Javascript json数据解析问题
我有以下数据:Javascript json数据解析问题,javascript,json,parsing,typescript,object,Javascript,Json,Parsing,Typescript,Object,我有以下数据: { “结果”:“{”绘图“:{”系列“:[{”颜色“:”绿色“,”标题“:”温度梯度“,”x单位“:”温度(°F)”,”y单位“:”深度(英尺“,”dashStyle“:”实心“,”线型“:”样条“,”x轴上顶点“:”真值“,”x单位“:”80.90.78101.56112.35123.33.91144.1994144.964166.96U值“:”26210.96U值0、-404.6、-809.2、-1213.8、-1618.4、-2023、-2427.6、-2832.2、-32
{
“结果”:“{”绘图“:{”系列“:[{”颜色“:”绿色“,”标题“:”温度梯度“,”x单位“:”温度(°F)”,”y单位“:”深度(英尺“,”dashStyle“:”实心“,”线型“:”样条“,”x轴上顶点“:”真值“,”x单位“:”80.90.78101.56112.35123.33.91144.1994144.964166.96U值“:”26210.96U值0、-404.6、-809.2、-1213.8、-1618.4、-2023、-2427.6、-2832.2、-3236.8、-3641.4、-4046、-4502、-4825\“}],“OperatingAgresult\”:null},
“状态代码”:200
}
我想得到两个新的变量
{Plot\\\:{“Series\”:[{“color\”:“green\”,“title\”:“Temperature Gradient\”,“x\u unit\”:“Temperature(°F)”,“y\u unit\”:“Depth(ft)”,“dashStyle\”:“Solid\”,“lineType\:“spline\”,“spline\”,“xAxisOnTop\”:true,“x\u值”:“80,90.78101.56112.35123.3133.91144.47155.96210.96u值:\“0,-404.6,-809.2,-1213.8,-1618.4,-2023,-2427.6,-2832.2,-3236.8,-3641.4,-4046,-4502,-4825\”},\“OperatingAgresult\”:null}
200
只需定义一个变量并使用点符号访问“结果”和“状态码”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
var jsonOBJ={
“结果”:“{”绘图“:{”系列“:[{”颜色“:”绿色“,”标题“:”温度梯度“,”x单位“:”温度(°F)”,”y单位“:”深度(英尺“,”dashStyle“:”实心“,”线型“:”样条“,”x轴上顶点“:”真“,”x单位“:”80.90.78101.56112.35123.33.91144.1994144.964166.116U值“:”8090.78101.78101.56112.35123.35123.3133.91144.96210.96210.24U值”\“0,-404.6,-809.2,-1213.8,-1618.4,-2023,-2427.6,-2832.2,-3236.8,-3641.4,-4046,-4502,-4825\”},\“OperatingAgresult\”:null}”,
“状态代码”:200
};
log(jsonOBJ.Result);
console.log(jsonOBJ.StatusCode);
首先为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);
在js中:
var obj=JSON.parse(JSON_字符串);变量1=目标结果;var variable2=obj.StatusCode
。什么是“问题”?这应该很简单!你有没有研究过JSON解析?有。但是我有一个错误:uncaughtsyntaxerror:JSON中位置1处的意外标记o:看起来您可以从JSON.parse(data.Result)
中获益,但是请发布一个,这样我们就可以看到您是如何读取字符串的