这是我的清单:
Name: Ben || Age: 5 || Group: 1
Name: Andy || Age: 6 || Group: 2
Name: Charlie || Age: 6 || Group: 2
Name: Ben || Age: 5 || Group: 1
Name: Andy || Age: 5 || Group: 2
Name: Charlie || Age: 5 || Group: 1
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我想对列表进行排序Group,如果Group等于那么Age,如果Age是等于那么Name.但到目前为止,我只能使用Lambda表达式按一个属性排序:
list.sort((Object o1, Object o2) -> o1.getGroup().compareTo(o2.getGroup()));
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如果我试试
o1.getGroup().compareTo(o2.getGroup()) && o1.getAge().compareTo(o2.getAge())
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事实证明错误......
And*_*eas 16
将lambda 表达式更改为lambda {block},您不必指定参数类型:
list.sort((o1, o2) -> {
int cmp = o1.getGroup().compareTo(o2.getGroup());
if (cmp == 0)
cmp = Integer.compare(o1.getAge(), o2.getAge());
if (cmp == 0)
cmp = o1.getName().compareTo(o2.getName());
return cmp;
});
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See*_*ose 13
您可以使用静态方法Comparator.comparing基于返回可比值的函数创建比较器.这种比较器可以与其他比较器链接.
假设您的类型名为Person,您将拥有:
Comparator<Person> c = Comparator
.comparing(p -> p.getGroup())
.thenComparing(p -> p.getAge())
.thenComparing(p -> p.getName())
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如果任何getter返回基本类型,则必须分别使用 - 例如 - comparingInt和thenComparingInt.您还可以使用方法引用:
Comparator<Person> c = Comparator
.comparing(Person::getGroup)
.thenComparing(Person::getAge)
.thenComparing(Person::getName)
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但是......如果你的类根据这些值有一个自然的顺序,你最好让它实现接口Comparable并在那里写比较逻辑:
class Person implements Comparable<Person> {
...
@Override
public int compareTo(Person other) {
int compare = Integer.compare(getGroup(), other.getGroup());
if (compare == 0) {
compare = Integer.compare(getAge(), other.getAge());
}
if (compare == 0) {
compare = getName.compareTo(other.getName());
}
return compare;
}
}
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此代码段也可用于lambda表达式:
list.sort((o1, o2) -> {
int compare = Integer.compare(o1.getGroup(), o2.getGroup());
if (compare == 0) {
compare = Integer.compare(o1.getAge(), o2.getAge());
}
if (compare == 0) {
compare = o1.getName.compareTo(o2.getName());
}
return compare;
});
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