JSON字符串到Java字符串

Question

我有这些 JSON 字符串：

{
    "Results": {
        "output1": {
            "type": "table",
            "value": {
                "ColumnNames": ["userId", "documentId", "Scored Labels", "Scored Probabilities"],
                "ColumnTypes": ["String", "String", "Boolean", "Double"],
                "Values": [["100213199594809000000", "1Ktol-SWvAh8pnHG2O7HdPrfbEVZWX3Vf2YIPYXA_8gI", "False", "0.375048756599426"], ["103097844766994000000", "1jYsTPJH8gaIiATix9x34Ekcj31ifJMkPNb0RmxnuGxs", "True", "0.753859758377075"]]
            }
        }
    }
}

我只想拥有ColumnNames 和Values。我已经尝试过这样的事情：

Map<String,Object> map = mapper.readValue(filename, Map.class);
String CN = (String) map.get("ColumnNames");

然后我收到以下错误：

Exception in thread "main" org.codehaus.jackson.JsonParseException: Unexpected character ('A' (code 65)): expected a valid value (number, String, array, object, 'true', 'false' or 'null')
 at [Source: java.io.StringReader@64232b15; line: 1, column: 2]`

我只用过几次 JSON。有人可以帮我吗？

对我来说最好的情况是这样的，我在另一种情况下做过：

String uId = (String) attr.get("userId");

有可能吗？

所以现在我已经这样做了：

我是这样尝试的：

public class ClientPOJO {

    private String userId;
    private String documentId;


    public String getuserId() {
        return userId;
    }

    public void setuserId(String userId) {
        this.userId = userId;
    }

    public String getdocumentId() {
        return documentId;
    }

    public void setdocumentId(String documentId) {
        this.documentId = documentId;
    }

}

然后：

ObjectMapper mapper = new ObjectMapper();
                    ClientPOJO clientes= mapper.readValue(filename, ClientPOJO.class);

String uid = clientes.getuserId();

但是现在当我制作一个 Prtinout 时，我会得到和以前一样的错误：

    Exception in thread "main" org.codehaus.jackson.JsonParseException: Unexpected character ('A' (code 65)): expected a valid value (number, String, array, object, 'true', 'false' or 'null')
 at [Source: java.io.StringReader@7a6eb29d; line: 1, column: 2]

Answer 1

Java-将JSON字符串转换成字符串/整数/对象

String jsonString = "{"username":"Gajender"}";
org.json.JSONObject jsonObj =new JSONObject(jsonString);
String name = (String) jsonObj.get("username").toString();

Answer 2

以下示例说明了解决问题的通用方法（基于 Jackson 库）。您可能希望增强解决方案以满足您的所有要求。

内嵌评论。

package com.stackoverflow;

import java.util.Iterator;
import java.util.List;
import java.util.Map;

import org.junit.Test;

import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.databind.ObjectMapper;

// Junit class
public class TableDeserExample {
    // sample input
    String inputJson = "{\n" +
            "    \"Results\": {\n" +
            "        \"output1\": {\n" +
            "            \"type\": \"table\",\n" +
            "            \"value\": {\n" +
            "                \"ColumnNames\": [\"userId\", \"documentId\", \"Scored Labels\", \"Scored Probabilities\"],\n" +
            "                \"ColumnTypes\": [\"String\", \"String\", \"Boolean\", \"Double\"],\n" +
            "                \"Values\": [[\"100213199594809000000\", \"1Ktol-SWvAh8pnHG2O7HdPrfbEVZWX3Vf2YIPYXA_8gI\", \"False\", \"0.375048756599426\"], [\"103097844766994000000\", \"1jYsTPJH8gaIiATix9x34Ekcj31ifJMkPNb0RmxnuGxs\", \"True\", \"0.753859758377075\"]]\n"
            +
            "            }\n" +
            "        }\n" +
            "    }\n" +
            "}";

    // POJO to map the Json structure. You may want to make it generalize based
    // on field "type"
    // (https://github.com/FasterXML/jackson-docs/wiki/JacksonPolymorphicDeserialization)
    public static class Result {
        private String type;
        private TableResult value;

        public String getType() {
            return this.type;
        }

        public void setType(String type) {
            this.type = type;
        }

        public void setValue(TableResult value) {
            this.value = value;
        }

        public TableResult getValue() {
            return this.value;
        }
    }

    // Pojo for table result
    public static class TableResult {
        private List<String> columnNames;
        private List<String> columnTypes;
        private List<Object[]> values;

        @JsonProperty("ColumnNames")
        public List<String> getColumnNames() {
            return this.columnNames;
        }

        public void setColumnNames(List<String> columnNames) {
            this.columnNames = columnNames;
        }

        @JsonProperty("ColumnTypes")
        public List<String> getColumnTypes() {
            return this.columnTypes;
        }

        public void setColumnTypes(List<String> columnTypes) {
            this.columnTypes = columnTypes;
        }

        @JsonProperty("Values")
        public List<Object[]> getValues() {
            return this.values;
        }

        public void setValues(List<Object[]> values) {
            this.values = values;
        }

    }

    // Top level Json POJO
    public static class ResultContainer {
        private Map<String, Result> results;

        @JsonProperty("Results")
        public Map<String, Result> getResults() {
            return this.results;
        }

        public void setResults(Map<String, Result> results) {
            this.results = results;
        }
    }

