douxiuyu2028 2016-08-23 01:20
浏览 6
已采纳

PHP尝试在函数中获取非对象错误的属性

Need some help please. I am getting a 'Trying to get property of non-object' error in a function call that looks at an array.

The array I am calling is

$var = array(
    "variableA" => "abc123",
    "variableB" => "123456789"
);

The function I am using is

  public function getJson($var)
    {
 $resource = sprintf("/info/%s/%s/json", $var->variableA, $var->variableB);
        return $this->_restCall('GET', $resource);
     }

I cant understand why the array values are not being passed through?

Could someone please help?

  • 写回答

1条回答 默认 最新

  • duanchao1002 2016-08-23 01:21
    关注

    $var is an array not an object. So you need to use array syntax, not object syntax:

    public function getJson($var)
    {
        $resource = sprintf("/info/%s/%s/json", $var['variableA'], $var['variableB']);
        return $this->_restCall('GET', $resource);
     }
    
    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 用vivado写数字逻辑实验报告撰写,FPGA实验
  • ¥15 为什么shp文件会有这种小方块?
  • ¥15 ecplise在连接数据库时显示加载驱动成功但是数据库连接失败
  • ¥15 visionmaster启动失败,提示为“机器不满足授权而被禁用”
  • ¥15 IDEA中圈复杂度如何具体设置
  • ¥50 labview采集不了数据
  • ¥15 Multisim红外倒车雷达仿真中距离问题
  • ¥15 请上面代码做什么处理或什么混淆
  • ¥15 英雄联盟自定义房间置顶
  • ¥15 W5500网线插上无反应