json数据解析的问题答案

【问题标题】：Issue with json data parsingjson数据解析的问题
【发布时间】：2018-04-15 03:47:39
【问题描述】：

我有以下数据：

{
  "Result": "{\"Plot\":{\"Series\":[{\"color\":\"green\",\"title\":\"Temperature Gradient\",\"x_unit\":\"Temperature (°F)\",\"y_unit\":\"Depth (ft)\",\"dashStyle\":\"Solid\",\"lineType\":\"spline\",\"xAxisOnTop\":true,\"x_values\":\"80,90.78,101.56,112.35,123.13,133.91,144.69,155.47,166.26,177.04,199.96,210.63,220\",\"y_values\":\"0,-404.6,-809.2,-1213.8,-1618.4,-2023,-2427.6,-2832.2,-3236.8,-3641.4,-4046,-4502,-4825\"}]},\"OperatingTagResult\":null}",
  "StatusCode": 200
}

我想得到两个新变量。

{\"Plot\":{\"Series\":[{\"color\":\"green\",\"title\":\"Temperature Gradient\",\"x_unit\":\"Temperature (°F)\",\"y_unit\":\"Depth (ft)\",\"dashStyle\":\"Solid\",\"lineType\":\"spline\",\"xAxisOnTop\":true,\"x_values\":\"80,90.78,101.56,112.35,123.13,133.91,144.69,155.47,166.26,177.04,199.96,210.63,220\",\"y_values\":\"0,-404.6,-809.2,-1213.8,-1618.4,-2023,-2427.6,-2832.2,-3236.8,-3641.4,-4046,-4502,-4825\"}]},\"OperatingTagResult\":null}
只需 200 来自“StatusCode”

我该如何实现？

使用 json 解析时出现下一个错误：

未捕获的 SyntaxError：位置 1 处 JSON 中的意外标记 o

【问题讨论】：

在js中：var obj = JSON.parse(JSON_STRING); var variable1 = obj.Result; var variable2 = obj.StatusCode.
什么是“问题”？这应该很简单！
你有没有尝试研究过 JSON 解析？
是的。但我有错误： Uncaught SyntaxError: Unexpected token o in JSON at position 1:
看起来您可以从JSON.parse(data.Result) 中受益，但请发布minimal reproducible example，以便我们了解您如何阅读字符串

标签： javascript json parsing typescript object

【解决方案1】：

只需定义一个变量并使用点符号来访问“Result”和“StatusCode”属性：

var income = {"Result":"{\"Plot\":{\"Series\":[{\"color\":\"green\",\"title\":\"Temperature Gradient\",\"x_unit\":\"Temperature (°F)\",\"y_unit\":\"Depth (ft)\",\"dashStyle\":\"Solid\",\"lineType\":\"spline\",\"xAxisOnTop\":true,\"x_values\":\"80,90.78,101.56,112.35,123.13,133.91,144.69,155.47,166.26,177.04,199.96,210.63,220\",\"y_values\":\"0,-404.6,-809.2,-1213.8,-1618.4,-2023,-2427.6,-2832.2,-3236.8,-3641.4,-4046,-4502,-4825\"}]},\"OperatingTagResult\":null}","StatusCode":200}

var result = income.Result; // {"Plot":{"Series":[{"color":"green", ...
var status = income.StatusCode; // 200

【讨论】：

【解决方案2】：

    var jsonOBJ = {
      "Result": "{\"Plot\":{\"Series\":[{\"color\":\"green\",\"title\":\"Temperature Gradient\",\"x_unit\":\"Temperature (°F)\",\"y_unit\":\"Depth (ft)\",\"dashStyle\":\"Solid\",\"lineType\":\"spline\",\"xAxisOnTop\":true,\"x_values\":\"80,90.78,101.56,112.35,123.13,133.91,144.69,155.47,166.26,177.04,199.96,210.63,220\",\"y_values\":\"0,-404.6,-809.2,-1213.8,-1618.4,-2023,-2427.6,-2832.2,-3236.8,-3641.4,-4046,-4502,-4825\"}]},\"OperatingTagResult\":null}",
      "StatusCode": 200
    };
    
    console.log(jsonOBJ.Result);
    console.log(jsonOBJ.StatusCode);

你不需要解析你的 json 因为已经是一个 jsonObject。尝试解析它 - JSON.parse(jsonString) 然后获取值，您会发现您遇到的错误

【讨论】：

【解决方案3】：

首先为你的 json 定义一个变量；

var Temperature = {"Result":"{\"Plot\":{\"Series\":[{\"color\":\"green\",\"title\":\"Temperature Gradient\",\"x_unit\":\"Temperature (°F)\",\"y_unit\":\"Depth (ft)\",\"dashStyle\":\"Solid\",\"lineType\":\"spline\",\"xAxisOnTop\":true,\"x_values\":\"80,90.78,101.56,112.35,123.13,133.91,144.69,155.47,166.26,177.04,199.96,210.63,220\",\"y_values\":\"0,-404.6,-809.2,-1213.8,-1618.4,-2023,-2427.6,-2832.2,-3236.8,-3641.4,-4046,-4502,-4825\"}]},\"OperatingTagResult\":null}","StatusCode":200}

then call the required filed you are required.

var income = {"Result":"{\"Plot\":{\"Series\":[{\"color\":\"green\",\"title\":\"Temperature Gradient\",\"x_unit\":\"Temperature (°F)\",\"y_unit\":\"Depth (ft)\",\"dashStyle\":\"Solid\",\"lineType\":\"spline\",\"xAxisOnTop\":true,\"x_values\":\"80,90.78,101.56,112.35,123.13,133.91,144.69,155.47,166.26,177.04,199.96,210.63,220\",\"y_values\":\"0,-404.6,-809.2,-1213.8,-1618.4,-2023,-2427.6,-2832.2,-3236.8,-3641.4,-4046,-4502,-4825\"}]},\"OperatingTagResult\":null}","StatusCode":200}

alert(Temperature.Result);
alert(Temperature.StatusCode);

【讨论】：