php分页donot显示下一个记录

Question

对不起，我把这个问题之前在wordpress错误。我正在分页工作，但我面临的一个问题是，当我点击下一个增量查询字符串中的数字为下一页，但数据保持不变，在页面上，我需要下四个记录显示，因为我已显示4记录每页。代码似乎很好，但我不知道为什么它不工作。下面是代码

$page = 0; 

if($page == 0){ 

    $page = "1"; 

}else{ 

    // If page is set, let's get it 
    echo $page = $_GET['page']; 

} 

// Now lets get all messages from your database 
$sql = "select * from about_us where sbmenu_id=9991"; 
$query = mysql_query($sql); 

// Lets count all messages 
$num = mysql_num_rows($query); 

// Lets set how many messages we want to display 
$per_page = "4"; 

// Now we must calculate the last page 
$last_page = ceil($num/$per_page); 

// And set the first page 
$first_page = "1"; 

// Here we are making the "First page" link 
echo "<a href='?page=".$first_page."'>First page</a> "; 
echo $page ; 
// If page is 1 then remove link from "Previous" word 
if($page == $first_page){ 

    echo "Previous "; 

    }else{ 

    if(!isset($page)){ 

      echo "Previous"; 

    }else{ 

      // But if page is set and it's not 1.. Lets add link to previous word to take us back by one page 
      $previous = $page-1; 
      echo "<a href='?page=".$previous."'>Previous</a> "; 

    } 

} 

// If the page is last page.. lets remove "Next" link 
if($page == $last_page){ 

    echo "Next "; 

    }else{ 

    // If page is not set or it is set and it's not the last page.. lets add link to this word so we can go to the next page 
    if(!isset($page)){ 

      $next = $first_page+1; 
      echo "<a href='?page=".$next."'>Next</a> "; 

    }else{ 

      $next = $page+1; 
      echo "<a href='?page=".$next."'>Next</a> "; 

    } 

} 

// And now lets add the "Last page" link 
echo "<a href='?page=".$last_page."'>Last page</a>"; 

// Math.. It gets us the start number of message that will be displayed 
$start = ($page-1)*$per_page; 

// Now lets set the limit for our query 
$limit = "LIMIT $start, $per_page"; 

// It's time for getting our messages 
$sql = "select * from about_us where sbmenu_id=9991 $limit"; 
$query = mysql_query($sql); 
?><table><?php 
echo "<br /><br /><tr style='height:70px;'>"; 

// And lets display our messages 
while($row = mysql_fetch_array($query) or die(mysql_error())) 
{....} 

Answer 1

变化前几行，然后尝试..

if(isset($_GET['page']) && $_GET['page'] > 0)  // If page is set, let's get it 
{ 
    echo $page = $_GET['page']; 
} 
else 
{ 
    $page = 1; 
} 

代替： -

$page = 0; 

if($page == 0){ 

$page = "1"; 

}else{ 

// If page is set, let's get it 
echo $page = $_GET['page']; 

}