我正在尝试根据成员资格级别选择数据库数据。如果成员为1级,他/她将显示1级数据,如果他/她是2级,则他们将看到1级和2级数据。
这是一个表格示例:
member| membership_level
------|----------------
john | 1
------|-----------------
andy | 2
然后我有PHP代码,它基于membership_level呈现数据,如下所示:
<?php if($userMember_level == 1) { ?>
Show data for membership level 1
<?php }elseif($userMember_level == 2) { ?>
Show data for membership level 1 And membership level 2
这是我的Mysqli声明:
//connect to database
require_once("../../db_query/connection.php");
$sql = "SELECT id,title,postdate FROM getting_started WHERE membership_level = ? AND active = ? AND membership_level = ? ORDER BY id DESC";
$stmt = $con->prepare($sql);
$stmt->bind_param('iii',$b=1,$a=1,$c=2);
$stmt->execute();
$stmt->store_result();
$numrow = $stmt->num_rows;
if($numrow >0){
$stmt->bind_result($id, $title, $postdate);
while($stmt->fetch()){ }
?>
<div class="resultWrap"> ...
我没有收到任何错误,但我也没有得到任何结果。
感谢您的帮助。
答案 0 :(得分:0)
您正在寻找的SQL是:
$sql = "
SELECT id,title,postdate
FROM getting_started
WHERE membership_level IN (?,?)
AND active = ?
ORDER BY id DESC
";
$stmt->bind_param('iii',1,2,1);