在Java

时间：2019-09-19 09:49:07

标签： java json jackson

要求是为输入的JSON对象到平化的JSON对象创建通用的平化工具。

示例JSON如下所示

{
  "Source": "source-1",
  "Rows": [
    {
      "Keys": {
        "device-id": "BC04-EBH-N3K-01",
        "interface-name": "TenGigE0/0/0/39",
        "node-name": "0/0/CPU0"
      },
      "Timestamp": 1567621527656,
      "inner": {
        "donm": {
          "id": "0062",
          "mol": {
            "rem": 30,
            "len": 11,
            "org": {
              "ldp": [
                {
                  "t": 486,
                  "o": 322
                },
                {
                  "t": 487,
                  "o": 32,
                  "twss": 1,
                  "tlv": "00:01"
                }
              ]
            },
            "chlen": 14,
            "poe": 5,
            "combs": 10,
            "chaype": 4,
            "rek": 0,
            "rem-um": 67
          },
          "detail": {
            "enas": "B,R",
            "systes": "B,R",
            "timng": 91,
            "syn": "C",
            "met-type": 0,
            "neses": {
              "lldEDIT": [
                {
                  "ium": 830,
                  "m": 1,
                  "ass": {
                    "ape": "ipv4",
                    "ipvs": "94"
                  }
                }
              ]
            },
            "pess": "0008",
            "por]d": 0,
            "pon": "BCtive",
            "sysme": "BC1"
          },
          "reme": "Bu1",
          "hean": 0,
          "porl": "Et1"
        }
      }
    }

  ],
  "Tey": {
    "epath": "Cgetail",
    "sustr": "MX",
    "coime": 1567621527653,
    "msp": 1567621527653,
    "come": 1567621527660,
    "nor": "BC5",
    "cid": 14789654
  }
}

我一直试图将其展平到3个级别，并提出了以下实用程序。但是，当我不得不处理数组和String，long，Timestamp等类型的值时，事情变得越来越复杂。而且，我无法理解如何维护嵌套键的唯一性。

public static Map<String,Object> flattenJson(JsonNode input){
        Map<String,Object> finalMap = new HashMap<>();

        ObjectMapper datamapper = new ObjectMapper();
        Map<String,Object> topLevelJsonMap = datamapper.convertValue(input,Map.class);
        Set<String> topLevelKeys = topLevelJsonMap.keySet();

        for(String topLevelKey : topLevelKeys){

            System.out.println("Key :::: "+topLevelKey);

            Object topLevelData = topLevelJsonMap.get(topLevelKey);
            System.out.println("value :::: "+topLevelData.toString());

            if(topLevelData instanceof ArrayNode){
                ArrayNode arrayOfData = (ArrayNode) topLevelData;
                for(JsonNode dataNode : arrayOfData){
                    flattenJson(input);
                }
            } else if(topLevelData instanceof JsonNode){
                Map<String,Object> innerLevelJsonMap = datamapper.convertValue(topLevelData,Map.class);
                Set<String> innerLevelKeys = innerLevelJsonMap.keySet();
                for(String innerLevelKey : innerLevelKeys){
                    System.out.println("inner key :::: "+innerLevelKey);
                    flattenJson((JsonNode) innerLevelJsonMap.get(innerLevelKey));
                }
            }else {
               finalMap.put(topLevelKey,topLevelData);
            }
        }
        return finalMap;
    }

非常感谢您的帮助。

3 个答案:

答案 0 :(得分：1)

尝试以下代码：

public static void flattenJson(JsonNode node, String parent, Map<String, ValueNode> map) {
    if (node instanceof ValueNode) {
        map.put(parent, (ValueNode)node);
    } else {
        String prefix = parent == null ? "" : parent + ".";
        if (node instanceof ArrayNode) {
            ArrayNode arrayNode = (ArrayNode)node;
            for(int i = 0; i < arrayNode.size(); i++) {
                flattenJson(arrayNode.get(i), prefix + i, map);
            }
        } else if (node instanceof ObjectNode) {
            ObjectNode objectNode = (ObjectNode) node;
            for (Iterator<Map.Entry<String, JsonNode>> it = objectNode.fields(); it.hasNext(); ) {
                Map.Entry<String, JsonNode> field = it.next();
                flattenJson(field.getValue(), prefix + field.getKey(), map);
            }
        } else {
            throw new RuntimeException("unknown json node");
        }
    }
}

public static Map<String, ValueNode> flattenJson(JsonNode input) {
    Map<String, ValueNode> map = new LinkedHashMap<>();
    flattenJson(input, null, map);
    return map;
}

然后您可以致电

    ObjectMapper om = new ObjectMapper();
    JsonNode jsonNode = om.readTree(json);
    Map<String, ValueNode> m = flattenJson(jsonNode);
    for (Map.Entry<String, ValueNode> kv : m.entrySet()) {
        System.out.println(kv.getKey() + "=" + kv.getValue().asText());
    }

