javascript - PHP: Json 获取个人数据

Question

我正在尝试从我的 JSON 数据中获取所有员工姓名。

这是我的 JSON 数据

[ { "0"              : "65"
  , "id"             : "65"
  , "1"              : "etertet"
  , "employee_name"  : "etertet"
  , "2"              : "1"
  , "employee_salary": "1"
  , "3"              : "2"
  , "employee_age"   : "2"
  } 
] 

这是我的 JSON 函数：

function get_employees($id=0)
  {
    global $connection;
    $query="SELECT * FROM employee";
    if($id != 0)
      {
        $query.=" WHERE id=".$id." LIMIT 1";
      }
    $response=array();
    $result=mysqli_query($connection, $query);
    while($row=mysqli_fetch_array($result))
      {
        $response[]=$row;
      }
    header('Content-Type: application/json');
    echo json_encode($response);
  }

这是我获取 JSON 数据的代码

<?php
  # An HTTP GET request example

  $url = 'http://localhost/cloud/v1/employees';
  $ch = curl_init($url);
  curl_setopt($ch, CURLOPT_TIMEOUT, 5);
  curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, 5);
  curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
  $data = curl_exec($ch);
  curl_close($ch);
  //echo $data;
  $obj = json_decode($data);
  echo $obj->employee_name;
?>

但我收到以下错误：

注意：尝试在第 14 行获取 C:\xampp\htdocs\cloud\v1\get.php 中非对象的属性**

谢谢

标签： javascriptphp

Answer 1

您可以访问如下

$ar = '[{"0":"65","id":"65","1":"etertet","employee_name":"etertet","2":"1","employee_salary":"1","3":"2","employee_age":"2"}]
';



$data = json_decode($ar);

echo  $data[0]->employee_name