A simple php question

[tonyalepski]

Hi,

I am trying to convert a php function into perl and need to understand the following line in red :

 

 

   function pagination($query, $per_page = 10,$page = 1, $url = '?'){       
     $query = "SELECT COUNT(*) as `num` FROM {$query}";
     $row = mysql_fetch_array(mysql_query($query));
     $total = $row['num'];
        $adjacents = "2";

     $page = ($page == 0 ? 1 : $page); 
     $start = ($page - 1) * $per_page;        
  
     $prev = $page - 1;       
     $next = $page + 1;
        $lastpage = ceil($total/$per_page);
     $lpm1 = $lastpage - 1;
     
     $pagination = "";
     if($lastpage > 1)
     { 

.........................................

 

 

What will be the output of this line above in red.

 

Thank you very much.

Regards.

Tonya

[Jacques1]

It's the ternary operator. Perl has it as well.

 

The expression $c ? $r_1 : $r_2 yields the value of $r_1 if $c evaluates to true, otherwise it yields the value of $r_2.

[tonyalepski]

Thank you Jacques,

 

Another one :

 

for ($counter = 1; $counter <= $lastpage; $counter++);

 

what will be equivalent  for it in perl as I am getting errors on it

 

Many thanks !!

[Jacques1]

What errors? This is a plain for loop which is one of the basic features of virtually every language.

 

No offense, but if you know neither PHP nor Perl nor the basics of programming, this task will be tough.

[ginerjm]

What errors are you getting on that line?  BTW - there is a semi at the end of the line - doesn't belong there.

[davidannis]

An example might look like this:

$lastpage=3;
for ($counter = 1; $counter <= $lastpage; $counter++){
    echo "We're on page $counter <br>";
}
echo "done";

which would output

We're on page 1
We're on page 2
We're on page 3
done

Edited August 25, 2014 by davidannis