也许你可以帮助我。数据库连接正常(Connected successfully
),表名正确(course_details
),但echo json_encode($data);
为NULL
<?php
include "db.php";
// Check connection
if ($con->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
$data=array();
$q=mysqli_query($con,"'SET CHARACTER SET utf8' select * from 'course_details'");
while ($row=mysqli_fetch_object($q)){
$data[]=$row;
}
echo json_encode($data);
$conn->close();
?>
提前致谢!
答案 0 :(得分:-1)
首先,我建议将查询包装在try catch块中
其次如上面的评论中所提到的,你在字符串中有2个查询。
then