Why is my php form not generating values? - php

Question

How do you create a form in PHP where the results entered in from the user will be generated in a table? When I test the pages, I only get the start of the table, key and results(values) fields but no values!
Here is my code (I'm shortening the number of fields to two since forms can be very long:
Order form:
<?php
foreach ($_POST as $key => $entry)
{
$totalfields++;
echo "<tr>\n";
echo "<td>" . $key . "</td>";
if (is_array ($entry)) {
$count = count($entry);
echo "<td>";
for ($i=0; $i<$count; $i++){
echo $entry[$i] . "<br />";
}
echo "</td>";
} else
{
echo "<td>$entry</td>";
}
echo "<tr>\n";
}
?>
<form method="post" action="address to the results form page">
<label for="fname">First Name:</label>
<input type="text" name="fname" id="fname"/><br/>
<br/>
<label for="lname">Last Name:</label>
<input type="text" name="lname" id="lname"/><br/>
<br/>
<p><input type="submit" value="Submit"/></p>
</form>
Results form page:
<div id="results table">
<table border='1'>
<tr>
<th>
Field
</th>
<th>
Results
</th>
</tr>
</table>
</div>

Answer 1

This actually posts fine for me (I used PHP_SELF for the form action). I would however suggest that you use $_SERVER[REQUEST_METHOD] to check for POST:
if( $_SERVER[REQUEST_METHOD] == 'POST' ) { ...
In addition, be explicit about calling the posted variables ($_POST[fname] & $_POST[lnane])

Answer 2

Your PHP generating the table should be in the resulting page, not in the form page. Or else, you can point the action of the form to the same page, to get the results posted and processed on the same page.
EDIT: What I mean is this: PHP — as you may know — is processed server-side. It means that your browser sends a request to the server, the server processes the PHP file and then it serve the client the processed version.
The problem in your code is that you have the PHP script processing the $_POST on the page in which the user is supposed to fill up the form. When a user fills up the form and hit submit, he will get directed to the action page, where the $_POST will be available and processed. Your misunderstanding is in the fact that you expect the $_POST to be available on the page where it's posted, and then to send it automagically to the result page. This is not what happens because the server doesn't have an opportunity to catch the information you sent on the same page, because it has already been served…
…unless you make the form action direct to the same page. Then the server catches the information and processes it.
You could go solving this problems in two ways really.
Point your action to the result page as you are doing now, but move your processing code to the resulting page.
Point your action to the same page.
You may even want to do something fancy, like check if there is $_POST information and, if there isn't, show a form. If there is, show the result. This is not at all difficult to implement as it follows:
<?php
if( $_POST ): ?>
<div id="results table">
<table border='1'>
<thead>
<tr>
<th>
Field
</th>
<th>
Results
</th>
</tr>
</thead>
<tbody>
<?php
foreach ($_POST as $key => $entry)
{
echo "<tr>\n";
echo "<td>" . $key . "</td>";
if (is_array ($entry)) {
$count = count($entry);
echo "<td>";
for ($i=0; $i<$count; $i++){
echo $entry[$i] . "<br />";
}
echo "</td>";
} else
{
echo "<td>$entry</td>";
}
echo "<tr>\n";
}
?>
</tbody>
</table>
</div>
<?php else: ?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<label for="fname">First Name:</label>
<input type="text" name="fname" id="fname"/><br/>
<br/>
<label for="lname">Last Name:</label>
<input type="text" name="lname" id="lname"/><br/>
<br/>
<p><input type="submit" value="Submit"/></p>
</form>
<?php endif; ?>
I made a working example for you. Just go to the link and click the Run [F9] button over the top to see it in action:
Working example