我正在做一个基于电子购物的大学项目，我尝试了这个代码来实现我的结果，这没有错误，但当我按下下一页链接或"按钮"时它不起作用。请提供任何帮助，将不胜感激
<?php
$con = mysqli_connect("localhost","root","","php184_proj_db");
// Check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if (!(isset($pagenum)))
{
$pagenum = 1;
}
//Here we count the number of results
//Edit $qry to be your query
$qry = "SELECT * FROM posts";
$result = mysqli_query($con,$qry);
$row = mysqli_num_rows($result);
//This is the number of results displayed per page
$page_rows = 5;
//This tells us the page number of our last page
$last = ceil($row/$page_rows);
//this makes sure the page number isn't below one, or more than our maximum pages
if ($pagenum < 1)
{
$pagenum = 1; //Pagination of MySQL Query Results Setting the Variables
}
elseif ($pagenum > $last)
{
$pagenum = $last;
}
//This sets the range to display in our query
$max = 'limit ' .($pagenum - 1) * $page_rows .',' .$page_rows;
$j=0;
$qry = "SELECT * FROM posts $max";
$result = mysqli_query($con,$qry);
while($row = mysqli_fetch_array($result))
{
$j++;
echo "<p>";
// This shows the user what page they are on, and the total number Query and Results of pages
echo " --Page $pagenum of $last--
<p>";
// First we check if we are on page one. If we are then we don't need a
//link to the previous page or the first page so we do nothing. If we aren't
//then we generate links to the first page, and to the previous page.
if ($pagenum == 1)
{
}
else
{
echo " <a
href='{$_SERVER['PHP_SELF']}?pagenum=1'>
<<-First</a>
";
echo " ";
$previous = $pagenum-1;
echo " <a
href='{$_SERVER['PHP_SELF']}?
pagenum=$previous'> <-Previous</a>
";
}
//just a spacer
echo " ---- ";
//This does the same as above, only checking if we are on the last page,
//and then generating the Next and Last links
if ($pagenum == $last)
{
}
else {
$next = $pagenum+1;
echo " <a
href='{$_SERVER['PHP_SELF']}?pagenum=$next'>Next
-></a> ";
echo " ";
echo " <a
href='{$_SERVER['PHP_SELF']}?pagenum=$last'>Last
->></a> ";
}
?>
如果您访问此URL:http://example.com/somepage.php?key=val，则不会在PHP中自动获得变量$key。相反，您必须使用$_GET['key']，它将保存该值。（在这种情况下："val"）
因此，在代码的开头添加以下内容：
if (isset($_GET['pagenum']) && $_GET['pagenum'] >= 1) {
$pagenum = (int) $_GET['pagenum'];
} else {
$pagenum = 1;
}
这不仅创建了$pagenum变量并为其提供URL中的值，还确保该值是一个有效的数字。如果没有，或者URL不包含pagenum，则将其设置为1。
可能的情况：
pagenum包含的不是整数，而是字符串（甚至可能是SQL注入尝试）pagenum是负数，或0- 根本没有设置
pagenum
在上述所有情况下，pagenum被设置为1。
如果pagenum包含浮点值（例如1.5.），则该值将强制转换为整数。在1.5的情况下，pagenum将变为1。
请记住，始终确保对用户输入进行消毒。