我正在尝试使用 php 和 mysqli 读取 mysql 中的列,但出现此错误。 注意:试图获取非对象的属性
PHP
<?php
$db = new mysqli(HOST, USER, PASSWORD, DATABASE);
global $db;
$subscribers = $db->query('SELECT * FROM subscribers');
foreach ($subscribers as $subscriber) {
$name = $subscriber->name; // name is the name of the column
echo $name;
}
?>
数据库
Name Type
ID int(10)
name varchar(50)
email varchar(150)
confirmed varchar(50)
date datetime
我漏了什么,说实话对象章节没看懂。
最佳答案
因为你应该获取结果:
<?php
$db = new mysqli(HOST, USER, PASSWORD, DATABASE);
global $db;
$subscribers = $db->query('SELECT * FROM subscribers');
while ($subscriber = $subscribers->fetch_object()) {
$name = $subscriber->name; // name is the name of the column
echo $name;
}
关于php - 从mysql读取列时尝试获取非对象的属性,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22835833/