Question

我有以下分页代码：

$rec_limit = 3;



/* Get total number of records */
$query = "SELECT count(id) FROM news";
$result = mysql_query($query);
if(!$result)
{
  die('Could not get data: ' . mysql_error());
}
$row = mysql_fetch_array($result, MYSQL_NUM );
$rec_count = $row[0];

if( isset($_GET{'page'} ) )
{
   $page = $_GET{'page'} + 1;
   $offset = $rec_limit * $page ;
}
else
{
   $page = 0;
   $offset = 0;
}
$left_rec = $rec_count - ($page * $rec_limit);

$query2 = "SELECT * ".
       "FROM news ".
       "LIMIT $offset, $rec_limit";

$result2 = mysql_query( $query2 );
if(! $result2 )
{
  die('Could not get data: ' . mysql_error());
}
while($row = mysql_fetch_array($result2, MYSQL_ASSOC))
{
    echo " ID :{$row['id']}  <br> ".
         " News : {$row['title']} <br> ".
         " Date : {$row['date']} <br> ".
         "--------------------------------<br>";
} 

if( $page > 0 )
{
   $last = $page - 2;
   echo "<a href=\"{$_SERVER['PHP_SELF']}?page=$last\">Last 3 Records</a> |";
   echo "<a href=\"{$_SERVER['PHP_SELF']}?page=$page\">Next 3 Records</a>";
}
else if( $page == 0 )
{
   echo "<a href=\"{$_SERVER['PHP_SELF']}?page=$page\">Next 3 Records</a>";
}
else if( $left_rec < $rec_limit )
{
   $last = $page - 2;
   echo "<a href=\"{$_SERVER['PHP_SELF']}?page=$last\">Last 3 Records</a>";
}

代码可以根据需要完美地工作，除了a href部分，即使没有新闻，它也会显示下一个3记录链接。我的意思是，它一直打开空记录的页面！ 如何在达到最后一条新闻ID时删除“下一条3条记录”链接。

谢谢

Answer 1

只需将最后一个条件置于其他两个条件之上并将<更改为<=：

if($left_rec <= $rec_limit)
{
    $last = $page - 2;
    echo "<a href=\"{$_SERVER['PHP_SELF']}?page=$last\">Last 3 Records</a>";
} else if ( $page > 0 )
{
    $last = $page - 2;
    echo "<a href=\"{$_SERVER['PHP_SELF']}?page=$last\">Last 3 Records</a> |";
    echo "<a href=\"{$_SERVER['PHP_SELF']}?page=$page\">Next 3 Records</a>";
}
else if( $page == 0 )
{
    echo "<a href=\"{$_SERVER['PHP_SELF']}?page=$page\">Next 3 Records</a>";
}

PHP分页最后的id记录

1 个答案: