Проблем със странициране

ty1s73r40

Registered
Немога да рзбера къде е грешката

<?php
$page = (int) $_GET['page'];
if (empty($page) or !isSet($page)) {
$page = 1;
}
$dbhost="";
$dbusername="";
$dbpassword="";
$dbname="";

$connect = mysql_connect($dbhost, $dbusername, $dbpassword);
$db = mysql_select_db($dbname,$connect) or die ("Could not select database");

$pp = 3;
$start = ($page * $pp) - $pp;

$result = mysql_query("SELECT * FROM XXXXX RIGHT JOIN XXXXX ON xxxx.product_code = xxxxxx.product_code WHERE xxxxxx.confirmed='0' LIMIT $start");
while($myrow = mysql_fetch_array($result))

{
echo "Резолтатите ";
}

function pageExists($page){
$next = mysqli_query("SELECT * FROM xxxxx RIGHT JOIN xxxxxx ON xxxxx.product_code = xxxxx.product_code WHERE xxxxx.confirmed='0' AND xxxxxx_id >'{$page}' ORDER BY id LIMIT 1");
$prev = mysqli_query("SELECT * FROM xxxxxx RIGHT JOIN xxxxxx ON xxxxxx.product_code = xxxxxx.product_code WHERE xxxxx.confirmed='0' AND xxxxx_id <'{$page}' ORDER BY id DESC LIMIT 1");
return array($prev,$next);

}
list($prev,$next) = pageExists($page);

if(isSet($prev)){
echo'<a href = "?page='.$prev.'">'.$prev.'</a>';
}
if(isSet($next)){
echo'<a href = "?page='.$next.'">'.$next.'</a>';
}
?>
 
Това долкото напълно автентично ли е?

PHP:
$result = mysql_query("SELECT * FROM XXXXX RIGHT JOIN XXXXX ON xxxx.product_code = xxxxxx.product_code WHERE xxxxxx.confirmed='0' LIMIT $start");
 
Да то попринцип кода е :
$result = mysql_query("SELECT * FROM xxxxx RIGHT JOIN xxxxxx ON xxxxx.product_code = xxxxx.product_code WHERE xxxxx.confirmed='0' OR xxxxx.status='В процес...'");
и си работеше но искам странициране да му сложа за това добавих другото но нещо се бугна
 
В последствие реших да ги изкарам оделно в две страници и затова оставих само $result = mysql_query("SELECT * FROM XXXXX RIGHT JOIN XXXXX ON xxxx.product_code = xxxxxx.product_code WHERE xxxxxx.confirmed='0' "); и добавих LIMIT $start
 
Забелязвам, че една от повтарящите се грешни е непълна или наналична употребата на дебъгване.

ty1s73r40 каза:
<?php
$page = (int) $_GET['page'];
if (empty($page) or !isSet($page)) {
$page = 1;
}
$dbhost="";
$dbusername="";
$dbpassword="";
$dbname="";

$connect = mysql_connect($dbhost, $dbusername, $dbpassword) or die(mysql_error());
$db = mysql_select_db($dbname,$connect) or die ("Could not select database");

$pp = 3;
$start = ($page * $pp) - $pp;

$result = mysql_query("SELECT * FROM XXXXX RIGHT JOIN XXXXX ON xxxx.product_code = xxxxxx.product_code WHERE xxxxxx.confirmed='0' LIMIT $start") or die(mysql_error());
while($myrow = mysql_fetch_array($result))

{
echo "Резолтатите ";
}

function pageExists($page){
$next = mysqli_query("SELECT * FROM xxxxx RIGHT JOIN xxxxxx ON xxxxx.product_code = xxxxx.product_code WHERE xxxxx.confirmed='0' AND xxxxxx_id >'{$page}' ORDER BY id LIMIT 1") or die(mysql_error());
$prev = mysqli_query("SELECT * FROM xxxxxx RIGHT JOIN xxxxxx ON xxxxxx.product_code = xxxxxx.product_code WHERE xxxxx.confirmed='0' AND xxxxx_id <'{$page}' ORDER BY id DESC LIMIT 1") or die(mysql_error());
return array($prev,$next);

}
list($prev,$next) = pageExists($page);

if(isSet($prev)){
echo'<a href = "?page='.$prev.'">'.$prev.'</a>';
}
if(isSet($next)){
echo'<a href = "?page='.$next.'">'.$next.'</a>';
}
?>

Ползвай echo(); var_dump(), print_r();, за да изведеш стойността на променливите....
 

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