sorry the correct code is:

in fooa.php for each  textfield

if (!empty($field1)) $value=$field1;
else $value="";

<input type="text" name="textfield" value="<?echo $value;?>">

so, when you come back from foob.php if a $field !empty the respective
textfiel will have a value in html form;


Best regards,
George Nicolae
IT Manager
___________________  - Professional Web Design

"Simos Varelakis" <[EMAIL PROTECTED]> wrote in message

Here is a problem :-)

A php post form fooa.php
Submit all form fields to a page foob.php
Page foob.php do some mysql db check and in one condition I want to
Return to fooa.php without losing any field value... I made this from
foob.php with
But this is more complex esspecially when in fooa.php there are radio
buttons + checkboxes.

I thing that if I could with Javascript in foob.php create an (invisble)
form and take all posted fields from
Fooa.php And resubmit it with form.submit  method to fooa.php I will
have best results But I donít know how to do this since I am a
begginer.. Anyone can help please send me an email

Thanks in advnace for your time & help and excuse me if is off topic

Best Regads


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