java 按属性在列表中查找特定对象
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Find specific object in a List by attribute
提问by user5725851
I have a list:
我有一个清单：
List<UserItem> userList = new ArrayList<>();
Where I add the following:
我添加以下内容：
User father = new User();
father.setName("Peter");
UserItem parent = new UserItem(father, null);
userList.add(parent);
I then create another user:
然后我创建另一个用户：
User daughter = new User();
daughter.setName("Emma");
UserItem child = new UserItem(daughter, <OBJECT IN LIST WHERE NAME IS "PETER">);
userList.add(child);
However, I need to change the text wrapped in <>above to the parent object I added before (the father), specified by the name ("Peter" in this case).
但是，我需要将<>上面包裹的文本更改为我之前添加的父对象（父亲），由名称（在本例中为“Peter”）指定。
How can I find an object in a List by a specific attribute?In my case, how can I find the object in the List that has the name "Peter"?
如何通过特定属性在列表中找到对象？就我而言，如何在 List 中找到名为“Peter”的对象？
Please note that I add hundreds, sometimes thousands, of different users like this to the list. Each "parent" has a unique name.
请注意，我在列表中添加了数百个，有时是数千个这样的不同用户。每个“父母”都有一个唯一的名字。
回答by Maroun
The obvious solution would be iterating on the list and when the condition is met, return the object:
显而易见的解决方案是在列表上迭代，当条件满足时，返回对象：
for (User user : userList) {
if ("peter".equals(user.getName()) {
return user;
}
}
And you can use filter(Java 8):
你可以使用filter（Java 8）：
List<User> l = list.stream()
.filter(s -> "peter".equals(s.getUser()))
.collect(Collectors.toList());
to get a list with all "peter" users.
获取所有“peter”用户的列表。
As suggested in comments, I think using Mapis a better option here.
正如评论中所建议的，我认为Map这里使用是更好的选择。
回答by maydos
Answer to your question is here: https://stackoverflow.com/a/1385698/2068880
您的问题的答案在这里：https: //stackoverflow.com/a/1385698/2068880
Stream peters = userList.stream().filter(p -> p.user.name.equals("Peter"))
However, as ruakh suggested, it's more reasonable to use Map<String, UserItem>to make it faster. Otherwise, it will iterate all the objects in the list to find users with name "Peter".
但是，正如 ruakh 所建议的那样，使用Map<String, UserItem>它来使其更快更合理。否则，它将遍历列表中的所有对象以查找名为“Peter”的用户。
回答by Viet
Other way with parallelStream with findAny
使用带有 findAny 的 parallelStream 的其他方式
Optional<UserItem> optional = userList.parallelStream().findAny(p -> p.user.getName().equalsIgnoreCase("Peter"));
UserItem user = optional.isPresent() ? optional.get() : null;
