Code to select a form and display it from option

evaoparah

Please I have a drop down of select option to chhose a form from two option but my code never works. I use "get" method and it didplays that select=feedback in the browser but never display the form. Whatr am I doing wron or what am I missing Please help. This is my code:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>

</head>
<body>

<div id="content" align= "center">
<h1>Access Login Form</h1>
Please make your choice from the menu below and fill the form.
</div>

<select name="select" value = "active" >
<option value="feedback5.php" selected>Feedback Form</option>
<option value="help5.php" selected>Help Request Form</option>
</select>

</body>
</html>

stolzyboy

where is your php code to include the form?

evaoparah

stolzyboy wrote:
where is your php code to include the form?

This is the whole code. The idea is for the form to submit the form selected back to the same page in the server for processing. By doing so the selected form will be dynamically be displayed.

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>

</head>
<body>

<div id="content" align= "center">
<h1>Access Login Form</h1>
Please make your choice from the menu below and fill the form.
</div>

<select name="select" value = "active" >
<option value="feedback5.php" selected>Feedback Form</option>
<option value="help5.php" selected>Help Request Form</option>
</select>

</body>
</html>

<?php

if($_SERVER['REQUEST_METHOD']=="GET") {

if($selected(feedback5.php) 
&selected == "feedback5.php";
}
elseif{
selected(help5.php){
$selected == "help.php";
}
$name= $_post['name'];
$email= $_post['email'];
$address= $_post['address'];
$msg= $_post['message'];

printf($name, $email, $address, $msg);
printf $formselected();

}else{
die('nothing was posted') ;
}
?>

rowanparker

Your if statements don't seem normal to me.

if ($selected = "feedback.php")

Piranha

rowanparker wrote:
Your if statements don't seem normal to me.
if ($selected = "feedback.php") 

If the if-statement should be like this you should have 2 = signs, otherwise it will assign "feedback.php" to the variable $selected and then the if-statement would always be true.

I think that evaoparah have to read up a bit on [man]if[/man]-statements and [man]operators[/man]. There are a few strange things in them.

rowanparker

Sorry, yes it would.

Just been at school for hours working with VBA (1 = sign) 🙁

evaoparah

I really need help I have written several codes in php for this but would not work out. I wrote with javascripts and it was ok but I need php

evaoparah

I have written another code but received an error on line 31 saying:

Fatal error: Function name must be a string in C:\Documents and Settings\Administrator\Desktop\Internet-PHP\forms-1-s.php on line 31
line 31 is: $formselect();

<?php
$formselect();
if($_SERVER['REQUEST_METHOD']=="GET") {

if($selected("feedback5.php")) {
feedback5.php == "$selected";
}
else{
echo "nothing is selected";
return false;
}
if($selected("help5.php")) {
help5.php == "$selected";
}
else{
echo "select something";
return false;
}

}else{
die('nothing was posted') ;
}
echo('formselect');
return true;
?>

Weedpacket

So what is $formselect supposed to be? As a variable, if it doesn't contain the name of a function, it's not going to work.

Same with $selected.

help5.php == "$selected";

This is meaningless.

I suggest you go to the PHP manual and go through its introductory [man]tutorial[/man] and find out a little about the language first instead of just making up random stuff without having the faintest idea what you're doing and then (a) expecting it to just magically work and (b) asking us to fix it when it doesn't.