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 $uploadedFileName, which is x.gif.
Your second page will know this, or may need to be passed the var
When the page knows this, you can then echo it in the form with something
<INPUT type="hidden" name="uploadedFileName" value="<?=$uploadedFileName?>">
Which will mean that the NEXT page knows, etc etc.
Easy enough, although hard to explain :)
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
> 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
> How do read the filename? I also want his filename in a hidden text field
> 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,
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