我有一個下面的下拉列表中的php塊下循環。另一個php塊是彈出窗口使用facebox。現在如何通過點擊下拉內容將值傳遞給第二個php塊。傳遞值到相同的PHP文件
<?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>
////另一個是
<?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 }?>
我想在點擊時使用not_id – iftekhar143
您必須通過ajax或yo將值傳遞給第二塊(將其更改爲函數)你將不得不點擊提交頁面。 – VipindasKS
「$( 「view_comments 」)點擊(函數() { \t 變種ID = $(本).attr(「 ID」); $就({ 類型: 「POST」, URL: 「notifications.php」, 數據: 「not_id =」 + ID, 緩存:假, 成功:功能(數據){ 警報( 「工作」);} }); 返回假; });' – iftekhar143