输出：

Source=source-1
Rows.0.Keys.device-id=BC04-EBH-N3K-01
Rows.0.Keys.interface-name=TenGigE0/0/0/39
Rows.0.Keys.node-name=0/0/CPU0
Rows.0.Timestamp=1567621527656
Rows.0.inner.donm.id=0062
Rows.0.inner.donm.mol.rem=30
Rows.0.inner.donm.mol.len=11
Rows.0.inner.donm.mol.org.ldp.0.t=486
Rows.0.inner.donm.mol.org.ldp.0.o=322
Rows.0.inner.donm.mol.org.ldp.1.t=487
Rows.0.inner.donm.mol.org.ldp.1.o=32
Rows.0.inner.donm.mol.org.ldp.1.twss=1
Rows.0.inner.donm.mol.org.ldp.1.tlv=00:01
Rows.0.inner.donm.mol.chlen=14
Rows.0.inner.donm.mol.poe=5
Rows.0.inner.donm.mol.combs=10
Rows.0.inner.donm.mol.chaype=4
Rows.0.inner.donm.mol.rek=0
Rows.0.inner.donm.mol.rem-um=67
Rows.0.inner.donm.detail.enas=B,R
Rows.0.inner.donm.detail.systes=B,R
Rows.0.inner.donm.detail.timng=91
Rows.0.inner.donm.detail.syn=C
Rows.0.inner.donm.detail.met-type=0
Rows.0.inner.donm.detail.neses.lldEDIT.0.ium=830
Rows.0.inner.donm.detail.neses.lldEDIT.0.m=1
Rows.0.inner.donm.detail.neses.lldEDIT.0.ass.ape=ipv4
Rows.0.inner.donm.detail.neses.lldEDIT.0.ass.ipvs=94
Rows.0.inner.donm.detail.pess=0008
Rows.0.inner.donm.detail.por]d=0
Rows.0.inner.donm.detail.pon=BCtive
Rows.0.inner.donm.detail.sysme=BC1
Rows.0.inner.donm.reme=Bu1
Rows.0.inner.donm.hean=0
Rows.0.inner.donm.porl=Et1
Tey.epath=Cgetail
Tey.sustr=MX
Tey.coime=1567621527653
Tey.msp=1567621527653
Tey.come=1567621527660
Tey.nor=BC5
Tey.cid=14789654

答案 1 :(得分：1)

为避免与键名冲突，可以使用JSON Pointer规范来创建它们。 Jackson库也支持它，因此以后可以使用它们遍历JsonNode节点。

简单的实现可能如下所示：

import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.node.ArrayNode;
import com.fasterxml.jackson.databind.node.ObjectNode;
import java.io.File;
import java.util.LinkedHashMap;
import java.util.Map;
import java.util.Objects;
import java.util.concurrent.atomic.AtomicInteger;

public class JsonApp {

  public static void main(String[] args) throws Exception {
    File jsonFile = new File("./test.json");

    ObjectMapper mapper = new ObjectMapper();
    JsonNode root = mapper.readTree(jsonFile);
    Map<String, JsonNode> map = new JsonFlattener(root).flatten();

    System.out.println("Use key-value pairs:");
    map.forEach(
        (k, v) -> {
          System.out.println(k + " => " + v);
        });

    System.out.println();
    System.out.println("Use pointers:");
    map.forEach(
        (k, v) -> {
          System.out.println(k + " => " + root.at(k));
        });
  }
}

class JsonFlattener {

  private final Map<String, JsonNode> json = new LinkedHashMap<>(64);
  private final JsonNode root;

  JsonFlattener(JsonNode node) {
    this.root = Objects.requireNonNull(node);
  }

  public Map<String, JsonNode> flatten() {
    process(root, "");
    return json;
  }

  private void process(JsonNode node, String prefix) {
    if (node.isObject()) {
      ObjectNode object = (ObjectNode) node;
      object
          .fields()
          .forEachRemaining(
              entry -> {
                  process(entry.getValue(), prefix + "/" + entry.getKey());
              });
    } else if (node.isArray()) {
      ArrayNode array = (ArrayNode) node;
      AtomicInteger counter = new AtomicInteger();
      array
          .elements()
          .forEachRemaining(
              item -> {
                process(item, prefix + "/" + counter.getAndIncrement());
              });
    } else {
      json.put(prefix, node);
    }
  }
}

