我有一個通過$ _ POST傳遞給我的搜索結果頁面,並存儲在一個變量名爲$項的值。然後,該變量在MySQL查詢中運行,以顯示搜索結果,如果有超過15個結果,我有一個分頁php腳本,將結果分成幾頁。PHP分頁失去在其他頁面變量
的第一頁是巨大的,結果的正確數量被發現和打印的頁數是可用的。如果我回顯該變量,則會在其中存儲正確的值。但是,如果我選擇查看下一批結果並刷新頁面的所有行(如果查詢只是SELECT *,而沒有參數)並且變量現在爲空
這裏是我的代碼(這是一個有點長,對不起,我打算把它的大吸盤遠在另一個頁面,包括它一旦它的正常工作)
<?php
$tbl_name="accounts";
$adjacents = 3;
if (isset($_POST['basic_search_submit'])){
$term = $_POST['item'];}
$query = "SELECT COUNT(*) as num FROM accounts LEFT JOIN accounts_cstm ON accounts.id = accounts_cstm.id_c WHERE accounts.name LIKE '%$term%' ORDER BY accounts.name";
$total_pages = mysql_fetch_array(mysql_query($query));
$total_pages = $total_pages[num];
$targetpage = "search_results_play.php";
$limit = 15;
$page = $_GET['page'];
if($page)
$start = ($page - 1) * $limit;
else
$start = 0;
$sql = "SELECT * FROM accounts LEFT JOIN accounts_cstm ON accounts.id = accounts_cstm.id_c WHERE accounts.name LIKE '%$term%' ORDER BY accounts.name LIMIT $start, $limit";
$result = mysql_query($sql);
if ($page == 0) $page = 1;
$prev = $page - 1;
$next = $page + 1;
$lastpage = ceil($total_pages/$limit);
$lpm1 = $lastpage - 1;
$pagination = "";
if($lastpage > 1)
{
$pagination .= "<div class=\"pagination\">";
if ($page > 1)
$pagination.= "<a href=\"$targetpage?page=$prev\">< < previous</a>";
else
$pagination.= "<span class=\"disabled\">< < previous</span>";
if ($lastpage < 7 + ($adjacents * 2))
{
for ($counter = 1; $counter <= $lastpage; $counter++)
{
if ($counter == $page)
$pagination.= "<span class=\"current\">$counter</span>";
else
$pagination.= "<a href=\"$targetpage?page=$counter\">$counter</a>";
}
}
elseif($lastpage > 5 + ($adjacents * 2)) //enough pages to hide some
{
if($page < 1 + ($adjacents * 2))
{
for ($counter = 1; $counter < 4 + ($adjacents * 2); $counter++)
{
if ($counter == $page)
$pagination.= "<span class=\"current\">$counter</span>";
else
$pagination.= "<a href=\"$targetpage?page=$counter\">$counter</a>";
}
$pagination.= "...";
$pagination.= "<a href=\"$targetpage?page=$lpm1\">$lpm1</a>";
$pagination.= "<a href=\"$targetpage?page=$lastpage\">$lastpage</a>";
}
elseif($lastpage - ($adjacents * 2) > $page && $page > ($adjacents * 2))
{
$pagination.= "<a href=\"$targetpage?page=1\">1</a>";
$pagination.= "<a href=\"$targetpage?page=2\">2</a>";
$pagination.= "...";
for ($counter = $page - $adjacents; $counter <= $page + $adjacents; $counter++)
{
if ($counter == $page)
$pagination.= "<span class=\"current\">$counter</span>";
else
$pagination.= "<a href=\"$targetpage?page=$counter\">$counter</a>";
}
$pagination.= "...";
$pagination.= "<a href=\"$targetpage?page=$lpm1\">$lpm1</a>";
$pagination.= "<a href=\"$targetpage?page=$lastpage\">$lastpage</a>";
}
else
{
$pagination.= "<a href=\"$targetpage?page=1\">1</a>";
$pagination.= "<a href=\"$targetpage?page=2\">2</a>";
$pagination.= "...";
for ($counter = $lastpage - (2 + ($adjacents * 2)); $counter <= $lastpage; $counter++)
{
if ($counter == $page)
$pagination.= "<span class=\"current\">$counter</span>";
else
$pagination.= "<a href=\"$targetpage?page=$counter\">$counter</a>";
}
}
}
if ($page < $counter - 1)
$pagination.= "<a href=\"$targetpage?page=$next\">next > ></a>";
else
$pagination.= "<span class=\"disabled\">next > ></span>";
$pagination.= "</div>\n";
}
?>
<? while ($row = mysql_fetch_array($result)){
//table displaying results
}
?>
<?=$pagination?>
任何幫助將不勝感激,謝謝
您的代碼對SQL注入開放。首先,你應該停止使用不推薦使用的'mysql_query()'。 PHP強烈建議這樣做:http://php.net/manual/en/function.mysql-query.php您還需要了解什麼是SQL注入以及如何在使用數據庫查詢之前對輸入進行清理。 – David 2012-07-24 20:50:43
請不要使用'mysql_ *'函數來編寫新代碼。他們不再維護,社區已經開始[棄用程序](http://goo.gl/KJveJ)。查看[*紅色框*](http://goo.gl/GPmFd)?相反,您應該瞭解[準備好的語句](http://goo.gl/vn8zQ)並使用[PDO](http://php.net/pdo)或[MySQLi](http://php.net/ mysqli的)。如果你不能決定哪些,[這篇文章](http://goo.gl/3gqF9)會幫助你。如果你選擇PDO,[這裏是很好的教程](http://goo.gl/vFWnC)。 – orourkek 2012-07-24 20:50:48
您好,我不得不承認,我很新的PHP所以像SQL注入的條件是相當新的給我,並請求mysql_query()是什麼,我是在大學只教3個月前。我感謝你的意見,我會看看這些教程和文章。我會誠實的說,我沒有看到一個紅色的盒子? – tatty27 2012-07-24 21:14:08