Forms

juliardiaz

I have to php files testA.php and testB.php
The testA.php contains form that will be submitted to testB.php
on the testB.php will have evaluation to the value of the submitted variables.
if it is not valid, i want it back to testA.php with the warning error.
if it is valid, will print "OK"
currently these are the codes... pls help me examine them, because they don't turn out right.

testA.php
<?
if ( $errCode ) {
print "Error";
}
?><!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML>
<HEAD>
<TITLE> New Document </TITLE>
<META NAME="Generator" CONTENT="EditPlus">
<META NAME="Author" CONTENT="">
<META NAME="Keywords" CONTENT="">
<META NAME="Description" CONTENT="">
</HEAD>

<BODY>

==================================================================

testB.php

<?
if (!$isi1 || !$isi2) {
header ("Location: TestA.php?errCode=1");
exit;
}
else {
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML>
<HEAD>
<TITLE> New Document </TITLE>
<META NAME="Generator" CONTENT="EditPlus">
<META NAME="Author" CONTENT="">
<META NAME="Keywords" CONTENT="">
<META NAME="Description" CONTENT="">
</HEAD>

<BODY>
<?
print "OK";
}

?>
</BODY>
</HTML>

Guys, please help me out ASAP.
Any help will be appreciated....
thank u very much

cahva

Change: if (!$isi1 || !$isi2) {

..to: if (!$POST['isi1'] || !$POST['isi2']) {

Should do the trick...

klansman

yes, that's it. you probably have register_variables = off and $isi1 does not exist.

[]

juliardiaz

Then it's solved.
Thanks guys
but, I still have... another problems...

I have this problem... it's enhancement from the first problem (the first on already solved)
so the enhancement are showing the old form (with the values already filled in before the submit button pressed)...
in the first one, if it is already submitted, and found some errors on the form/values on the form, it should be returned to testA.php. (returning : the form and the error message)
and what i want to have now is testA.php containing : the form, values filled in the form and the error message.
can somebody give me a clue ... how to do it? (on php or javascript maybe..)

klansman

that's easy... you have to create a form in formB.php with hidden-type inputs back to formA.php. now, i don't know if you will need a submit button or not.

If you do, that's not really a problem: you can show an error message in formB.php with the submit button saying: "go back" or something like that.

If you don't, just make the form and find a way to refresh the page back to formA.php... i'm not quite sure on how that is done...

[]

juliardiaz

Originally posted by klansman
that's easy... you have to create a form in formB.php with hidden-type inputs back to formA.php. now, i don't know if you will need a submit button or not.

hi klansman
thanks for ur comments..
but i think that if we're using hidden type..
it will be many... (well.. the real form consists of many fields)

If you do, that's not really a problem: you can show an error

i guess... there's somthin.. like in javascript.. we can use onclick... u can do the examination of all fields.. but i don't know how...

cahva

You dont have to write every posted variable. Just fetch them from $_POST array.

<form name='form1' action='testA.php' method='post'>
<?php
while(list($key,$val)=each($_POST))
{
echo "<input type='hidden' name='$key' value='$val'>";
}
?>
<a onClick="document.form1.submit()">Error. Go back to form</a>
</form>

klansman

hello again!

i forgot to explain... mine is a different approach... i keep things in the same script, so that when you click the submit button, the script get's reloaded (but heading in a different direction - think of a flowchart) and the form variables are checked. The form code lies in a function and that function can take as arguments the values of the fields...

If the form is correctly filled, some action or decision is made. If the form is incorrectly filled, there is an error message, and the form display function gets called again with the arguments being the fields correctly filled in the previous form.

i hope you understand. english is not my mother-tongue.

[]

juliardiaz

Originally posted by cahva
You dont have to write every posted variable. Just fetch them from $_POST array.
<form name='form1' action='testA.php' method='post'>
<?php
while(list($key,$val)=each($_POST))
{
echo "<input type='hidden' name='$key' value='$val'>";
}
?>
<a onClick="document.form1.submit()">Error. Go back to form</a>
</form>
[/B]

Umm.. Okay.. I'll try that. Thanks cahva..