我正在研究Java Spring REST API,它提供了类似这样的JSON输出:
{
"speed1mbps": null,
"speed10mbps": null,
"speed100mbps": null,
"speed1000mbps": null,
"phone": null,
"NAME": "some name",
"STREET": "ST address",
"city": "somecity",
"state": "some state",
"zip": "232409",
}
但是我需要格式化的JSON输出,如下所示。
{
speed:[
1mbps:null
10mbps:null
100mbps:null
1000mbps:null
1000mbps:null
],
"phone": null,
"NAME": "some name",
"STREET": "ST address",
"city": "somecity",
"state": "some state",
"zip": "232409",
}
是否可以格式化JSON?请告诉我如何格式化JSON。
答案 0 :(得分:0)
如果要根据需要格式化JSON,则需要创建自定义dto,请参阅下面的示例:
class Dto {
private Map<String, String> speeds;
private String phone;
private String name;
private String street;
private String city;
private String state;
private String zip;
}
然后您可以以...等速度添加条目。
speeds.put("1mbps", null);
speeds.put("10mbps", null);
speeds.put("100mbps", null);
speeds.put("1000mbps", null);
然后有效的JSON结果将是....
{
speeds:{
1mbps:null,
10mbps:null,
100mbps:null,
1000mbps:null,
1000mbps:null
},
"phone": null,
"NAME": "some name",
"STREET": "ST address",
"city": "somecity",
"state": "some state",
"zip": "232409",
}
这是您应该返回REST响应的类。