您好我正在尝试从PHP中的字符串处理和读取/解析数据。 var转储时的字符串看起来像这样
@Html.DropDownList("Status", (SelectList)ViewBag.NominationStatuses)
我希望能够存储
string(266) "{ "address": { "address_line1": "2391 US HIGHWAY 22 W", "address_line2": "", "address_city": "UNION", "address_state": "NJ", "address_zip": "07083-8517", "address_country": "US", "object": "address" } }"
,address_line1
,address_line2
,address_city
,address_state
和address_zip
变量,以便我可以随意操作它们。< / p>
我该怎么做呢。
感谢您的时间。
答案 0 :(得分:1)
您必须使用json_decode()
功能来检索您的数据:
$input = '{ "address": { "address_line1": "2391 US HIGHWAY 22 W", "address_line2": "", "address_city": "UNION", "address_state": "NJ", "address_zip": "07083-8517", "address_country": "US", "object": "address" } }';
$result = json_decode($input, true);
var_dump($result); // see whole array
echo $result['address']['address_line1']; // specific data
您可以从manual获取更多信息。
答案 1 :(得分:0)
您可以使用以下指令将字符串转换为JSON
$obj = json_decode($json);
然后您可以从JSON获取值,如下所示
$obj->address->address_line1;