在PHP中讀取和存儲JSON數據中的變量

Question

我有以下ID號為JSON的數據。 在末尾：

[{"Name":"malek","Email":"malek@bar.com","Donation":"5","Type":"Cash","id":0},{"Name":"donnie","Email":"donnie@bar.com","Donation":"6","Type":"Cheque","id":1},{"Name":"ramanuj","Email":"ramanuj@bar.com","Donation":"3","Type":"NEFT","id":2},{"Name":"neha","Email":"neha@bar.com","Donation":"2","Type":"RTGS","id":3},{"Name":"aasha","Email":"aasha@bar.com","Donation":"1","Type":"CC","id":4},{"Name":"zia","Email":"zia@bar.com","Donation":"6","Type":"PayU","id":5},{"Name":"John","Email":"John@bar.com","Donation":"7","Type":"Instamojo","id":6}]

我要特別獲取John的Email：並將其存儲在變量$ johnemail中，以便$ johnemail =='John@bar.com'

最簡單的方法是什么？

Answer 1

使用json_decode ， array_column （從PHP 5.5開始可用）和array_search函數的簡單解決方案：

$json_data = '[{"Name":"malek","Email":"malek@bar.com","Donation":"5","Type":"Cash","id":0},{"Name":"donnie","Email":"donnie@bar.com","Donation":"6","Type":"Cheque","id":1},{"Name":"ramanuj","Email":"ramanuj@bar.com","Donation":"3","Type":"NEFT","id":2},{"Name":"neha","Email":"neha@bar.com","Donation":"2","Type":"RTGS","id":3},{"Name":"aasha","Email":"aasha@bar.com","Donation":"1","Type":"CC","id":4},{"Name":"zia","Email":"zia@bar.com","Donation":"6","Type":"PayU","id":5},{"Name":"John","Email":"John@bar.com","Donation":"7","Type":"Instamojo","id":6}]';

$search_name = 'John';
$items = json_decode($json_data, TRUE);
$emails = array_column($items, 'Name', 'Email');   // email is good for using as array key, as it should be unique
$johnemail = array_search($search_name, $emails);  // note that this is case-sensitive search 

print_r($johnemail);  // "John@bar.com"