Ok
problem solved
thank you everybody
----- Original Message -----
From: "Richard Davey" <[EMAIL PROTECTED]>
To: "Dominique ANOKRE" <[EMAIL PROTECTED]>
Cc: "Php List" <[EMAIL PROTECTED]>
Sent: Friday, February 20, 2004 11:35 AM
Subject: Re[2]: [PHP] display a hiddenfield
> Hello Dominique,
>
> Friday, February 20, 2004, 11:29:21 AM, you wrote:
>
> DA> I 've given a value se the code :
> DA> $_POST["hiddenField"] = $image_avant;
> DA> where $image_avant is a variable which contains a value !!
>
> No, you've overwritten whatever might have been in the $_POST array
> with the image_avant value.
>
> Also even if this would work (which it doesn't) you are still
> outputting the HTML to the browser BEFORE you do this anyway.
>
> Try like this (as has already been suggested):
>
> <?
> print("<form method=\"post\" action=\"image.php\">\n");
> print("<input type=\"hidden\" name=\"hiddenField\"
> value=\"$image_avant\">\n");
> print("<input type=\"submit\" name=\"avant\" value=\"Avant\">\n");
> print("</form>\n");
> ?>
>
> --
> Best regards,
> Richard Davey
> http://www.phpcommunity.org/wiki/296.html
>
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