Getting array values into a form

[lerro]

How do i get an array of data retrieved from the database to display its output on the screen within a form so that a user can select an option, which can then be passed back in a $_POST array and processed again for storage in the database?

The function that retrieves data for display looks like this

function display_output()
{

	$output = mysql_query("SELECT field1, field2, field3, FROM tableName") or die(mysql_error()); 



	Print "<table border >";

	Print "<tr>";		
	Print "<th>Field1:</th>"; 
	Print "<th>Field2:</th>"; 
	Print "<th>Field3:</th>"; 
	Print "</tr>";

	while($info = mysql_fetch_array($output)) 
	{ 
	Print "<tr>"; 
	Print "<td>".$info['field1'] . "</td> "; 
	Print "<td>".$info['field2'] . "</td> ";
	Print "<td>".$info['field3'] . "</td> ";		
                    Print "</tr>";

	} 

	Print "</table>"; 

}

The form selections for each line of the array that need to be output with the function above are

<label>
<input type="radio" name="Choice" value="yes" />
Approved</label>
<br />
<label>
<input type="radio" name="Choice" value="no" />
Declined</label>

What the user should be able to see is on each line is first the option to select followed by the array information.
Then when the submit buttton is selected, this information is sent through the $_POST array

[anakadote]

If I understand correctly, then you'll also need to get the row's primary key from the database so that you can update that specific row:

$output = mysql_query("SELECT primaryKey, field1, field2, field3, FROM tableName") or die(mysql_error());

and then:

while($info = mysql_fetch_array($output))
{
Print "<tr>";
Print "<td>".$info['field1'] . "</td> ";
Print "<td>".$info['field2'] . "</td> ";
Print "<td>".$info['field3'] . "</td> ";
Print '<td><label><input type="radio" name="Choice_'.$info['primaryKey'].'" value="1" />Approved</label>
<br />
<label><input type="radio" name="Choice_'.$info['primaryKey'].'" value="0" />Declined</label></td>';
Print "</tr>";

}

And then, when you submit the form, you loop back through the results to find the selected value:

while($info = mysql_fetch_array($output))
{
if($_POST['Choice_'.$info['primaryKey']] == 1){
// approved
} else {
// declined
}
}

[lerro]

Thanks

How would i display the form tag as part of the print output? as i have to show the opening and closing form tag as well as the submit button so as to get the data passed back as a "POST" to be read.

I tried

Print "<form action="<?php echo $PHP_SELF;?>" method="POST" >";

but there seems to be an error as it's a php tag embedded in another php tag
and is this the correct way to print?

[anakadote]

I'd do it like:

echo '<form action="' . $_SERVER['PHP_SELF'] . '" method="POST">';

[lerro]

I used

$_SERVER['PHP_SELF']

and it worked well.

My challenge now is when i try to enter options in the "if" statement. I ran it but no update is done in the database. This is the example.

<?php
while($info = mysql_fetch_array($output))
{
if($POST['Choice'.$info['app_no']] == 1){

$sql = "UPDATE tableName
SET field1 = 'NewChoice' WHERE primaryKey = '".$info['primaryKey']."' ";
mysql_query($sql);
} else {

}
}

?>