如果我已经将票证表设置为拆分为多个页面，如何使其底部具有页面链接？

Question

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编辑：

我仍然不知道该怎么做。有人可以帮忙吗？我知道它被称为“分页”，但我不知道如何让它在底部显示页面链接。我希望有人能够提供帮助。

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I have a table of tickets and all of the data is loaded in from a database to the table. I have pages on it (using ?page=1, ?page=2, etc.) but, how can I make it where it has "Next Page", "Previous Page", "Last Page", "Page something of something", etc. at the bottom?

完成这一切的代码是：

<table id="tickets">
 <tr>
 <th>Name</th>
 <th>Email</th>
 <th>Subject</th>
 <th>Created on</th>
 <th style="width:65px;">Status</th>
 <th>Actions</th>
 </tr><?php
$startpoint = $_GET["page"];
if (isset($_GET["page"])) {
    $startpoint = $_GET["page"];
}
else {
    $startpoint = 1;
}
$startpoint = $startpoint - 1;
$startpoint = $startpoint * 10;
$endpoint = $startpoint + 10;

function addEllipsis($string, $length) {
    $end = "...";
    if (strlen($string) > $length) {
        $length -= strlen($end); // $length =  $length – strlen($end);
        $string = substr($string, 0, $length);
        $string .= $end; //  $string =  $string . $end;
    }
    return $string;
}
$x = 1;
if ($show == "active") {
    $result = mysql_query("SELECT * FROM tickets WHERE is_active='1'");
    while ($info = mysql_fetch_array($result)) {
        $x++;
        $name = $info['name'];
        $email = $info['email'];
        $subject = $info['subject'];
        $ticketid = $info['ticket'];
        $isActive = $info['is_active'];
        $created = $info['created'];
        if ($isActive == '1') {
            $status = "<span class=\"open\">Open</span>";
        }
        if ($isActive == '0') {
            $status = "<span class=\"closed\">Closed</span>";
        }
        if ($x > $startpoint && $x < $endpoint) {
            echo "
 <tr>
 <td>".$name."</td>
 <td>".$email."</td>
 <td title=\"".$subject."\">".addEllipsis($subject, 16)."</td>
 <td>".$created."</td>
 <td>".$status."</td>
 <td><a href=\"/employee/employee.php?ticket=".$ticketid."\">View ticket &raquo;</a></td>
 </tr>";
        }
    }
}
else if ($show == "closed") {
    $result = mysql_query("SELECT * FROM tickets WHERE is_active='0'");
    while ($info = mysql_fetch_array($result)) {
        $x++;
        $name = $info['name'];
        $email = $info['email'];
        $subject = $info['subject'];
        $ticketid = $info['ticket'];
        $isActive = $info['is_active'];
        $created = $info['created'];
        if ($isActive == '1') {
            $status = "<span class=\"open\">Open</span>";
        }
        if ($isActive == '0') {
            $status = "<span class=\"closed\">Closed</span>";
        }
        if ($x > $startpoint && $x < $endpoint) {
            echo "
 <tr>
 <td>".$name."</td>
 <td>".$email."</td>
 <td title=\"".$subject."\">".addEllipsis($subject, 16)."</td>
 <td>".$created."</td>
 <td>".$status."</td>
 <td><a href=\"/employee/employee.php?ticket=".$ticketid."\">View ticket &raquo;</a></td>
 </tr>";
        }
    }
}
else {
    $result = mysql_query("SELECT * FROM tickets");
    while ($info = mysql_fetch_array($result)) {
        $x++;
        $name = $info['name'];
        $email = $info['email'];
        $subject = $info['subject'];
        $ticketid = $info['ticket'];
        $isActive = $info['is_active'];
        $created = $info['created'];
        if ($isActive == '1') {
            $status = "<span class=\"open\">Open</span>";
        }
        if ($isActive == '0') {
            $status = "<span class=\"closed\">Closed</span>";
        }
        if ($x > $startpoint && $x < $endpoint) {
            echo "
 <tr>
 <td>".$name."</td>
 <td>".$email."</td>
 <td title=\"".$subject."\">".addEllipsis($subject, 16)."</td>
 <td>".$created."</td>
 <td>".$status."</td>
 <td><a href=\"/employee/employee.php?ticket=".$ticketid."\">View ticket &raquo;</a></td>
 </tr>";
        }
    }
}
  ?>

 </table>

我怎样才能让它在底部显示每个页面的链接（只是一个简单的“1”、“2”等）？我无法自己创建它们，因为我不知道有多少页面之类的东西。如果当您位于某个页面时链接可以变成灰色的不可点击文本，那就太好了。有人帮助我使用页面功能，但他们不在线，所以他们无法帮助我。 （我还不太擅长 PHP）有人知道如何实现这一点吗？

Answer 1

-编辑-
聊天后，解决方案包括向查询添加 LIMIT 而不是使用 $x 来防止显示某些行，如下所示：https://gist.github.com/1233850

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运行以下代码，因为它可以单独工作，所以可能会给您一个想法：

<style>span{background:#ccc;padding:2px 6px;margin:0 2px;}</style>

<?php
    $currentPage = isset($_GET['page']) ? $_GET['page'] : 1;
    $totalPages = 5; //this must be fetched from the db

    //define previous and next
    $previous = ($currentPage > 1) ? $currentPage - 1 : false; 
    $next = ($currentPage < $totalPages) ? $currentPage + 1 : false;

    //here we'll save the resulting code
    $pagination = array();
    //first
    $pagination[] = ($currentPage == 1) ? "First" : "<a href='?page=1'>First</a>";
    //previous
    $pagination[] = $previous ? "<a href='?page=$previous'>Previous</a>" : "Previous";
    //current/total
    $pagination[] = "Page $currentPage / $totalPages";
    //next
    $pagination[] = $next ? "<a href='?page=$next'>Next</a>" : "Next";
    //last
    $pagination[] = ($currentPage == $totalPages) ? "Last" : "<a href='?page=$totalPages'>Last</a>";

    //show it
    foreach($pagination as $part){
        echo "<span>$part</span>";
    }
?>

Answer 2

好的，一般可以这样完成：

if (isset($_GET["page"])) {
     $currentPage = $_GET["page"];
}
else {
     $currentPage = 1;
}

Good start。要获取页数，您需要计算每页 db/items 中的记录数。

$count = mysql_query("SELECT COUNT(id) FROM tickets"); //yop, it's kind of shortcut
$pages = ceil($count/$itemsPerPage);

好的，生成菜单的页码（假设我们想要像“< 1 ... 3 4 5 ...最后>”->其中$curretnPage = 4）：

$menu = array();  //array indexes are strings to make this clearer, it normally should be $menu[] = $sth; to preserve adding order which is important here
if($currentPage > 1) {
     $menu['previous'] = $currentPage - 1;
}
$menu['first'] = 1;
if($currentPage > 3) {
     $menu['dots'] = 0;         
}
else if($currentPage == 3) {
     $menu['oneBeforeCurrent'] = $currentPage - 1;
     $menu['current'] = $currentPage;
}
else if($currentPage == 2) {         
     $menu['current'] = $currentPage;
}
$menu['oneAfterCurrent'] = $currentPage + 1;

...它变得有点复杂，也许有更好、更快的方法，但你可以看到它是如何完成的（不知道它“应该如何完成”）。它并不完整，但是......你知道该怎么做:)
我希望你的意思是，你不能创建菜单...如果你有带有数字的数组，只需循环并创建带有数字的html链接，然后将“...”插入0。

干杯