PHP尝试在函数中获取非对象错误的属性

问题描述：

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?

需要一些帮助。我在查看数组的函数调用中得到“试图获取非对象属性”错误。 p>

我调用的数组是 p>

  $ var = array（
“variableA”=＆gt;“abc123”，
“variableB”=＆gt;“123456789”
）; 
  code>  pre> 
  
 我正在使用的函数是 p> 
 
 
 公共函数getJson（$ var）
 {
 $ resource = sprintf（“/ info /％s /％  s / json“，$ var-＆gt; variableA，$ var-＆gt; variableB）; 
返回$ this-＆gt; _restCall（'GET'，$ resource）; 
} 
  code>  pre  > 
 
 我无法理解为什么没有传递数组值？ p> 
 
 
有人可以帮忙吗？ p> 
  div>

答

$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);
 }

PHP尝试在函数中获取非对象错误的属性

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