如何在 json 对象中创建数组

Question

我正在使用这样的东西 -String Number1=to_1;        String Number2=to_2;        String[] arrayNumbers = new String[] {(char)34+Number1+(char)34,(char)34+Number2+(char)34};        System.out.println(Arrays.toString(arrayNumbers));        JSONObject jsonObj = new JSONObject();        jsonObj.put("to",Arrays.toString(arrayNumbers));        jsonObj.put("type",type);        jsonObj.put("callback",callbackUrl);        JSONArray array = new JSONArray();        JSONObject Array_item = new JSONObject();        jsonObj.put(type, array);        Array_item.put("caption",captionName);        array.add(Array_item);        System.out.println(jsonObj.toString());预期-{  "to":["91890xx", "91890xx"], "type": "document", "document" : {"caption" : "doc"},"callback":"{{callback}}"}实际-{"document":[{"caption":"hello"}],"callback":"{{callback}}","to":"[\"91890xxx\", \"91890xx\"]","type":"document"}我不知道任何更多的逻辑来删除到数字的双引号，因为它考虑到一个字符串，其中两个数字都应该是数组格式，如预期中所述。

Answer 1

首先，我向您展示正确的代码：

import org.json.JSONArray;

import org.json.JSONException;

import org.json.JSONObject;

public class Test {

  /**

   * <pre>

   *  {

   *    "to":["91890xx", "91890xx"],

   *    "type": "document",

   *    "document" : {"caption" : "doc"},

   *    "callback":"{{callback}}"

   *  }

   * </pre>

   *

   * @param args

   * @throws JSONException

   */

   public static void main(String[] args) throws JSONException {

    String number1 = "91890";

    String number2 = "91890";

    String[] numbers = new String[]{number1, number2};

    JSONArray toNode = new JSONArray();

    for (String number : numbers) {

      toNode.put(number);

    }

    JSONObject jsonObj = new JSONObject();

    jsonObj.put("to", toNode);

    jsonObj.put("type", "document");

    jsonObj.put("document", new JSONObject().put("caption", "doc"));

    jsonObj.put("callback", "{{callback}}");

    System.out.println(jsonObj.toString());

  }

}

结果：

{"document":{"caption":"doc"},"callback":"{{callback}}","to":["91890","91890"],"type":"document"}

如果要创建一个 josn 数组节点，请显示 use 和 use 方法来添加元素。JSONArrayJSONArray#put(*)

将字符串放入 或 中，不需要用引号（“） 包装字符串。此外，你应该写，而不是在Java中有点晦涩难懂。JSONArrayJSONObject\"(char) 34

以下情况用于回复注释。

import org.json.JSONArray;

import org.json.JSONException;

import org.json.JSONObject;

import java.math.BigDecimal;

import java.net.URI;

import java.net.URL;

import java.util.*;

public class Test1 {

  public static void main(String[] args) throws JSONException {

    Map<String, Object> map = new HashMap<>();

    map.put("key1", "value1");

    map.put("key2", new Date());

    map.put("key3", 1);

    map.put("key4", null);

    map.put("key5", Collections.singletonMap("key5-key1", "value"));

    map.put("key6", Arrays.asList(1, 2, 3, 4));

    map.put("key7", BigDecimal.TEN);

    map.put("key8", new String[]{"a", "b", "c"});

    map.put("key9", TestEnum.A);

    map.put("key10", new TestEnum[]{TestEnum.A, TestEnum.B, TestEnum.C});

    Object json = buildJsonObj(map);

    System.out.println(json);

  }

  private static Object buildJsonObj(Object source) throws JSONException {

    if (source == null) {

      return null;

    }

    if (isSimpleValueType(source.getClass())) {

      return source;

    }

    if (source instanceof Map) {

      Map<Object, Object> map = (Map<Object, Object>) source;

      JSONObject jsonObject = new JSONObject();

      for (Map.Entry<Object, Object> entry : map.entrySet()) {

        Object key = entry.getKey();

        if (!(key instanceof String)) {

          throw new IllegalArgumentException("key must be string.");

        }

        jsonObject.put((String) key, buildJsonObj(entry.getValue()));

      }

      return jsonObject;

    }

    if (source instanceof Iterable) {

      Iterable<Object> iterable = (Iterable<Object>) source;

      JSONArray jsonArray = new JSONArray();

      for (Object value : iterable) {

        jsonArray.put(buildJsonObj(value));

      }

      return jsonArray;

    }

    if (source.getClass().isArray()) {

      Object[] array = (Object[]) source;

      JSONArray jsonArray = new JSONArray();

      for (Object value : array) {

        jsonArray.put(buildJsonObj(value));

      }

      return jsonArray;

    }

    throw new IllegalArgumentException("Unsupported type: " + source + ".");

  }

  private static boolean isSimpleValueType(Class<?> clazz) {

    return (Enum.class.isAssignableFrom(clazz) ||

        CharSequence.class.isAssignableFrom(clazz) ||

        Number.class.isAssignableFrom(clazz) ||

        Date.class.isAssignableFrom(clazz) ||

        URI.class == clazz || URL.class == clazz ||

        Locale.class == clazz);

  }

  public enum TestEnum {

    A, B, C

  }

}

结果：

{"key1":"value1","key2":"Thu Mar 14 20:20:49 CST 2019","key5":{"key5-key1":"value"},"key6":[1,2,3,4],"key3":1,"key9":"A","key7":10,"key8":["a","b","c"],"key10":["A","B","C"]}

Answer 2

替换行

jsonObj.put（“to”，Arrays.toString（arrayNumbers））;

自

jsonObj.put（“to”， Arrays.asList（arrayNumbers））;

Answer 3

let a = {"document":[{"caption":"hello"}],"callback":"{{callback}}","to":"[\"91890xxx\", \"91890xx\"]","type":"document"};

    a = JSON.parse(JSON.stringify(a));

    let b = JSON.parse(a.to);

    a.to = b;

    console.log(a);

Answer 4

取代

jsonObj.put("to",Arrays.toString(arrayNumbers));

跟

jsonObj.put("to",arrayNumbers);