dongwei5740 2019-08-20 04:20
浏览 64
已采纳

使用PHP从Google表格JSON数据中回显特定值

I'm trying to figure out "how-to" echo out the "tbenrud@gushd.net" (red arrow pointing to it) value from this JSON file. I can echo $data[feed][entry][1][id], but I run into trouble when I get to gsx$emailaddress. I assume it is because of the $ in the label. I do not have control over the label name (created by Google Sheets).

Any help is greatly appreciated.

enter image description here

My Code

$urlSheet = 'https://spreadsheets.google.com/feeds/list/1s7UCtbRY2dU3YdDu_kjnr-_y5Lh49yMHQRzgiDt8D4c/2/public/values?alt=json';
$jsonSheet = file_get_contents($urlSheet);
$data = json_decode($jsonSheet, true);

Thank you!

Todd

  • 写回答

2条回答 默认 最新

  • doukuo9116 2019-08-20 04:27
    关注

    If you use single quotes, the $ won't trigger PHP variable interpolation.

    echo $data['feed']['entry'][1]['gsx$emailaddress']['$t'];
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(1条)

报告相同问题?

悬赏问题

  • ¥20 Intellij实现web登录界面
  • ¥15 IPQ5018制作烧录固件,boot运行失败(操作系统-linux)(相关搜索:操作系统)(相关搜索:操作系统)
  • ¥20 icefall在librispeech基础上加入个人数据集
  • ¥30 keepalive高可用故障运维配置询问
  • ¥15 求帮助!国家电网内网u盘突然识别不出来了。
  • ¥15 matlab语音变速变调同时实现
  • ¥15 如何用Thoony写ESP32温湿度检测无源蜂鸣器报警代码?
  • ¥20 部件内部的CT图像数据集
  • ¥15 Visual studio调用动态库
  • ¥15 双目摄像头标定后的校准文件