上面的代码显示：

Use key-value pairs:
/Source => "source-1"
/Rows/0/Keys/device-id => "BC04-EBH-N3K-01"
/Rows/0/Keys/interface-name => "TenGigE0/0/0/39"
/Rows/0/Keys/node-name => "0/0/CPU0"
/Rows/0/Timestamp => 1567621527656
/Rows/0/inner/donm/id => "0062"
/Rows/0/inner/donm/mol/rem => 30
/Rows/0/inner/donm/mol/len => 11
/Rows/0/inner/donm/mol/org/ldp/0/t => 486
/Rows/0/inner/donm/mol/org/ldp/0/o => 322
/Rows/0/inner/donm/mol/org/ldp/1/t => 487
/Rows/0/inner/donm/mol/org/ldp/1/o => 32
/Rows/0/inner/donm/mol/org/ldp/1/twss => 1
/Rows/0/inner/donm/mol/org/ldp/1/tlv => "00:01"
/Rows/0/inner/donm/mol/chlen => 14
/Rows/0/inner/donm/mol/poe => 5
/Rows/0/inner/donm/mol/combs => 10
/Rows/0/inner/donm/mol/chaype => 4
/Rows/0/inner/donm/mol/rek => 0
/Rows/0/inner/donm/mol/rem-um => 67
/Rows/0/inner/donm/detail/enas => "B,R"
/Rows/0/inner/donm/detail/systes => "B,R"
/Rows/0/inner/donm/detail/timng => 91
/Rows/0/inner/donm/detail/syn => "C"
/Rows/0/inner/donm/detail/met-type => 0
/Rows/0/inner/donm/detail/neses/lldEDIT/0/ium => 830
/Rows/0/inner/donm/detail/neses/lldEDIT/0/m => 1
/Rows/0/inner/donm/detail/neses/lldEDIT/0/ass/ape => "ipv4"
/Rows/0/inner/donm/detail/neses/lldEDIT/0/ass/ipvs => "94"
/Rows/0/inner/donm/detail/pess => "0008"
/Rows/0/inner/donm/detail/por]d => 0
/Rows/0/inner/donm/detail/pon => "BCtive"
/Rows/0/inner/donm/detail/sysme => "BC1"
/Rows/0/inner/donm/reme => "Bu1"
/Rows/0/inner/donm/hean => 0
/Rows/0/inner/donm/porl => "Et1"
/Tey/epath => "Cgetail"
/Tey/sustr => "MX"
/Tey/coime => 1567621527653
/Tey/msp => 1567621527653
/Tey/come => 1567621527660
/Tey/nor => "BC5"
/Tey/cid => 14789654

Use pointers:
/Source => "source-1"
/Rows/0/Keys/device-id => "BC04-EBH-N3K-01"
/Rows/0/Keys/interface-name => "TenGigE0/0/0/39"
/Rows/0/Keys/node-name => "0/0/CPU0"
/Rows/0/Timestamp => 1567621527656
/Rows/0/inner/donm/id => "0062"
/Rows/0/inner/donm/mol/rem => 30
/Rows/0/inner/donm/mol/len => 11
/Rows/0/inner/donm/mol/org/ldp/0/t => 486
/Rows/0/inner/donm/mol/org/ldp/0/o => 322
/Rows/0/inner/donm/mol/org/ldp/1/t => 487
/Rows/0/inner/donm/mol/org/ldp/1/o => 32
/Rows/0/inner/donm/mol/org/ldp/1/twss => 1
/Rows/0/inner/donm/mol/org/ldp/1/tlv => "00:01"
/Rows/0/inner/donm/mol/chlen => 14
/Rows/0/inner/donm/mol/poe => 5
/Rows/0/inner/donm/mol/combs => 10
/Rows/0/inner/donm/mol/chaype => 4
/Rows/0/inner/donm/mol/rek => 0
/Rows/0/inner/donm/mol/rem-um => 67
/Rows/0/inner/donm/detail/enas => "B,R"
/Rows/0/inner/donm/detail/systes => "B,R"
/Rows/0/inner/donm/detail/timng => 91
/Rows/0/inner/donm/detail/syn => "C"
/Rows/0/inner/donm/detail/met-type => 0
/Rows/0/inner/donm/detail/neses/lldEDIT/0/ium => 830
/Rows/0/inner/donm/detail/neses/lldEDIT/0/m => 1
/Rows/0/inner/donm/detail/neses/lldEDIT/0/ass/ape => "ipv4"
/Rows/0/inner/donm/detail/neses/lldEDIT/0/ass/ipvs => "94"
/Rows/0/inner/donm/detail/pess => "0008"
/Rows/0/inner/donm/detail/por]d => 0
/Rows/0/inner/donm/detail/pon => "BCtive"
/Rows/0/inner/donm/detail/sysme => "BC1"
/Rows/0/inner/donm/reme => "Bu1"
/Rows/0/inner/donm/hean => 0
/Rows/0/inner/donm/porl => "Et1"
/Tey/epath => "Cgetail"
/Tey/sustr => "MX"
/Tey/coime => 1567621527653
/Tey/msp => 1567621527653
/Tey/come => 1567621527660
/Tey/nor => "BC5"
/Tey/cid => 14789654

答案 2 :(得分：0)

只需通过将每个级别的密钥附加到更高级别的密钥中来创建密钥的名称空间。 换句话说，扁平化的JSON密钥为：

{
 "L1key::L2key": "L2val",
 "L1key": "L1val",
 "L1key::L2key::L3key": "L3val"
}

这样，您可以保证唯一性，但也可以从中创建原始json。最后，请确保键中不存在级别分隔符（此处为::）。

HTH

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