Question

从以以下格式存储的缓存中检索

{address={address1=string, address2=string, addressId=2045, city=string, fxGeocode=string, 
houseNumber=string, isActive=true, postalCode=string, state=string, streetName=string, zip=string}, 
externalIds=[{externalId=string, externalIdDocId=38915437-69d7-449a-9bd8-9832c78b2010, 
partyRoleExternalIdType=string}], name=string, organizationRoleId=990}

我必须将其转换回json对象。最好的方法是什么？

转换后的json对象采用以下格式

{
"address": {
    "address1": "string",
    "address2": "string",
    "addressId": "2045",
    "city": "string",
    "fxGeocode": "string",
    "houseNumber": "string",
    "isActive": "true",
    "postalCode": "string",
    "state": "string",
    "streetName": "string",
    "zip": "string"
},
"externalIds": [
    {
        "externalId": "string",
        "externalIdDocId": "38915437-69d7-449a-9bd8-9832c78b2010",
        "partyRoleExternalIdType": "string"
    }
],
"name": "string",
"organizationRoleId": "990"
}

Answer 1

我会用Jackson JSON库来做。

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-core</artifactId>
    <version>2.10.1</version>
</dependency>

这应该为您提供所需的输出：

ObjectMapper mapper = new ObjectMapper();

Map<String, Map<String, String>> myMap = new HashMap<>();

// add stuff to your map of maps

// pretty print
String json = mapper.writerWithDefaultPrettyPrinter().writeValueAsString(myMap);

System.out.println(json);

Answer 2

您可以使用com.fasterxml.jackson.core库。

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
    <version>2.9.5</version>
</dependency>

     and use objectMapper to convert this object into JSON.
     //Object to JSON in String

       ObjectMapper mapper = new ObjectMapper();
      String jsonInString = mapper.writeValueAsString(object);

      // beautify JSON
      String json = mapper.writerWithDefaultPrettyPrinter().writeValueAsString(object);

Answer 3

您可以使用#include <string> using namespace std; class MyFloat { protected: char* mData; public: MyFloat(float v) { cout << "using constructer\n"; mData = new char[10]; int k=snprintf(mData, sizeof(mData),"%f", v); } operator char *() const { cout << "using convert\n"; return _strdup(mData); } ~MyFloat() { cout << "using des\n"; delete[] mData; } MyFloat(const MyFloat& a) { cout << "using copy\n"; mData = new char[strlen(a.mData) + 1]; mData = _strdup(a.mData); } MyFloat& operator=(const MyFloat& a) { cout << "using operator\n"; mData = new char[strlen(a.mData) + 1]; mData = _strdup(a.mData); return *this; } friend istream& operator>>(istream& in, MyFloat& a) { float t; in >> t; snprintf(a.mData, sizeof(a.mData), "%f", t); return in; } friend ostream& operator<<(ostream& out, const MyFloat& a) { out.write(a.mData, sizeof(a.mData)); return out; } static MyFloat ValueOf(float value) { cout << "using static\n"; MyFloat temp(value); return temp; } }; int main() { char* sValue = MyFloat::ValueOf(123.45); cout << sValue; MyFloat temp(2.3); cin >> temp; cout << temp; getchar(); return 0; }类将Java类对象转换为jason，将json转换为Java类对象

ObjectMapper

输出

{“ address”：{“ address1”：“ string”，“ address2”：“ string”}}

Answer 4

Gson是Google提供的用于json序列化/反序列化的优秀库。 Maven依赖项：

<dependency>
  <groupId>com.google.code.gson</groupId>
  <artifactId>gson</artifactId>
  <version>2.8.6</version>
</dependency>

示例序列化：

Gson gson = new Gson();

Map<String, Map<String, String>> someMap = new HashMap<>();

try (FileWriter writer = new FileWriter("C:\\folder\\file.json")) {
    gson.toJson(someMap, writer);
} catch (IOException e) {
    e.printStackTrace();
}

如何将Map <String，TreeMap <String，String >>字符串转换为Java中的Json对象？

4 个答案: