When the file is uploaded to the server (I assume you've got that far), typically you KNOW the name of the file, or have renamed it to a specific file name to suit your methods (otherwise people may upload different pictures called me.gif which overwrite each other).
So you may have a var $uploadedFileName, which is x.gif. Your second page will know this, or may need to be passed the var "file.php?uploadedFileName=<?=$uploadedFileName?>". When the page knows this, you can then echo it in the form with something like: <INPUT type="hidden" name="uploadedFileName" value="<?=$uploadedFileName?>"> Which will mean that the NEXT page knows, etc etc. Easy enough, although hard to explain :) Justin French -------------------- Creative Director http://Indent.com.au -------------------- on 02/05/02 10:29 PM, r ([EMAIL PROTECTED]) wrote: > Greetings All, > Special greeting to all my new pals on the list, you know who you are. > > First let me tell you what i have done then what i need. > > I have made a program to upload a picture (gif or jpeg, for clearness lets > call the picture "x.jpg") and then the page will change and a new page > will > display asking the user/client what kind of a picture was it > (eg: cartoon/family pic/portfolio pic or adult pic) thats it. > > > My doubt is here: > In the second page I want to ask the client if "x.jpg" is the picture > he/she > uploaded. > How do read the filename? I also want his filename in a hidden text field > on > the same page so that i may enter it into the database. > > > Also can anybody recomend a good online manual for PHP, or a downloadable > version with LOTS of GOOD examples. > ( Note : I am running on windows) > > Any help appreciated, > > Cheers, > -Ryan. > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
