使用PHP在JSON中细分数据

Question

如何优雅地将下面的内容转换为JSON，以便“data”（即1,2,3,4和5）中的每个元素都是独立的，并且可以在不使用str_replace或preg_replace的情况下轻松遍历？

{
   "status":"success",
   "data":{
      "1":"this is an element",
      "2":"this is an element",
      "3":"this is an element",
      "4":"this is an element",
      "5":"this is an element",
    }
}

Answer 1

假设您已经拥有JSON字符串，请使用json_decode()，并将第二个参数提供为true，以获取已解码JSON的关联数组。

$str = '{"status":"success","data":{"1":"this is an element","2":"this is an element","3":"this is an element","4":"this is an element","5":"this is an element"}}'
$json = json_decode($str, true);

变量$json将是一个关联数组。