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我有一個搜索表單,我想分頁,但是當我按分頁搜索的下一個按鈕或第二頁,所有的結果搞砸了,沒有數據被取出來自查詢。我不知道確切的問題可能是,但是這是我的搜索代碼:分頁搜索php
include ('paginate.php'); //include of paginat page
$per_page = 5;
$find = $_POST['find'];
$find = strtoupper($find);
$find = strip_tags($find);
$find = trim ($find); // number of results to show per page
$result = mysql_query("SELECT * FROM projects WHERE p_name LIKE '%$find%'");
$total_results = mysql_num_rows($result);
$total_pages = ceil($total_results/$per_page);//total pages we going to have
//------------if page is setcheck-----------------//
if (isset($_GET['page'])) {
$show_page = $_GET['page']; //it will telles the current page
if ($show_page > 0 && $show_page <= $total_pages) {
$start = ($show_page - 1) * $per_page;
$end = $start + $per_page;
} else {
// error - show first set of results
$start = 1;
$end = $per_page;
}
} else {
// if page isn't set, show first set of results
$start = 0;
$end = $per_page;
$show_page=1;
}
// display pagination
if(isset($_GET['page'])){
$page = intval($_GET['page']);
}else{
$page =1;
}
$tpages=$total_pages;
if ($page <= 0)
$page = 1;
?>
<?php
$reload = $_SERVER['PHP_SELF'] . "?tpages=" . $tpages;
echo '<div class="pagination"><ul>';
if ($total_pages > 1) {
echo paginate($reload, $show_page, $total_pages);
}
echo "</ul></div>";
// display data in table
echo "<table class='table table-bordered'>";
echo "<thead><tr><th>Project</th> <th>Country</th> <th>Active</th>
</tr></thead>";
for ($i = $start; $i < $end; $i++) {
// make sure that PHP doesn't try to show results that don't exist
if ($i == $total_results) {
break;
}
?><form name="frmactive" method="POST" action="">
<input type="hidden" name="id" value="<?php echo mysql_result($result, $i, 'p_id');?>" />
<?php
// echo out the contents of each row into a table
echo '<tr><td>' . mysql_result($result, $i, 'p_name') . '</td>';
echo '<td>' . mysql_result($result, $i, 'p_country') . '</td>';
if (mysql_result($result, $i, 'p_isActive')=='1'){
echo '<td>Active</td>';
echo '<td align="center"><a href="activeproject.php?id=' . mysql_result($result, $i, 'p_id') . '">Edit</a></td>';
}
else{
echo'<td>Inactive</td> ';
echo '<td align="center"><a href="activeproject.php?id=' . mysql_result($result, $i, 'p_id') . '">Edit</a></td>';
echo "</tr>";
}
// close table>
}
echo "</table>";
// pagination
?>
,這是paginate.php
function paginate($reload, $page, $tpages) {
$adjacents = 2;
$prevlabel = "‹ Prev";
$nextlabel = "Next ›";
$out = "";
// previous
if ($page == 1) {
$out.= "<span>" . $prevlabel . "</span>\n";
} elseif ($page == 2) {
$out.= "<li><a href=\"" . $reload . "\">" . $prevlabel . "</a>\n</li>";
} else {
$out.= "<li><a href=\"" . $reload . "&page=" . ($page - 1) . "\">" . $prevlabel . "</a>\n</li>";
}
$pmin = ($page > $adjacents) ? ($page - $adjacents) : 1;
$pmax = ($page < ($tpages - $adjacents)) ? ($page + $adjacents) : $tpages;
for ($i = $pmin; $i <= $pmax; $i++) {
if ($i == $page) {
$out.= "<li class=\"active\"><a href=''>" . $i . "</a></li>\n";
} elseif ($i == 1) {
$out.= "<li><a href=\"" . $reload . "\">" . $i . "</a>\n</li>";
} else {
$out.= "<li><a href=\"" . $reload . "&page=" . $i . "\">" . $i . "</a>\n</li>";
}
}
if ($page < ($tpages - $adjacents)) {
$out.= "<a style='font-size:11px' href=\"" . $reload . "&page=" . $tpages . "\">" . $tpages . "</a>\n";
}
// next
if ($page < $tpages) {
$out.= "<li><a href=\"" . $reload . "&page=" . ($page + 1) . "\">" . $nextlabel . "</a>\n</li>";
} else {
$out.= "<span style='font-size:11px'>" . $nextlabel . "</span>\n";
}
$out.= "";
return $out;
}
可否請你指出其中的錯誤正在發生?感謝您的時間
使用極限。 –
我以前使用過限制但我有同樣的問題 –