Question

我有一个包含JSON数据的字符串变量，如下所示。

var jsonstring = [{"latitude":"51.5263","longitude":"-0.120285","altitude":"","device":"123","rating":"5","region":"Europe","customer":"","time":"1-2 Weeks","error":"Error 1","application":"Phone","age":"< 1 Year"}, {"latitude":"27.58","longitude":"23.43","altitude":"","device":"Asc 140","rating":"4","region":"Africa","customer":"","time":"< 1 Week","error":"Error 1","application":"PCB","age":"1-3 Years"}, {"latitude":"39.55","longitude":"116.25","altitude":"","device":"CVB","rating":"4","region":"China","customer":"","time":"1-2 Weeks","error":"Error 2","application":"Sorting","age":"3-5 Years"}]

我想获取此字符串并将其转换为对象数组（，这将是$ scope.variable ），以便我能够单独访问每个对象。我尝试使用 JSON.parse（），但它将整个字符串放入一个对象而不是多个对象。

请帮助我。

Answer 1

您要使用JSON.parse解析整个字符串。

然后可以像任何其他阵列一样到达阵列中的每个对象，例如myArray[index]，myArray.map() / myArray.forEach()等

Answer 2

var jsonstring = '[{"latitude":"51.5263","longitude":"-0.120285","altitude":"","device":"123","rating":"5","region":"Europe","customer":"","time":"1-2 Weeks","error":"Error 1","application":"Phone","age":"< 1 Year"}, {"latitude":"27.58","longitude":"23.43","altitude":"","device":"Asc 140","rating":"4","region":"Africa","customer":"","time":"< 1 Week","error":"Error 1","application":"PCB","age":"1-3 Years"}, {"latitude":"39.55","longitude":"116.25","altitude":"","device":"CVB","rating":"4","region":"China","customer":"","time":"1-2 Weeks","error":"Error 2","application":"Sorting","age":"3-5 Years"}]';
$scope.variable = JSON.parse(jsonstring);

Answer 3

[{ “纬度”： “51.5263”， “经度”： “ - 0.120285”， “高度”： “”， “设备”： “123”， “等级”： “5”， “区域”：”欧洲“，”客户“：”“，”时间“：”1-2周“，”错误“：”错误1“，”申请“：”电话“，”年龄“：”＆lt; 1 Year“}， {“纬度”：“27.58”，“经度”：“23.43”，“海拔”：“”，“设备”：“Asc 140”，“评级”：“4”，“地区”：“非洲”，“ customer“：”“，”time“：”＆lt; 1 Week“，”error“：”Error 1“，”application“：”PCB“，”age“：”1-3 Years“}， { “纬度”： “39.55”， “东经”： “116.25”， “高度”： “”， “设备”： “CVB”， “等级”： “4”， “区”： “中国”，“客户“：”“，”time“：”1-2周“，”错误“：”错误2“，”申请“：”排序“，”年龄“：”3 - 5年“}]

这是一个数组对象。它不是一个字符串对象。

你可以这样再试一次：

var jsonString =“[]”; var json = JSON.parse（jsonString）;

Answer 4

JSON.parse()方法将字符串解析为JSON。您的问题中的代码显示您正在尝试解析JS对象而不是字符串。

在下面的示例中，如果您尝试解析JS对象，则会收到错误。

var jsonstring = [{},{}];
JSON.parse(jsonstring); // ERROR Uncaught SyntaxError: Unexpected token o

以下工作（请注意jsonstring是一个字符串，而不是这里的对象）：

var jsonstring = '[{},{}]';
JSON.parse(jsonstring); // OK

从String变量解析JSON数据并将其转换为$ scope.variable中的对象

4 个答案: