如何在php中回显json对象属性？ [重复]

This question already has an answer here:

这个问题在这里已有答案:

• How can I access an array/object? 4 answers

如何访问数组/对象? 4个答案

JSON data

"orders":[  
     "billing_details":{  
        "company":"Test Company",
        "firstname":"Munadil",
        "postcode":"5000",
        "street":"Dhaka, Bangladesh",
        "email":"[e-mail]",
        "lastname":"Fahad",
        "ph_number":"880191111111",
        "city":"Dhaka",
        "state":"Mirpur",
        "country_code":"BN",
        "user_id":16003511,
        "salutation":null
     }]

In PHP

$json_output = json_decode($response);
foreach ( $json_output->orders as $orders ){
foreach ($orders->billing_details as $billing_details) {echo "<b>Name:</b><br>".$billing_details->firstname." ".$billing_details->lastname."<br>";}
}

But I am getting below error message,

但我收到以下错误信息,

Notice: Trying to get property of non-object in ....

注意:试图获取非对象的属性....

How can I echo data inside "billing_details" object under array "orders" ?

如何在数组“orders”下的“billing_details”对象内回显数据?

2 个解决方案

#1

Try this

$json_output = json_decode($response['orders']);

$ json_output = json_decode($ response ['orders']);

echo $json_output;

#2

Try this:

$json_output = json_decode($response);

foreach ($json_output['orders'] as $billing_details) {
echo "<b>Name:</b><br>$billing_details['firstname'] $billing_details['lastname']<br>";}