php json_解码删除具有空值的属性

我有一个Json字符串，我正在使用php的Json_decode对其进行解码

弦

            "address": {
                "address": null,
                "postalCode": null,
                "phoneNumber": "",
                "city": null
            }

当我解码字符串时，我得到

            ["address"]=>
                  array(1) {
                  ["phoneNumber"]=>
                       string(0) ""

它本质上是以null作为值剥离属性，即地址、城市。我能阻止这种事情发生吗

完全JSON

---

属性没有被剥离，您可能正在自己剥离它，执行如下所述的操作：

请参见您的代码示例：

$test = '{"address": {
            "address": null,
            "postalCode": null,
            "phoneNumber": "",
            "city": null
        }}';

$test_decoded = json_decode($test,true);
print_r($test_decoded);

//outputs as expected:
//Array ( [address] => Array ( [address] => [postalCode] => [phoneNumber] => [city] => ) )

---

无论如何，理论上，

null

与最初没有定义相同。这就是为什么未定义的变量在被访问时被视为null。提供完整的JSON。添加完整的JSON@Dharam@h2ooooooo你说得对，有别的东西在吞噬我的空值。

$test = '{"address": {
            "address": null,
            "postalCode": null,
            "phoneNumber": "",
            "city": null
        }}';

$test_decoded = json_decode($test,true);
print_r($test_decoded);

//outputs as expected:
//Array ( [address] => Array ( [address] => [postalCode] => [phoneNumber] => [city] => ) )