    // A contract to map the result "values" to the expected object
    public static interface ResultMapper<T> {
        T map(TableResult map, Object[] row);
    }

    // Basic implementation for mapping user object from json "values[i]" array
    public static class UserTableResultMapper implements ResultMapper<User> {

        @Override
        public User map(TableResult result, Object[] row) {
            User user = new User();
            // Here use any mapper logic based on column name
            // Retrieved from result object.
            // Below are for illustration only
            user.setId(String.valueOf(row[0]));
            user.setDocumentId(String.valueOf(row[1]));
            return user;
        }

    }

    // A result reader class
    public static class ResultReader<T> implements Iterable<T> {
        private TableResult result;
        private ResultMapper<T> mapper;

        public ResultReader(TableResult result, ResultMapper<T> mapper) {
            this.result = result;
            this.mapper = mapper;
        }

        @Override
        public Iterator<T> iterator() {
            final Iterator<Object[]> itr = result.getValues().iterator();
            return new Iterator<T>() {

                @Override
                public void remove() {
                    throw new UnsupportedOperationException();
                }

                @Override
                public T next() {
                    Object[] values = itr.next();
                    return mapper.map(result, values);
                }

                @Override
                public boolean hasNext() {
                    return itr.hasNext();
                }
            };
        };
    }

    public static class User {
        private String id;
        private String documentId;
        // and others

        public String getId() {
            return this.id;
        }

        public void setDocumentId(String documentId) {
            this.documentId = documentId;
        }

        public void setId(String id) {
            this.id = id;
        }

        public String getDocumentId() {
            return this.documentId;
        }

    }

    @Test
    public void simpleTest() throws Exception {
        ObjectMapper mapper = new ObjectMapper();
        ResultContainer file = mapper.readValue(inputJson, ResultContainer.class);
        Result result = file.getResults().get("output1");
        ResultReader<User> userResultReader = new ResultReader<>(result.getValue(), new UserTableResultMapper());
        for (User user : userResultReader) {
            System.out.println(user.getId() + " : " + user.getDocumentId());
        }
    }
}

Answer 3

如果您确切知道 json 的结构（例如您发布的 json），那么您可以使用 Gson 来获取您的对象，如下所示：

JsonParser parser = new JsonParser();
JsonObject json = (JsonObject) parser.parse("your_json_string_here");
String column = json.get("Results").getAsJsonObject().get("output1").getAsJsonObject().get("value").getAsJsonObject().get("ColumnNames").getAsJsonArray().toString();
String value = json.get("Results").getAsJsonObject().get("output1").getAsJsonObject().get("value").getAsJsonObject().get("Values").getAsJsonArray().toString();    
System.out.println(column);
System.out.println(value);

如果您需要一些更通用的东西，那么您可以将 json 字符串解析为 HashMap<String, Object>，然后使用递归读取 HashMap 并获得您想要的值。 示例（在我的代码中，Map的类型将对应一个Json Object，List的类型将对应Json字符串中的Array）：

Type type = new TypeToken<HashMap<String, Object>>() {}.getType();
Gson gson = new Gson();
HashMap<String, Object> map = gson.fromJson("your_json_string_here", type);
    for (String key : map.keySet()) {
        Object obj = map.get(key);
        if (obj instanceof List) {
            for (Object o : (List) obj) {
                if (o instanceof Map) {
                    loop((Map) o);
                } else {
                    System.out.println(key + " : " + o);
                }
            }
        } else if (obj instanceof Map) {
            loop((Map) obj);
        } else {
            System.out.println(key + " : " + obj);
        }
    }
}

private static void loop(Map<String, Object> map) {
    for (String key : map.keySet()) {
        Object obj = map.get(key);
        if (obj instanceof List) {
            for (Object o : (List) obj) {
                if (o instanceof Map) {
                    loop((Map) o);
                } else {
                    System.out.println(key + " : " + o);
                }
            }
        } else if (obj instanceof Map) {
            loop((Map) obj);
        } else {
            System.out.println(key + " : " + obj);
        }
    }
}

Answer 4

Jackson 和任何其他库都不会将 Values 数组解析为具有客户端数据的对象，例如您的 POJO。您可以通过获取此 JSON 中的原始数据树并通过迭代此树中的 Values 数组来构造对象来实现此目的。假设ColumnNames 的顺序是固定的，那么您可以像这样使用Jackson 进行解析：

final ObjectMapper mapper = new ObjectMapper();
final JsonNode tree = mapper.readTree(json);
final JsonNode values = tree.findValue("Values");

final List<ClientPOJO> clients = new ArrayList<>();
for (JsonNode node : values) {
    final ClientPOJO client = new ClientPOJO();
    client.setUserId(node.get(0).asText());
    client.setDocumentId(node.get(1).asText());
    client.setScoredLabels(node.get(2).asBoolean());
    client.setScoredProbabilities(node.get(3).asDouble());
    clients.add(client);
}

JsonNode 的文档。基本上，findValue 可以让另一个节点深入到树中，get 可以通过索引获取数组元素，asText 等可以将 JSON 中的值解析为 Java 中的适当类型。

由于您似乎可以灵活地选择 JSON 解析库，我建议您尝试使用来自 com.fasterxml 的 Jackson 2 而不是来自 org.codehaus 的 Jackson 1。