我怎样才能优雅地将以下内容转换为JSON,以便"data"中的每个元素(即1,2,3,4和5)都是独立的,并且在不使用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",
}
}
假设你的意思是你已经有一个 JSON 字符串,请使用 json_decode()
,并提供第二个参数作为true
来获取解码 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。