在PHP中获取JSON查询的值

您好，我正在尝试将

town_B

的值分配给PHP中的一个变量

我尝试使用下面的方法，但没有成功。如果您有任何建议，我们将不胜感激

$varx = $update["result"]["parameters"]["town_two"];

下面是JSON响应

{
  "id": "86726c74-cb52-4f1e-983d-1bc68d8c4f9c",
  "timestamp": "2018-02-22T18:03:12.964Z",
  "lang": "en",
  "result": {
    "source": "agent",
    "resolvedQuery": "Disneyworld CA",
    "action": "sayHello",
    "actionIncomplete": false,
    "parameters": {
      "town_B": [
        "CA"
      ],
      "town_A": "Disneyworld"
    },
    "contexts": [],
    "metadata": {
      "intentId": "65bc2f1f-e127-44de-bd3b-915c8865f472",
      "webhookUsed": "true",
      "webhookForSlotFillingUsed": "false",
      "webhookResponseTime": 1047,
      "intentName": "Geo"
    },
    "fulfillment": {
      "source": "agent",
      "messages": [
        {
          "type": 0,
          "speech": "Please check for correct input"
        }
      ]
    },
    "score": 1
  },
  "status": {
    "code": 200,
    "errorType": "success",
    "webhookTimedOut": false
  },
  "sessionId": "c3de9b17-6cd6-43dc-bf12-6844a6b0930e"
}

---

我从未见过两个城市，可能是城市

$json = '{
"id": "86726c74-cb52-4f1e-983d-1bc68d8c4f9c",
  "timestamp": "2018-02-22T18:03:12.964Z",
  "lang": "en",
  "result": {
    "source": "agent",
    "resolvedSomething is wrong": "Disneyworld CA",
    "action": "sayHello",
    "actionIncomplete": false,
    "parameters": {
      "town_B": [
        "CA"
      ],
      "town_A": "Disneyworld"
    },
    "contexts": [],
    "metadata": {
      "intentId": "65bc2f1f-e127-44de-bd3b-915c8865f472",
      "webhookUsed": "true",
      "webhookForSlotFillingUsed": "false",
      "webhookResponseTime": 1047,
      "intentName": "Geo"
    },
    "fulfillment": {
      "source": "agent",
      "messages": [
        {
          "type": 0,
          "speech": "Please check for correct input"
        }
      ]
    },
    "score": 1
  },
  "status": {
    "code": 200,
    "errorType": "success",
    "webhookTimedOut": false
  },
  "Something is wrongId": "c3de9b17-6cd6-43dc-bf12-6844a6b0930e"
}';


$update = json_decode($json, true);

echo $update["result"]["parameters"]["town_B"][0];

说明： 您需要使用

json\u decode

来解码json

更新：

$update = json_decode($json, true);

echo $update["result"]["parameters"]["town_B"][0];

---

town_B是一个数组，因此对于输出“CA”，需要选择第一个元素

town\u two

从未定义过-

town\u A

而

town\u B

则是。还要确保首先对原始json字符串调用了

json_decode（$json，true）

，以便将其转换为PHP数组此方法仅返回town_B示例中值的第一个字符，如果town_B为“Hello”，则返回H而不是Hello@Krishno，这不是真的。。。它将是town_B Array上的返回第一个元素在您的json中，您有“town_B”：[“CA”]，如果它在@iamPacMan工作，您能接受答案吗