Question

JSON看起来像这样

{
    73: {
        id: 74,
        title: "39",
        body: "11100000000000000000#IS#2.5",
        created_at: "2018-04-23 21:19:19",
        updated_at: "2018-04-23 21:19:19"
    }
}

我需要在不知道73是什么的情况下访问73的值。

echo $studentData->;

Answer 1

您可以使用reset()获取对象的第一个成员：

$json = '{"73":{"id":74,"title":"39","body":"11100000000000000000#IS#2.5","created_at":"2018-04-23 21:19:19", "updated_at":"2018-04-23 21:19:19"}}';

$data = json_decode($json);
$first = reset($data);
var_dump($first);

输出：

object(stdClass)#1 (5) {
    ["id"] => int(74)
    ["title"] => string(2) "39"
    ["body"] => string(27) "11100000000000000000#IS#2.5"
    ["created_at"] => string(19) "2018-04-23 21:19:19"
    ["updated_at"] => string(19) "2018-04-23 21:19:19"
}

Answer 2

一个选项是使用json_decode($jsonData, true)将json解码为数组，其中$ jsonData是您的json对象。这将为您提供一个数组，然后您可以获得该数组的第73个索引。

Answer 3

使用正则表达式的替代解决方案

$str = '{"73":{"id":74,"title":"39","body":"11100000000000000000#IS#2.5","created_at":"2018-04-23 21:19:19", "updated_at":"2018-04-23 21:19:19"}}';

preg_match('/{.*?({.*?}).*?}/s', $str, $matches, PREG_OFFSET_CAPTURE, 0);

$result = json_decode($matches[1][0]);

var_dump($result);

结果：

object(stdClass)[1]
public 'id' => int 74
public 'title' => string '39' (length=2)
public 'body' => string '11100000000000000000#IS#2.5' (length=27)
public 'created_at' => string '2018-04-23 21:19:19' (length=19)
public 'updated_at' => string '2018-04-23 21:19:19' (length=19)

使用php访问json而不知道元素值

3 个答案: