Javascript PHP:Json获取单个数据
我试图从JSON数据中获取所有员工的姓名 这是我的JSON数据Javascript PHP:Json获取单个数据,javascript,php,Javascript,Php,我试图从JSON数据中获取所有员工的姓名 这是我的JSON数据 [ { "0" : "65" , "id" : "65" , "1" : "etertet" , "employee_name" : "etertet" , "2" : "1" , "employee_salary": "1" , "3" : "2" , "employee_a
[ { "0" : "65"
, "id" : "65"
, "1" : "etertet"
, "employee_name" : "etertet"
, "2" : "1"
, "employee_salary": "1"
, "3" : "2"
, "employee_age" : "2"
}
]
这是我的JSON的功能:
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 $obj->employee_name;
?>
但我得到的错误如下:
注意:试图在第14行的C:\xampp\htdocs\cloud\v1\get.php中获取非对象的属性**
谢谢您需要循环JSON中的数组以获取所有名称:
foreach ($obj as $emp) {
echo $emp->employee_name . "<br>";
}
foreach($obj作为$emp){
echo$emp->employee_name.“
”;
}
您可以通过以下方式访问
$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
你能在DevsiOdedra上显示json数据吗。JSON数据位于图像上方。谢谢,我的意思是你可以在这里发布同样的代码。嗨@devsiodera,我已经添加了代码。非常感谢。