Spring 查詢轉換為嵌套的 JSON 結構

Question

I'm new to spring and Java and trying to figure out how to go about formatting the json response into the desired structure.

我有一個 spring 查詢，它從如下表中返回 2 列，這確實是 json 結構所需的鍵和值：

名稱	價值觀
車	豐田
自行車	施溫
小型摩托車	Razor
A0	11
A1	12
A2	13
B0	2000
B1	4000
B2	22000

來自 controller 的當前 json output 是這樣的：

[{
        "names": "Car",
        "values": "Toyota"
    },
    {
        "names": "Bike",
        "values": "Schwinn"
    },
    {
        "names": "Scooter",
        "values": "Razor"
    },
    {
        "names": "A0",
        "values": "11"
    },
    {
        "names": "A1",
        "values": "12"
    },
    {
        "names": "A2",
        "values": "13"
    },
    {
        "names": "B0",
        "values": "2000"
    },
    {
        "names": "B1",
        "values": "4000"
    },
    {
        "names": "B2",
        "values": "22000"
    }
]

所需的 json 格式是刪除表列名稱的地方，而是使用鍵的名稱列創建 json 結構：

{
    "Car": "Toyota",
    "Bike": "Schwinn",
    "Scooter": "Razor",
    "Data": [{
        "A0": "11",
        "B0": "2000"
    }, {
        "A1": "12",
        "B1": "4000"
    }, {
        "A2": "13",
        "B2": "22000"
    }]
}

存儲庫

   @Query (value = "Select names, values ... :id")
    List<Data> findData(@Param("id") Long id) ;

      interface Data {
        String getnames();
        String getvalues();
    }

服務

    public List<Data> getData(Long id) {return repo.findData(id);}

Controller

    @GetMapping("/getdata/{id}")
    public ResponseEntity<List<Data>> getData(@PathVariable Long id) {
            List<Data> c = service.getData(id);
            return new ResponseEntity<>(c, HttpStatus.OK);

    }

似乎我需要處理結果集並需要遍歷它們以創建所需的結構，但不確定如何繼續進行，或者可能有更簡單的方法來獲得所需的結構。 任何指導將不勝感激。

Answer 1

所以返回一個 ResponseEntity<Map<String, Object>> 而不是一個 List 來模擬一個 Json object。

List<Data> c = service.getData(id);

Map<String, Object> map = new HashMap<>();
map.put("Key", "Value");
map.put("Car", c.get(0).getvalues());
map.put("Entire List", c);

return new ResponseEntity<>(c, HttpStatus.OK);

顯然，您必須編寫自己的邏輯，但它應該非常簡單。 或者，更好的是，如果您要經常使用它，請考慮為返回的 object 制作一個 class，然后返回 ResponseEntity< YourCustomObject >

Answer 2

這看起來有點復雜，我認為您應該為 A0 B0 之類的值設置主鍵關聯

import com.black_dragon.utils.JacksonUtils;

import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.regex.Pattern;
import java.util.stream.Collectors;

import static java.util.stream.Collectors.groupingBy;

/**
 * @author black_dragon
 * @version V1.0
 * @Package com.black_dragon.swing
 * @date 2022/9/6 10:35
 * @Copyright
 */
public class ConvertToMap {
    String names;
    String values;

    public String getNames() {
        return names;
    }

    public void setNames(String names) {
        this.names = names;
    }


    public String getValues() {
        return values;
    }

    public void setValues(String values) {
        this.values = values;
    }

    private static String DIGIT_REGEX = "[^0-9]";

    private static String LETTER_DIGIT_REGEX = "[a-zA-Z]+";

    public static Integer getDigit(String str){
        Pattern pattern = Pattern.compile(DIGIT_REGEX);
        if(!isLetterDigit(str)){
            String[] keySet = pattern.split(str);
            if(keySet.length > 0){
                return Integer.valueOf(keySet[1]);
            }
        }
        return -1;
    }

    public static boolean isLetterDigit(String str){
        return str.matches(LETTER_DIGIT_REGEX);
    }

    private static String fetchGroupKey(ConvertToMap convertToMap){
        return String.valueOf(getDigit(convertToMap.names));
    }

