job using php? Thanks in advance!

Ok I will thanks

Thanks