Question

好的我就是这个我正在为一个网站创建一个脚本设计，它将使用分页从数据库中绘制引号，

我已经很好地完成了这一部分，设置了分页类并实际打印出结果，但我一直在尝试使用编号系统进行打印 - 就像每个都一样。

我希望每个报价都有自己的编号，如：

“一个人不能超越他的影子”;
“foo不是傻瓜，只是在php中使用”

“等等”;

看看我的班级：

class pagination extends cleanPost{

var $p=1, $max_r,$limits;
var $count_all=0,$sql,$total,$table,$totalres,$totalpages;
var $r,$i;
var $show=10;



function setMax($max_r){

       $this->p = $this->clean($_GET['p']);
       $this->max_r = $max_r;

        if(empty($this->p))
        {
             $this->p = 1;
        }

       $this->limits = ($this->p - 1) * $this->max_r;

    }  
function setData($table){

   $this->table = $table;
   $this->sql = "SELECT * FROM ".$this->table." LIMIT ".$this->limits.",".$this->max_r."";
   $this->sql = mysql_query($this->sql) or die(mysql_error());
   $this->total = "SELECT * FROM ".$this->table."";
   $this->totalres =  mysql_query($this->total) or die(mysql_error());
   $this->count_all = mysql_num_rows($this->totalres); // count all the rows
   $this->totalpages = ceil($this->count_all / $this->max_r); // work out total pages 
}  



function displayLinks($show){

    $this->show = $show; // How many links to show

    echo "<br><br>";

    if($this->p > 1) // If p > then one then give link to first page
    {
        $pagination .= "<div class=\"pagination\">
         <a href=?p=1> [first] </a>  ";    
    }
    else
    { // else show nothing
        $pagination .= "<div class=\"pagination\">"."";
}
if($this->p != 1)
    { // if p aint equal to 1 then show previous text

        $previous = $this->p-1;

    $pagination .= "<a href=?p=$previous>«prev</a>";

}
else
    { //else show nothing
    $pagination.= "<span class=\"disabled\">first</span>"."";
} 
for($i =1; $i <= $this->show; $i++) // show ($show) links
{

        if($this->p > $this->totalpages)
            { // if p is greater then totalpages then display nothing
            echo "";

    }
    else if($_GET["p"] == $this->p)
            { //if p is equal to the current loop value then dont display that value as link
               $pagination .= $this->p ;
    }
    else{
               $pagination .= " <a href=?p=".$this->p."> ".$this->p." </a>"; // else display the rest as links
    }

    $this->p++; //increment $p  
}



if($_GET["p"] == $this->totalpages)
    {// if page is equal to totalpages then  dont display the last page at the end of links
   $pagination.= "<span class=\"disabled\">last</span>"."";
}
else  if($_GET["p"] != $this->totalpages)

{
// $pagination .= "..."; // display dots   
#   $pagination .= "<a href=?p=".$this->totalpages.">  ".$this->totalpages." </a>"; 
}
else { // else display the last page link after other ones

}
if($_GET["p"] < $this->totalpages)// if p is less then total pages then show next link
{
    $next = $_GET["p"] + 1;
    $pagination .= "<a href=?p=$next> next»</a>"; 
    $pagination.= "<a href=?p=".$this->totalpages."> [last] </a>"."";   
}
  $pagination.= "</div>\n";            
echo $pagination;

}

////////// 这是我的报价显示页面

$page= new pagination;
  $page->setMax(5);//set the maximum quote shown per page
  $page   ->setData("quotes");
   $page  ->display();

echo "<table border=1 width=100%>";
  while ($row = mysql_fetch_array($page->sql)) 
  { 

    echo  $num_results = mysql_num_rows($page->sql);
    echo "<tr>";

  for ($i=0; $i <$num_results; $i++)
{
echo "<td>";
 $num_found = $i + 1;

 echo $num_found.$row['ngwaquote'];
  echo "</td>";
 }
echo "</tr>";

  }
  echo "</table><p>";
   $page  ->displayLinks(4) ;

我得到的结果是，如果我点击分页链接中的第一个数字给我 quote.html？p = 1时

我让他们连续列出来页面：1 1 [引用]滚石没有苔藓。[/ quote]
2 [引用]滚石没有苔藓。[/ quote]
3 [引用]滚石没有苔藓。[/ quote]
4 [引用]滚石没有苔藓。[/ quote]
5 [引用]滚石没有苔藓。[/ quote]

但是另一页quote.html？p = 2 而不是从开始第2页 6 [qoute]无[/报价] 7 [引用]滚石没有苔藓。[/ quote]

而非再次从一个开始计数页面：1 1 [qoute]无[/报价] 2 [引用]滚石没有苔藓。[/ quote]

我希望它看起来像我们在BBC FOOTBALL博客中看到的那样评论页面2从101开始而不是从1再次开始

我希望我的问题不会太久理解，如果我很抱歉只是为了更好地理解而发布这些问题。

Answer 1

我认为你可能会使问题复杂化。一些伪代码：

Current number = Number of items per page * Current page number

Current page number来自您的网址，其中包含?p= GET参数。

然后在每次迭代中回显引用，向Current number添加一个引号。

使用分页从数据库中抽取的编号结果 - PHP

1 个答案: