Mandrill应用程序返回此：[{"email":"steven@gmail.com","status":"sent","_id":"234we4fvba4a3e8517d7a9","reject_reason":null}]
我只需要得到"状态"的值。类似这样的东西：echo $result['status'];我如何在PHP中做到这一点？
使用json_decode()获取状态。。像这样：
<?php
$str = '[{"email":"steven@gmail.com","status":"sent","_id":"234we4fvba4a3e8517d7a9","reject_reason":null}]';
$json = json_decode($str, true);
echo $json[0]['status'];
?>
您可以像一样使用
$json = '[{"email":"steven@gmail.com","status":"sent","_id":"234we4fvba4a3e8517d7a9","reject_reason":null}]';
$json_array = json_decode($json);
print "<pre>";print_r($json_array[0]->status);