与数据库的连接成功,但是,当尝试实现我返回的结果时,我遇到了一些意外错误,不确定自己在做什么错
文件1
<?php
include 'dbConnect.php';
$conn = OpenCon();
echo "Connected Successfully";
//CloseCon($conn);
?>
<html>
<body>
<br>
Welcome <?php echo $_GET["name"]; ?><br>
Your email address : <?php echo $_GET["Email"]; ?>
<?php
$sql = "SELECT title, date, time FROM Performance p JOIN
Production r
ON p.Title=r.Title";
$result = $conn->query($sql); --- this is line 30
if ($result->num_rows > 0) { --- this is line 32
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<br> Title: ". $row["title"]. " - Date: ".
$row["date"]. " - Time: " . $row["time"] . "<br>";
}
} else {
echo "0 results";
}
?>
我得到的错误是
连接成功
警告:mysqli :: query():无法在第30行的C:\ xampp2 \ htdocs \ adam \ perf.php中获取mysqli
注意:尝试获取非对象的属性“ num_rows” 第32行的C:\ xampp2 \ htdocs \ adam \ perf.php 0个结果
对于我在哪里出错的任何帮助表示赞赏
编辑:
名为dbConnect的单独文件中与数据库的连接,该文件链接到
function OpenCon()
{
dbhost = "localhost";
$dbuser = "root";
$dbpass = "1234";
$db = "adam";
$conn = new mysqli($dbhost, $dbuser, $dbpass,$db) or die("Connect failed: %s\n". $conn -> error);
return $conn;
}
function CloseCon($conn)
{
$conn -> close();
}