如何读取 json_decode 结果并将其转储？

Question

我的 json 回复有点问题。

我将它转储到我的 php 代码中，结果就是我在此处附加的内容

如何转储 statusText？

我已经尝试解码了，我也试过这个：

dump($myVar['statusText']);

或类似的东西

（为了获取我发布的 json 的转储，我刚刚做了 dump($myVar); )

JsonResponse {#325
  #data: "{"code":"OK","status":"ok","data":{"UUID":"f239ae18-98af-4224-8b4f-7713c71a5576","order":{something here },"orderRows":[something else here}}"
  #callback: null
  #encodingOptions: 271
  +headers: ResponseHeaderBag {#326
    #computedCacheControl: array:2 [
      "no-cache" => true
      "private" => true
    ]
    #cookies: []
    #headerNames: array:4 [
      "content-type" => "Content-Type"
      "access-control-allow-origin" => "Access-Control-Allow-Origin"
      "cache-control" => "Cache-Control"
      "date" => "Date"
    ]
    #headers: array:4 [
      "content-type" => array:1 [
        0 => "application/json; charset=utf-8"
      ]
      "access-control-allow-origin" => array:1 [
        0 => "*"
      ]
      "cache-control" => array:1 [
        0 => "no-cache, private"
      ]
      "date" => array:1 [
        0 => "Mon, 06 May 2019 08:15:28 GMT"
      ]
    ]
    #cacheControl: []
  }
  #content: "{"code":"OK","status":"ok","data":{"UUID":"f239ae18-98af-4224-8b4f-7713c71a5576","order":{something},"orderRows":[something else}}"
  #version: "1.0"
  #statusCode: 200
  #statusText: "OK"
  #charset: null
}

只是想看看 "status" => 'ok' 我的生活会很完美 :D

Answer 1

查看此输出，代码似乎也存储在内容中。

"status":"ok"

因此，在 getData() 方法中遵循 this documentation，您应该能够检索状态：

$data = $myVar->getData();
var_dump($data->status);

预计这将 return 一串 "ok"。

如何读取 json_decode 结果并将其转储？

How to read a json_decode result and dump it?

php

json

jsonresponse