php 分页 - 每页 10 页
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pagination - 10 pages per page
提问by arthur
I have a pagination script that displays a list of all pages like so:prev [1][2][3][4][5][6][7][8][9][10][11][12][13][14] next
But I would like to only show ten of the numbers at a time:prev [3][4][5][6][7][8][9][10][11][12] next
我有一个分页脚本，显示所有页面的列表，如下所示：prev [1][2][3][4][5][6][7][8][9][10][11][12][13][14] next
但我想一次只显示十个数字：prev [3][4][5][6][7][8][9][10][11][12] next
How can I accomplish this? Here is my code so far:
我怎样才能做到这一点？到目前为止，这是我的代码：
<?php
/* Set current, prev and next page */
$page = (!isset($_GET['page']))? 1 : $_GET['page'];
$prev = ($page - 1);
$next = ($page + 1);
/* Max results per page */
$max_results = 2;
/* Calculate the offset */
$from = (($page * $max_results) - $max_results);
/* Query the db for total results.
You need to edit the sql to fit your needs */
$result = mysql_query("select title from topics");
$total_results = mysql_num_rows($result);
$total_pages = ceil($total_results / $max_results);
$pagination = '';
/* Create a PREV link if there is one */
if($page > 1)
{
$pagination .= '< a href="?page='.$prev.'">Previous</a> ';
}
/* Loop through the total pages */
for($i = 1; $i <= $total_pages; $i++)
{
if(($page) == $i)
{
$pagination .= $i;
}
else
{
$pagination .= '< a href="index.php?page='.$i.'">'.$i.'</a>';
}
}
/* Print NEXT link if there is one */
if($page < $total_pages)
{
$pagination .= '< a hr_ef="?page='.$next.'"> Next</a>';
}
/* Now we have our pagination links in a variable($pagination) ready to
print to the page. I pu it in a variable because you may want to
show them at the top and bottom of the page */
/* Below is how you query the db for ONLY the results for the current page */
$result=mysql_query("select * from topics LIMIT $from, $max_results ");
while ($i = mysql_fetch_array($result))
{
echo $i['title'].'<br />';
}
echo $pagination;
?>
回答by Svisstack
10 next pages
下一页 10
for($i = $page + 1; $i <= min($page + 11, $total_pages); $i++)
or if you want 5 prev and 5 next
或者如果你想要 5 个上一个和 5 个下一个
for($i = max(1, $page - 5); $i <= min($page + 5, $total_pages); $i++)
回答by Will
I've just been looking for an answer to the same original question, and couldn't find it, so this is what I came up with. I hope someone else finds it useful.
我一直在寻找同一原始问题的答案，但找不到，所以这就是我想出的。我希望其他人觉得它有用。
$totalPages = 20;
$currentPage = 1;
if ($totalPages <= 10) {
$start = 1;
$end = $totalPages;
} else {
$start = max(1, ($currentPage - 4));
$end = min($totalPages, ($currentPage + 5));
if ($start === 1) {
$end = 10;
} elseif ($end === $totalPages) {
$start = ($totalPages - 9);
}
}
for ($page = $start; $page <= $end; $page++) {
echo '[' . $page . ']';
}
Results:
结果：
$currentPage = 1; // [1][2][3][4][5][6][7][8][9][10]
$currentPage = 4; // [1][2][3][4][5][6][7][8][9][10]
$currentPage = 10; // [6][7][8][9][10][11][12][13][14][15]
$currentPage = 17; // [11][12][13][14][15][16][17][18][19][20]
$currentPage = 20; // [11][12][13][14][15][16][17][18][19][20]
回答by Carson Myers
If you just want a quick fix, you might try modifying your forloop a little. For example, you won't want to start at 1, and you won't necessarily want to loop while $i <= $total_pages.
如果您只想快速修复，您可以尝试for稍微修改您的循环。例如，您不想从 1 开始，也不一定要循环 while $i <= $total_pages。
Displaying an odd number of pagination links might make more sense: you would display the current page, then four to the left of it, and four to the right. Something like this:
显示奇数个分页链接可能更有意义：您将显示当前页面，然后在它的左侧显示四个，在右侧显示四个。像这样的东西：
for($i = $page_number - 4; $i <= $page_number + 4; $i++) {
but you would obviously need to do a little bit more to ensure you weren't displaying negative numbers, or displaying more links than there are pages.
但您显然需要做更多的工作，以确保您不会显示负数，或显示的链接数量不会超过页面数量。
回答by Awais Mustafa
$page = 3;
$totalPages = 33;
$count = 9;
$startPage = max(1, $page - $count);
$endPage = min( $totalPages, $page + $count);
if($page-1 > 0){
echo '<a class="btn btn-default" href="/search-results?page="'.($page-
1).'"><< Prev</a>';
}
for($i = $startPage; $i < $endPage; $i++): if($i <= $totalPages):
echo '<a class="btn btn-<?=$i == $page || $i == 1 && $page == "" ?
'success' : 'primary';?>"style="margin-right:2px;" href="/search-
results?page="'.$i.'">'.$i.'</a>';
endif; endfor;
if($page < $totalPages){
echo '<a class="btn btn-default" href="/search-results?page="'.
($page+1).'">Next >></a>';
}
