从php中的多级json获取字符串

我试图从json中获取一些字符串，但尚未成功：(

我尝试了许多代码，但仍然没有运气，不能完全理解这一点

json示例是：

{
    "nginx_version": "1.7.11.3 Gryphon",
    "nginx_rtmp_version": "1.1.4",
    "built": "Mar 19 2015 20:36:41",
    "pid": "4680",
    "uptime": "1876",
    "naccepted": "3",
    "bw_in": "404904",
    "bytes_in": "83699680",
    "bw_out": "404912",
    "bytes_out": "83700478",
    "server": {
        "application": [{
            "name": "msdk",
            "live": {
                "stream": {
                    "name": "xkiz531",
                    "time": "1854519",
                    "bw_in": "399752",
                    "bytes_in": "82628640",
                    "bw_out": "399752",
                    "bytes_out": "82628588",
                    "bw_audio": "178480",
                    "bw_video": "221272",
                    "client": [{
                        "id": "2788",
                        "address": "example.com\/live\/live",
                        "time": "1854519",
                        "flashver": "ngx-local-relay",
                        "dropped": "0",
                        "avsync": "-14",
                        "timestamp": "1853566",
                        "active": []
                    }, {
                        "id": "2787",
                        "address": "197.14.103.17",
                        "time": "1854987",
                        "flashver": "FMLE\/3.0 (compatible; FMSc\/1.0)",
                        "swfurl": "rtmp:\/\/example.com\/msdk",
                        "dropped": "0",
                        "avsync": "-14",
                        "timestamp": "1853566",
                        "publishing": [],
                        "active": []
                    }],
                    "meta": {
                        "video": {
                            "width": "1280",
                            "height": "720",
                            "frame_rate": "30",
                            "codec": "H264",
                            "profile": "High",
                            "compat": "0",
                            "level": "3.1"
                        },
                        "audio": {
                            "codec": "AAC",
                            "profile": "LC",
                            "channels": "2",
                            "sample_rate": "44100"
                        }
                    },
                    "nclients": "2",
                    "publishing": [],
                    "active": []
                },
                "nclients": "2"
            }
        }, {
            "name": "test",
            "live": {
                "nclients": "0"
            }
        }]
    }
}

我现在的代码是：

$array = json_decode($json,TRUE);

var_dump($array);
foreach($array['items'] as $item) {
    echo $item['server']['application']['live']['stream']['name'];
}

我想获取字符串上这些键的值：

---

bw_音频、bw_视频、宽度、高度、帧速率、编解码器、配置文件、音频编解码器和采样率

您在JSON中没有

项

键，

应用程序是一个数组，因此您应该执行以下操作：

$array = json_decode($json,TRUE);
echo $array['server']['application'][0]['live']['stream']['name'];

注意：如果可能有多个应用程序，请在数组上迭代
foreach
，而不是使用
[0]
我在JSON中没有看到
项。
application
是一个数组。您需要使用
['application'][0]
或对其进行迭代。您想要获取键
bw_音频
，
bw_视频
，还是想要获取这些键的值？键的值…………很抱歉，我将更新我的问题我仍然看不到
$array['items]的位置
是。我假设您显示的JSON是
$array
@chedlywalid您在JSON中没有
项
键。我编辑了我的答案。