Question

我需要PHP方式才能在PHP中显示store_name

我尝试了以下但没有成功

    <?php
require_once("admin/system/core.php");
$fetch = new server();
$fetch->connect();
$store = $fetch->getstore("movies");

print $store;
?>

JSON看起来像这样 [{"_id":{"$id":"4f67da1538fc5d7347000000"},"store_name":"movies","categories":{"name":"hoyts","products":{"name":"GoldClass","Price":"12.00","CashBack":"2.00"}}}]

Answer 1

我假设返回的值是json字符串。所以你必须decode从字符串到object json_decode函数

$store = $fetch->getstore("movies");
$data  = json_decode($store);

foreach($data as $d) {
    echo $d->store_name;
}

Answer 2

假设你的数组将是

$array = array('id'=>'4f67da1538fc5d7347000000',
        'store_name'=>'movies');
echo $json = json_encode($array);

它回声

{"id":"4f67da1538fc5d7347000000","store_name":"movies"}

然后你可以使用

$data = json_decode($json);
echo $data->store_name;

echo的movies

仅打印JSON字符串的一部分

2 个答案: