从php中的json

本文介绍了从php中的json_decode()获取值时出错?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个示例代码:

$description = '{"2G Network":"GSM 850 / 900 / 1800 / 1900 ","3G Network":"HSDPA 850 / 900 / 1700 / 1900 / 2100 "}'; $data = json_decode($description); echo $data->2G Network; // OR echo $data['2G Network'];

结果是错误，如何解决！

result is error, how to fix it !

推荐答案

尝试一下:

echo $data->{'2G Network'};

问题不在于JSON，而是您尝试访问的对象属性中有一个空格.如果使用花括号{ }，则可以使用字符串来命名要获取/设置的属性.

The problem wasn't with JSON, but that you had a space in the object property you were trying to access. If you use curly braces { }, then you can use strings to name the property you want to get/set.