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Imam mali problem:
- Imam jednu php stranicu i ona je u vezi linkova (prikazuje linkove iz baze). Moje pitanje je kako da napravim da ako u bazi ima npr. vise od 10 linkova da se napravi druga stranica tipa: linkovi.php?strana=2 ? ?? ?? ?
 
Kod:
<?php 

//This code created by Angela Bradley on php.About.com

// Connects to your Database 
mysql_connect("your.hostaddress.com", "username", "password") or die(mysql_error()); 
mysql_select_db("address") or die(mysql_error()); 

//This checks to see if there is a page number. If not, it will set it to page 1 
if (!(isset($pagenum))) 
{ 
$pagenum = 1; 
} 

//Here we count the number of results 
//Edit $data to be your query 
$data = mysql_query("SELECT * FROM topsites") or die(mysql_error()); 
$rows = mysql_num_rows($data); 

//This is the number of results displayed per page 
$page_rows = 4; 

//This tells us the page number of our last page 
$last = ceil($rows/$page_rows); 

//this makes sure the page number isn't below one, or more than our maximum pages 
if ($pagenum < 1) 
{ 
$pagenum = 1; 
} 
elseif ($pagenum > $last) 
{ 
$pagenum = $last; 
} 

//This sets range that we will display in our query 
$max = 'limit ' .($pagenum - 1) * $page_rows .',' .$page_rows;

//This is your query again, the same one... the only difference is we add $max into it
$data_p = mysql_query("SELECT * FROM topsites $max") or die(mysql_error()); 

//This is where you display your query results
while($info = mysql_fetch_array( $data_p )) 
{ 

Print $info['Name']; 
echo "<br>";
} 

echo "<p>";

// This shows the user what page they are on, and the total number of pages
echo " --Page $pagenum of $last-- <p>";

// First we check if we are on page one. If we are then we don't need a link to the previous page or the first page so we do nothing. If we aren't then we generate links to the first page, and to the previous page.
if ($pagenum == 1) 
{
} 
else 
{
echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=1'> <<-First</a> ";
echo " ";
$previous = $pagenum-1;
echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$previous'> <-Previous</a> ";
} 

//just a spacer
echo " ---- ";

//This does the same as above, only checking if we are on the last page, and then generating the Next and Last links
if ($pagenum == $last) 
{
} 
else {
$next = $pagenum+1;
echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$next'>Next -></a> ";
echo " ";
echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$last'>Last ->></a> ";
} 
?>

Ovde imas ceo kod za tvoju stranicu.
 
Ako ti je suvise komplikovano pokusaj sa ovim uproscenim (skracenim) kodom koji sam koristio ranije za neke slicne probleme.
Kod:
//Konekcija sa DB
include ("konekcija.php");
//query
$result1 = mysql_query("SELECT count(*) FROM baza");
if (!$result1) {
echo("<P>Greska: " . mysql_error() . "</P>");
exit();
}
// koliko redova ima u bazi smestas u $brredova
$row = mysql_fetch_array($result1) ;
$brredova = ($row["count(*)"]);

//******************************
//sta raditi sa podacima iz baze
if ($brredova < 10) {
//ispis prvih 10 linkova
}
else  {
//ispis prvih 10 linkova
//pravljenje linka na sledecu stranu
}

Tvoje pitanje je bilo kako da napravim da ako u bazi ima npr. vise od 10 linkova da se napravi druga stranica tipa: linkovi.php?strana=2 ? ?? ?? ?
Prvi deo odgovora imas ispisan u kodu, a sto se tice linkova switch bi tu mogao da odradi posao:

Kod:
switch ($strana) { 
case "1":echo("<P><a href='linkovi.php?strana=1' >" . "pogledaj! </a></P>"); break;
case "2":echo("<P><a href='linkovi.php?strana=2' >" . " pogledaj! </a></P>") ; break;
//itd...
}

Pre toga bi islo if(!isset($strana)) itd...
Nadam se da te nisam jos vise zabunio poz!
 

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