Limiting Output

[Sharif]

Okay, first off this is what I am doing: I use links like this:
http://www.domain.com/page.php?pageid=1
to show some pages. page.php contains this example code that uses the "pageid" variable":

<?php
$display_title = mysql_query("SELECT pagetitle FROM legal HERE pageid=$pageid")
or die ("MySQL Error - Please report this to the webmaster");
while ($row = mysql_fetch_array($display_title))
{
$legal = $row["pagetitle"];
echo $legal;
}
?>

Now you can see that this outputs whatever is in the pagetitle field in the "legal" table. In the "legal" table, The highest number of pageid I have is 6. So for the last page, the URL would be:
http://www.domain.com/page.php?pageid=6. I wanted to know how I can prevent the output of a blank page if say someone enters pageid=7 manually in the URL, 7 or whatever number 7 or higher since they don't exist in the database. How can I do this? Thank you.

[ir4z0r]

<?php 
  $display_title = mysql_query("SELECT pagetitle FROM legal WHERE pageid=" . $_GET['pageid'])
or die ("MySQL Error - Please report this to the webmaster"); 
  if ($display_title)
  {
    while ($row = mysql_fetch_array($display_title)) 
    { 
      $legal = $row["pagetitle"]; 
      echo $legal; 
    }
  }
  else
  {
     echo "not available";
  }
?>

if your PHP-Version is rather old, ignore the $_GET stuff and use it like before

[Sharif]

Thank you for the code but can you please explain what you added? I have no clue at all about what you did - newbie brain!

[ir4z0r]

here it comes 😉

if the query doesn't return anything, $display_title will be 0 or false

using mysql_fetch_array() then would cause an error, hence we check
it before:

if ($display_title)
{

is short for:

if ($display_title == true)
{

but if it's empty we echo some message, saying nothing has been found

[Sharif]

Okay I tried your code exactly as you gave it, and I purposely entered a pageid in the URL that doesn't exist (ex: http://domain.com/page.php?pageid=7) and I still just get a blank page instead of a message that says "This page does not exist.". What am I doing wrong?

[Sharif]

I am so proud of myself today, this is my best breakthrough in PHP and MySQL yet - I must be the "Most Developed Newbie" of the month:

Here is the code that perfectly works and fixes my problems:
<?php
$display_title = mysql_query("SELECT pagetitle FROM legal WHERE pageid=$pageid")
or die ("MySQL Error - Please report this to the webmaster");
if ($pageid >6)
{
echo("No Such Page");
}
else
{
while ($row = mysql_fetch_array($display_title))
{
$legal = $row["pagetitle"];
echo $legal;
}
}
?>

Now let me explain this to you younguns. I thought, why not just put my while statement as the FALSE return. That is if the pageid is bigger than 6, it will proceed as normal. Just look and be enflamed in jealousy - I am so proud today - || PAT ON THE BACK ||. Thanks ir4z0r for starting me off though 😉

[ir4z0r]

strange, the other way should have worked, too 😕

but if it works ... thumbs up

just one note: you might want to put the query in the else {}
cause otherways it would query unnessecarily if an invalid id has been entered

[Sharif]

Hey one more thing, say there was no pageid entered in the URL, how can I display the error messange for that too?

[stolzyboy]

if ($pageid >6 || $pageid == "")

[Sharif]

stolozyboy, sorry, that doesn't work. I also tried putting an elseif in there to see maybe if that would help but didn't.

[stolzyboy]

if ($pageid >6 || !$pageid)

[Sharif]

Hey, I tried that too - doesn't work.

[ir4z0r]

try it vice versa then:

if (($pageid < 7) && ($pageid > 0) && is_numeric($pageid))
{

}

[Sharif]

Tried that too, don't know why this isn't working at all...

[ir4z0r]

would you mind posting the whole code and describing what happens [not]?