连接List中所有Maps的String值 [英] Concatenate the String values of all Maps in a List
问题描述
我正在尝试调整Lambda功能,但是在这里和那里很少有人挣扎。
I am trying to adapt Lambda features, however few struggles here and there.
List<Map<String, String>> list = new LinkedList<>();
Map<String, String> map = new HashMap<>();
map.put("data1", "12345");
map.put("data2", "45678");
list.add(map);
我只想以逗号分隔格式打印值,如 12345,45678
I just want to print the values in comma separated format like 12345,45678
所以这是我的试用
list.stream().map(Map::values).collect(Collectors.toList()) //Collectors.joining(",")
,输出为 [[12345,45678]]
。这意味着,有一个列表和内部列表,它在 0
索引处创建逗号分隔值。我确实理解它为什么会这样做。
and the output is [[12345,45678]]
. It means, there's a list and inside list it's creating the comma separated value at 0
index. I do understand why it's doing though.
但我没有通过如何提取我想要的结果,除非我调用 .get(0)
在该表达式的末尾。
But I didn't get through how to extract my desired result unless I call .get(0)
in the end of that expression.
任何有关如何更好地使用lambdas的帮助/更多见解将会有所帮助
Any help/some more insights on how to use lambdas better will be helpful
推荐答案
试试这个:
list.stream().map(Map::values).flatMap(Collection::stream).collect(Collectors.joining(","))
flatMap
方法将集合< Stream< String>>
展平为一个单个流< String>
。
The flatMap
method flattens out a Collection<Stream<String>>
into a single Stream<String>
.
或者,您可以尝试@Holger在评论中建议的内容。
Alternately, you can try what @Holger suggested in the comments.
请注意,由于您使用的是 Map
,因此无法保证订单会被保留。您可能会在最后看到输出为 45678,12345
。如果您希望保留订单,可以使用 LinkedHashMap
而不是 HashMap
。
Note that since you are using a Map
, there is no guarantee that the order will be preserved. You might see the output as 45678,12345
at your end. If you wish to preserve the order, you can use a LinkedHashMap
instead of a HashMap
.
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