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我正在开发一个基于电子购物的大学项目，我尝试使用此代码来实现我的结果，这没有错误但是当我按下一页链接或“按钮”时它不起作用请帮助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> ";

            }

            ?>

Answer 1

如果您访问此网址：http://example.com/somepage.php?key=val，则不会在PHP中自动获取变量$key。相反，您必须使用$_GET['key']，它将保留该值。 （在这种情况下：'val'）

因此，在代码开头的某处，添加以下内容：

if (isset($_GET['pagenum']) && $_GET['pagenum'] >= 1) {
    $pagenum = (int) $_GET['pagenum'];
} else {
    $pagenum = 1;
}

这不仅会创建$pagenum变量并从URL中提供值，还会确保该值是有效数字。如果没有，或者该网址不包含pagenum，则会将其设置为1。

可能的情况：

• pagenum不包含整数，而是包含字符串（甚至可能是SQL injection次尝试）
• pagenum是一个负数，或0
• pagenum未设置

在上述所有情况下，pagenum都设置为1。

如果pagenum包含一个浮点数（例如1.5.），则该值将转换为整数。在1.5的情况下，pagenum将变为1。

请记住，始终确保您清理用户输入。