    public static void main(String[] args) {
        String jsonString = "[{\n" +
                "        \"names\": \"Car\",\n" +
                "        \"values\": \"Toyota\"\n" +
                "    },\n" +
                "    {\n" +
                "        \"names\": \"Bike\",\n" +
                "        \"values\": \"Schwinn\"\n" +
                "    },\n" +
                "    {\n" +
                "        \"names\": \"Scooter\",\n" +
                "        \"values\": \"Razor\"\n" +
                "    },\n" +
                "    {\n" +
                "        \"names\": \"A0\",\n" +
                "        \"values\": \"11\"\n" +
                "    },\n" +
                "    {\n" +
                "        \"names\": \"A1\",\n" +
                "        \"values\": \"12\"\n" +
                "    },\n" +
                "    {\n" +
                "        \"names\": \"A2\",\n" +
                "        \"values\": \"13\"\n" +
                "    },\n" +
                "    {\n" +
                "        \"names\": \"B0\",\n" +
                "        \"values\": \"2000\"\n" +
                "    },\n" +
                "    {\n" +
                "        \"names\": \"B1\",\n" +
                "        \"values\": \"4000\"\n" +
                "    },\n" +
                "    {\n" +
                "        \"names\": \"B2\",\n" +
                "        \"values\": \"22000\"\n" +
                "    }\n" +
                "]";

        List<ConvertToMap> convertToMaps = JacksonUtils.toJavaList(jsonString, ConvertToMap.class);

        // Extract a string that does not contain numbers and convert it to a map
        Map<String, Object> result = convertToMaps.stream()
                .filter(x -> isLetterDigit(x.names))
                .collect(Collectors.toMap(ConvertToMap::getNames, ConvertToMap::getValues));
        List<Map<String, String>> mapList = new ArrayList<>();
        // Group by string numbers containing numbers
        Map<String, List<ConvertToMap>> stringListMap = convertToMaps.stream().collect(groupingBy(convertToMap -> fetchGroupKey(convertToMap)));
        for (String key : stringListMap.keySet()) {
            if(Integer.valueOf(key) >= 0){
                mapList.add(stringListMap.get(key)
                        .stream()
                        .collect(Collectors.toMap(ConvertToMap::getNames, ConvertToMap::getValues)));
            }
        }
        result.put("Data", mapList);
        System.out.println(JacksonUtils.toJSONString(result));
    }
}

Answer 3

假設您的數據密鑰名稱模式是一個非數字后跟數字。

https://github.com/octomix/josson

反序列化

Josson josson = Josson.fromJsonString(
    "[" +
    "    {" +
    "        \"names\": \"Car\"," +
    "        \"values\": \"Toyota\"" +
    "    }," +
    "    {" +
    "        \"names\": \"Bike\"," +
    "        \"values\": \"Schwinn\"" +
    "    }," +
    "    {" +
    "        \"names\": \"Scooter\"," +
    "        \"values\": \"Razor\"" +
    "    }," +
    "    {" +
    "        \"names\": \"A0\"," +
    "        \"values\": \"11\"" +
    "    }," +
    "    {" +
    "        \"names\": \"A1\"," +
    "        \"values\": \"12\"" +
    "    }," +
    "    {" +
    "        \"names\": \"A2\"," +
    "        \"values\": \"13\"" +
    "    }," +
    "    {" +
    "        \"names\": \"B0\"," +
    "        \"values\": \"2000\"" +
    "    }," +
    "    {" +
    "        \"names\": \"B1\"," +
    "        \"values\": \"4000\"" +
    "    }," +
    "    {" +
    "        \"names\": \"B2\"," +
    "        \"values\": \"22000\"" +
    "    }" +
    "]");
    

轉型

JsonNode node = josson.getNode(
    "toObject('o')" +
    ".collect(o[names !=~ '\\D\\d+']*" +
    "         .map(names::values)" +
    "        ,o[names =~ '\\D\\d+']*" +
    "         .group(names.substr(1), map(names::values))@" +
    "         .elements" +
    "         .mergeObjects()" +
    "         .@toObject('Data')" +
    ")" +
    ".flatten(1)" +
    ".mergeObjects()");
System.out.println(node.toPrettyString());

Output

{
  "Car" : "Toyota",
  "Bike" : "Schwinn",
  "Scooter" : "Razor",
  "Data" : [ {
    "A0" : "11",
    "B0" : "2000"
  }, {
    "A1" : "12",
    "B1" : "4000"
  }, {
    "A2" : "13",
    "B2" : "22000"
  } ]
}