dongling3243 2015-08-19 13:47
浏览 576
已采纳

警告:json_decode()期望参数1为字符串

I insert PHP array to mySql using json_encodemethod like this :

["11","10","4"]

Now I need to convert to php array:

$me = ["11","10","4"];
$you = json_decode($me, true);
echo $you;

But in result I see : Warning: json_decode() expects parameter 1 to be string, array given in C:\xampp\htdocs\test\test.php on line 5

How do fix this?!

  • 写回答

1条回答 默认 最新

  • douhong6187 2015-08-19 13:49
    关注

    Your problem is that $me isn't a string. You should simply encapsulate it in single-quotes to change this.

    $me = '["11","10","4"]';
    $you = json_decode($me);
    print_r($you);  // becasue its now a PHP array, 
                    // copy/paste will get you every time
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥20 matlab计算中误差
  • ¥15 对于相关问题的求解与代码
  • ¥15 ubuntu子系统密码忘记
  • ¥15 信号傅里叶变换在matlab上遇到的小问题请求帮助
  • ¥15 保护模式-系统加载-段寄存器
  • ¥15 电脑桌面设定一个区域禁止鼠标操作
  • ¥15 求NPF226060磁芯的详细资料
  • ¥15 使用R语言marginaleffects包进行边际效应图绘制
  • ¥20 usb设备兼容性问题
  • ¥15 错误(10048): “调用exui内部功能”库命令的参数“参数4”不能接受空数据。怎么解决啊