Question

我在html下面有一个用于下拉列表的php块，它在循环下。另一个php块用于使用facebox的弹出窗口。现在如何通过单击下拉内容将值传递给第二个php块。

<?php


         $msql="select notification_id,applicant_id,notification_dis from app_notification where applicant_id=$app_id order by notification_id desc " ;
        $result_msql=$bd->query($msql);
        while($messagecount=$result_msql->fetch_assoc())

        {
        $not_id=    $messagecount['notification_id'];
        $comment=$messagecount['notification_dis'];


        ?>




            <div  class="comment_ui" >



        <a href="#info"  class='view_comments' rel='facebox' id="<?php echo $not_id; ?>"></br>   <?php echo "check1---$not_id".$comment."";?>
                </a>                  

            </div>

////另一个是

      <?php



        echo $_POST['not_id'];  
       $mysql_hostname = "localhost";
      $mysql_user = "root";
      $mysql_password = "";
   $mysql_database = "test";
   $prefix = "";

        $bd = new mysqli($mysql_hostname, $mysql_user,           $mysql_password,$mysql_database) or die("Opps some thing went wrong");


            echo "check2---$not_id";

            $msql_show="select  header_type,notification_main,notification_dis from app_notification where notification_id=$not_id";
            $show_msql=$bd->query($msql_show);

            while($message_show=$show_msql->fetch_assoc())
            {
                $header=$message_show['header_type'];
                $main=$message_show['notification_main'];


            $msql_multi="SELECT file_name FROM biz_logo WHERE    w_id=(SELECT w_id FROM post_job
            WHERE job_id=(SELECT job_id FROM job_notice WHERE 
            notice_id=(SELECT notice_id FROM app_notification where notification_id=$not_id)))";
            }
            $msql_multi=$bd->query($msql_multi);
            while($message_multi=$msql_multi->fetch_assoc())
            {

              $file_name=$message_multi['file_name'];



            }



            print "<center><h1>".$header."$not_id</h1></center>";
    print"<span style='margin-right:50em'><table> 
                 <p><t>".$file_name."</t></p><tr><td>".$main."</td>
    </tr> <tr><td></td></tr></table><span> ";              
            ?>





        <?php }?>

Answer 1

in php manually you can use sessions or if you need

$_SESSION["favcolor"] = "green";
$_SESSION["favanimal"] = "cat";


<?php
// Echo session variables that were set on previous page
echo "Favorite color is " . $_SESSION["favcolor"] . ".<br>";
echo "Favorite animal is " . $_SESSION["favanimal"] . ".";
?>

or if you need use html you can use inputbox and post get method

将值传递给相同的php文件

1 个答案: