如何将所有Java的hashMap内容放在一起，而不是替换现有的键和值？(How to putAll on Java hashMap contents of one to another, but n

如何将所有Java的hashMap内容放在一起，而不是替换现有的键和值？(How to putAll on Java hashMap contents of one to another, but not replace existing keys and values?)

我需要将所有的键和值从一个A HashMap复制到另一个B上，而不是替换现有的键和值。

最好的方法是什么？

我在想，而不是迭代keySet和checkig是否存在，我会的

Map temp = new HashMap(); // generic later temp.putAll(Amap); A.clear(); A.putAll(Bmap); A.putAll(temp);

I need to copy all keys and values from one A HashMap onto another one B, but not to replace existing keys and values.

Whats the best way to do that?

I was thinking instead iterating the keySet and checkig if it exist or not, I would

Map temp = new HashMap(); // generic later temp.putAll(Amap); A.clear(); A.putAll(Bmap); A.putAll(temp);

最满意答案

看起来你愿意创建一个临时的Map ，所以我会这样做：

Map tmp = new HashMap(patch); tmp.keySet().removeAll(target.keySet()); target.putAll(tmp);

在这里， patch是要添加到target地图的地图。

感谢Louis Wasserman，这是一个利用Java 8中新方法的版本：

patch.forEach(target::putIfAbsent);

It looks like you are willing to create a temporary Map, so I'd do it like this:

Map tmp = new HashMap(patch); tmp.keySet().removeAll(target.keySet()); target.putAll(tmp);

Here, patch is the map that you are adding to the target map.

Thanks to Louis Wasserman, here's a version that takes advantage of the new methods in Java 8:

patch.forEach(target::putIfAbsent);