php报错,怎么解决undefined index这些呢?还有就是最下面为什么YEAR后面是null啊?

Question

<?php require("db_config.php"); $conn=mysqli_connect($mysql_server_name,$mysql_username,$mysql_password) or die("error connecting") ; mysqli_query($conn,"set names 'utf8'"); //数据库输出编码 mysqli_select_db($conn,$mysql_database); //打开数据库 $result = mysqli_query($conn,"select * from GDP"); $data=""; $array= array(); class User{ public $year; public $GDP; } while($row = mysqli_fetch_array($result,MYSQLI_ASSOC)){ $user=new User(); $user->year = $row['year']; $user->GDP = $row['GDP']; $array[]=$user; } $data=json_encode($array); // echo "{".'"user"'.":".$data."}"; echo $data;?>怎么解决notice的这些呢?还有就是最下面为什么YEAR后面是null啊?

Answer 1

$user->year=$row['year'];你这里循环的$row里面有year这个东西吗？报错说没有你查询的GDP表里里面应该没有这个字段吧

Answer 2

看看你数据库中 有year这个栏位吗?本回答被提问者采纳