Question

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＆＃34; StatusCode＆＃34;

我该如何实现？

使用json parse时出现了下一个错误：

未捕获的SyntaxError：位置1的JSON中的意外标记o

Answer 1

只需定义一个变量并使用点表示法来访问“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

Answer 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 beacuse已经是一个jsonObject了。尝试解析它 - JSON.parse(jsonString)，然后获取值，你会发现你有错误

Answer 3

首先为你的json定义一个变量;

json数据解析问题

3 个答案: