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Need help to improve pagination

by terrykhatri (Acolyte)
on Jul 21, 2014 at 15:14 UTC ( #1094521=perlquestion: print w/replies, xml ) Need Help??
terrykhatri has asked for the wisdom of the Perl Monks concerning the following question:

Hi Poj,

I have created pagination using the following code it works but does not look graceful as it displays links for all 103 pages what I need help for is to make it compact like : [First][Prev][1][2][3][4][5][6]...[103][Next][last], I have looked at perl pagination modules @ cpan but there aren't enough examples for a guy of my level to use them

# Count how many rows are there in a table, so that we can use it for +$pagenum. my $sql = qq!SELECT COUNT(*) from "Orders" !; my $sth = $dbh->prepare ("$sql"); $sth->execute() || quit(); my @row = $sth->fetchrow_array; $sth->finish; # Setting offset, limit and page number my $offset = 0; my $limit = 8; my $pagenum = ceil($row[0]/$limit); # Assigning value to $offset as 0 or whatever will be the $pagenum i.e +. 1 or 2 .... $offset=param('page')? $limit*param('page') :0; # Get the data $sql = qq!SELECT a."OrderID", b."CompanyName" AS "CustomerName", c."FirstName"::text || ' ' ||c."LastName"::text AS "E +mployeeName", a."OrderDate"::DATE, a."RequiredDate"::DATE, a."Shipp +edDate"::DATE, d."CompanyName" AS "ShipVia", a."Freight", a."ShipNam +e", a."ShipAddress", a."ShipCity", a."ShipRegion", a."ShipPostalCode", a."ShipCountry" FROM "Orders" a, "Customers" b, "Employees" c, "Shippers" +d WHERE a."CustomerID" = b."CustomerID" AND a."EmployeeID" = c."EmployeeID" AND a."ShipVia" = d."ShipperID" ORDER BY 1 LIMIT $limit OFFSET $offset !; $sth=$dbh->prepare("$sql"); $sth->execute() || quit(); .... then html stuff skiped.... # Showing page number with link my $first_page = $pagenum - $pagenum ; my $last_page = $pagenum - 1; $pagenum = $pagenum - 1; print q(<ul class="tsc_pagination tsc_paginationA tsc_paginationA09">) +; print qq(<li><a href='vieword.pl?page=$first_page'>First>); for my $i (0 .. $pagenum) { print qq(<li><a href='vieword.pl?page=$i'>$i>); } print qq(<li><a href='vieword.pl?page=$last_page'>Last>); print q(</ul>);

Your help as usual will be much appreciated.

Many thanks !

Terry

Replies are listed 'Best First'.
Re: Need help to improve pagination
by Laurent_R (Canon) on Jul 21, 2014 at 17:42 UTC
Re: Need help to improve pagination (stand up)
by Anonymous Monk on Jul 21, 2014 at 17:45 UTC
Re: Need help to improve pagination
by Anonymous Monk on Jul 21, 2014 at 15:26 UTC
    Nice SQLi vector there, but you've been warned about this already.
Re: Need help to improve pagination
by Anonymous Monk on Aug 08, 2014 at 00:04 UTC
Re: Need help to improve pagination
by Anonymous Monk on Jul 23, 2014 at 21:30 UTC

    perlintro#Perl variable types, Writing subroutines

    #!/usr/bin/perl -- ## ## ## perltidy -olq -csc -csci=10 -cscl="sub : BEGIN END if " -otr -opr +-ce -nibc -i=4 -pt=0 "-nsak=*" #!/usr/bin/perl -- use strict; use warnings; Main( @ARGV ); exit( 0 ); sub Main { print join "\n", pageLinksHtml( 170, 11, 10, 5 ); } sub pageNumbers { return { first => 1 }, 9, 10, { curr => 11 }, 12, 13, { last => 17 }, ; } ## end sub pageNumbers sub pageLinksHtml { my( $total, $curr, $rate, $links ) = @_; my @nums = pageNumbers( $total, $curr, $rate, $links ); for my $num ( @nums ) { if( ref $num ) { if( my $first = $$num{first} ) { $num = sprintf '<li><a href="?;page=%d">{%d</a>...</li +>', $first, $first; } elsif( my $curr = $$num{curr} ) { $num = sprintf '<li><b>%d</b></li>', $curr; } elsif( my $last = $$num{last} ) { $num = sprintf '<li>...<a href="?;page=%d">%d}</a></li +>', $last, $last; } } else { $num = sprintf '<li><a href="?;page=%d">%d</a></li>', $num +, $num; } } return '<ul>', @nums, '</ul>'; } ## end sub pageLinksHtml
      Great, many many thanks, I will now try and figure out how to integrate your code with my script. Best regards. Terry

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