<?php
include '../db_mysql/myqsl_conn.php';
session_start();
$comm_ID = $_SESSION['fromid'];
$sql = "SELECT * FROM ` comment` WHERE ` comment`.commodities_ID = '$comm_ID'";
$my = mysqli_query($db,$sql);
if($my){echo "ok";}
while($row = mysqli_fetch_assoc($my)){
echo $row['comment_work'];
var_dump($row);
}
?>输出结果:数据库:可以输出 commodities_ID 的值,但是就是输出不了comment_work 和 comment_name的值。
添加回答
举报
0/150
提交
取消