Question

我一直在使用JSON，我在文档（JAVA）中看到，JSONObject的put（）和accumulate（）几乎做同样的事情？

那是什么？

Answer 1

我看到了JSONObject的Java源代码，accumulate和put之间的区别在于使用accumulate（String key，Object Value），如果“key”存在一些值，则检查Object是否为数组，如果它是一个数组，然后将“value”添加到数组中，否则将为此键创建一个数组。

然而，在put中，如果密钥存在，则其值将替换为值 - “value”

这是JSONObject accumulate的源代码（String key，Object Value）

/**
 * Appends {@code value} to the array already mapped to {@code name}. If
 * this object has no mapping for {@code name}, this inserts a new mapping.
 * If the mapping exists but its value is not an array, the existing
 * and new values are inserted in order into a new array which is itself
 * mapped to {@code name}. In aggregate, this allows values to be added to a
 * mapping one at a time.
 *
 * <p> Note that {@code append(String, Object)} provides better semantics.
 * In particular, the mapping for {@code name} will <b>always</b> be a
 * {@link JSONArray}. Using {@code accumulate} will result in either a
 * {@link JSONArray} or a mapping whose type is the type of {@code value}
 * depending on the number of calls to it.
 *
 * @param value a {@link JSONObject}, {@link JSONArray}, String, Boolean,
 *     Integer, Long, Double, {@link #NULL} or null. May not be {@link
 *     Double#isNaN() NaNs} or {@link Double#isInfinite() infinities}.
 */

  public JSONObject accumulate(String name, Object value) throws JSONException {
    Object current = nameValuePairs.get(checkName(name));
    if (current == null) {
        return put(name, value);
    }

    if (current instanceof JSONArray) {
        JSONArray array = (JSONArray) current;
        array.checkedPut(value);
    } else {
        JSONArray array = new JSONArray();
        array.checkedPut(current);
        array.checkedPut(value);
        nameValuePairs.put(name, array);
    }
    return this;
}

以下是JSONObject put（String key，Object value）

的代码

 /**
 * Maps {@code name} to {@code value}, clobbering any existing name/value
 * mapping with the same name.
 *
 * @return this object.
 */
public JSONObject put(String name, boolean value) throws JSONException {
    nameValuePairs.put(checkName(name), value);
    return this;
}

JSONObject accumulate和put之间有什么区别？

1 个答案: