传递带有分页的$_GET（或$_POST)变量（会话不工作）

Question

我知道这是一个常见的问题，因为我在决定发帖之前一直在寻找答案，但我似乎无法找到一个解决方案。

问题：我有一个分页脚本(PHP)用于我的搜索结果。显然常见的是，第一页的结果显示良好，然后在移动到第2、3页等时失败。

当点击页面2、3等时，我在搜索中使用的每个变量都有一个“未知索引”错误。

因此,I$_从表单中获得以下变量：

$_SESSION['var1']= $_GET['var1'];
$_SESSION['var2']= $_GET['var2'];
$_SESSION['var3']= $_GET['var3'];

$limit = 4; 

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

$stages = 3;
$page = mysql_escape_string(isset($_GET['page'])) ? (int)$_GET['page'] : 1;
if($page) {
  $start = ($page - 1) * $limit; 
} else {
  $start = 0;   
}   

// Get page data
$query1 = "SELECT * FROM $tableName ORDER BY joindate DESC 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 = 'hp1col'><div class='paginate'>";
$pagetotal = $total_pages.' Results';

// 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>";}                 
}
}

// Next
if ($page < $counter - 1){ 
$paginate.= "<a href='$targetpage?page=$next'>Next</a>";
} else {
$paginate.= "<span class='disabled'>Next</span>";
}
$paginate.= "</div></div>";     
}

while($row = mysql_fetch_array($result))
{
if (($part1 >= $_SESSION['var1']) AND  ($part2 <= $_SESSION['var2']) AND ($part3 <= $_SESSION['var3'])) {
echo 
"[Results]  
}
}
echo $paginate;

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

谢谢

Answer 1

因此您需要将这些查询参数传递到下一页。如果您的页面希望$_get['var1']出现，但您的URL中没有？var1=foo，那么它显然不会工作。处理此问题的最简单方法是http_build_query:

printf('<a href="%s?%s">Next</a>',
       $targetpage,
       http_build_query(array('page' => 2) + $_GET));

这将保留$_get中的所有当前值，并向其添加一个page=2参数。根据需要修改您的案例。