need help making default display only on one page

[CodeMama]

Here is my script, but I want if the page isn't browse.php that no default records display ....please help.

            

                                                       //if the page is browse default to A's if on other pages just show pagination links and not default listings
                                                       if ($_SERVER['SCRIPT_FILENAME'] = "browse.php" ) {
                                                         $selectedLetter = "A";
                                                       }

                                                else {
                                                     $selectedLetter = ""; 
                                                }


					 //Create array with letters AND number sign
					$letters = range('A','Z');
					array_push($letters, 'nums');

					$menu = '';
					$selectedLetter = isset($_GET['letter']) ? $_GET['letter'] : null;
					foreach($letters as $letter) 
					{
						if($letter == $selectedLetter && $selectedLetter != 'nums')
						{
							$menu .= sprintf('%s&nbsp', $letter);
						}
						else if($letter == $selectedLetter && $selectedLetter == 'nums')
						{
							$menu .= sprintf('%s&nbsp', '#');
						}
						else
						{
							if($letter == 'nums')
							{
								$menu .= sprintf('<a href="browseMatt.php?letter=%s">%s</a>&nbsp;', 'nums', '#');
							}
							else
							{
								$menu .= sprintf('<a href="browseMatt.php?letter=%s">%s</a>&nbsp;', $letter, $letter);
							}
						}

					}

					echo "<div align=\"center\"><b>{$menu}</b><br /></div>";







					//Show all restaurants that start with $letter  not between "A" and "Z"
					if ($selectedLetter == "nums") 
					{
						for($i = 0; $i <= 9; $i++)
						{
							$sql = "SELECT DISTINCT ID, name, address FROM restaurants WHERE name LIKE '$i%'";


							$result = mysql_query($sql) or die(mysql_error());

							while($row = mysql_fetch_assoc($result))
							{
								$name = $row['name'];
								 printf('<a href="view.php?ID=%s"><b>%s</b><br />%s<br /><br /></a>', $row['ID'], $row['name'], $row['address']);

							}                    
						}


					}
				    else
				    {
					 	$sql = "SELECT DISTINCT ID, name, address FROM restaurants WHERE name LIKE '$selectedLetter%'";

						$result = mysql_query($sql) or die(mysql_error());

						while($row = mysql_fetch_assoc($result))
						{
							$name = $row['name'];
							 printf('<a href="view.php?ID=%s"><b>%s</b><br />%s<br /><br /></a>', $row['ID'], $row['name'], $row['address']);

						}                          	
				    }

                                                  

[ixalmida]

Change from assignment operator "=" to conditional operator "==". The assignment operator creates a condition that is ALWAYS true unless there is a type mismatch or something.

if ($_SERVER['SCRIPT_FILENAME'] == "browse.php" )

[CodeMama]

thanks but I had tried that too, it doesn't work...

[Bjom]

do a var_dump($_SERVER['SCRIPT_FILENAME']); and see what you get. Always make sure you actually know what is going on.

this:

if ($_SERVER['SCRIPT_FILENAME'] == "browse.php" )

can never be true.

this:

if (basename($_SERVER['SCRIPT_FILENAME']) == "browse.php" )

can.