Question

我正在尝试显示来自6个不同表格的数据并使用分页，以便用户可以滚动浏览类似于购物网站的项目。

我正在使用以下代码在10页的页面调用请求（作为测试）中显示一个名为request id的字段，该字段工作正常。

<?php
include('connect.php'); 
$targetpage = "page.php";    
$limit = 10; 

$query = "SELECT COUNT(*) as num FROM request";
$total_pages = mysql_fetch_array(mysql_query($query));
$total_pages = $total_pages[num];

$stages = 3;
$page = mysql_escape_string($_GET['page']);
if($page){
$start = ($page - 1) * $limit; 
}else{
$start = 0; 
    }       
// Get page data
$query1 = "SELECT * FROM request LIMIT $start, $limit";
$result = mysql_query($query1);

// Initial page num setup
if ($page == 0){$page = 1;}
$prev = $page - 1;  
$next = $page + 1;                          
$lastpage = ceil($total_pages/$limit);      
$LastPagem1 = $lastpage - 1;                    

$paginate = '';
if($lastpage > 1)
{   
$paginate .= "<div class='paginate'>";
// Previous
if ($page > 1){
$paginate.= "<a href='$targetpage?page=$prev'>previous</a>";
}else{
$paginate.= "<span class='disabled'>previous</span>";   }

// Pages    
if ($lastpage < 7 + ($stages * 2))  // Not enough pages to breaking it up
{   
for ($counter = 1; $counter <= $lastpage; $counter++)
{
if ($counter == $page){
$paginate.= "<span class='current'>$counter</span>";
}else{
$paginate.= "<a href='$targetpage?page=$counter'>$counter</a>";}                    
}
}
elseif($lastpage > 5 + ($stages * 2))   // Enough pages to hide a few?
{
// Beginning only hide later pages
if($page < 1 + ($stages * 2))       
{
for ($counter = 1; $counter < 4 + ($stages * 2); $counter++)
{
if ($counter == $page){
$paginate.= "<span class='current'>$counter</span>";
}else{
$paginate.= "<a href='$targetpage?page=$counter'>$counter</a>";}                    
}
$paginate.= "...";
$paginate.= "<a href='$targetpage?page=$LastPagem1'>$LastPagem1</a>";
$paginate.= "<a href='$targetpage?page=$lastpage'>$lastpage</a>";       
}
// Middle hide some front and some back
elseif($lastpage - ($stages * 2) > $page && $page > ($stages * 2))
{
$paginate.= "<a href='$targetpage?page=1'>1</a>";                          
$paginate.= "<a href='$targetpage?page=2'>2</a>";
$paginate.= "...";
for ($counter = $page - $stages; $counter <= $page + $stages; $counter++)
{
if ($counter == $page){
$paginate.= "<span class='current'>$counter</span>"; 
}else{
$paginate.= "<a href='$targetpage?page=$counter'>$counter</a>";}                    
}
$paginate.= "...";
$paginate.= "<a href='$targetpage?page=$LastPagem1'>$LastPagem1</a>";
$paginate.= "<a href='$targetpage?page=$lastpage'>$lastpage</a>";       
}
// End only hide early pages
else
{
$paginate.= "<a href='$targetpage?page=1'>1</a>";
$paginate.= "<a href='$targetpage?page=2'>2</a>";
$paginate.= "...";
for ($counter = $lastpage - (2 + ($stages * 2)); $counter <= $lastpage; $counter++)
{
if ($counter == $page){
$paginate.= "<span class='current'>$counter</span>";
}else{
$paginate.= "<a href='$targetpage?page=$counter'>$counter</a>";}                    
}
}
}
// Next
if ($page < $counter - 1){ 
$paginate.= "<a href='$targetpage?page=$next'>next</a>";
}else{
$paginate.= "<span class='disabled'>next</span>";
}
$paginate.= "</div>";       
}
echo $total_pages.' Results';
// pagination
echo $paginate;
?>
<ul>
<?php 
while($row = mysql_fetch_array($result))
{       
echo '<li>'.$row['requestid'].'</li>';
}
?>

我现在正尝试使用以下查询显示总共6个表中的数据，但它不起作用？

我认为我没有在查询中正确使用COUNT命令？

$query = "SELECT COUNT as num r.*, m.*, u.*, a.*, i.*, b.*, r.*
FROM request r INNER JOIN movie m ON m.movieid = r.movieid 
INNER JOIN actor a ON a.actorid = r.actorid 
INNER JOIN users u ON u.userid = r.userid
INNER JOIN item i ON i.itemid = r.itemid
INNER JOIN brand b ON b.brandid = r.brandid
WHERE gender = 'male'";

Answer 1

这是您使用COUNT功能的方式，您必须指定您正在计算的内容。

SELECT COUNT(*) as num r.*, m.*, u.*, a.*, i.*, b.*, r.*
FROM request r INNER JOIN movie m ON m.movieid = r.movieid 
INNER JOIN actor a ON a.actorid = r.actorid 
INNER JOIN users u ON u.userid = r.userid
INNER JOIN item i ON i.itemid = r.itemid
INNER JOIN brand b ON b.brandid = r.brandid
WHERE gender = 'male'";

php分页查询多个表

1 个答案: