在 Java 中用 HashMap 中的匹配项填充数组

if(zooAnimals .get(key).getColor().equals("Brown"){
    for(int i=0; i < animalArray.length; i++) {
        animalArray[i] = zooAnimals.get(key);
    }
}

您在颜色检查条件中定义了 for 循环。

1. 最后匹配的动物仅在每次满足条件时覆盖数组时才会被存储。示例：每次都会被赋值为 trueanimalArray[0]zooAnimals.get(key).getColor().equals("Brown")
2. 每次满足条件时，for 循环都会迭代 10 次，将值分配给索引 0-9 处的数组

int index=0;
for(String key: zooAnimals .keySet()){
    if(zooAnimals .get(key).getColor().equals("Brown"){
        animalArray[index] = zooAnimals.get(key);
        index++;
    }
}       

你可以更简单地使用流来解决这个问题：

String color = "brown";
Animal[] animals = zooAnimals
        .values()  // get a collection of the map's values 
        .stream()  // convert it into a stream      
        // filter stream to gather only animals of a certain color
        .filter(animal -> animal.getColor().equalsIgnoreCase(color))
        .toArray(Animal[]::new); // collect the stream into an array of animals

下面是一个可运行的示例：

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

public class TestAnimals {
    public static void main(String[] args) {
        Map<String, Animal> zooAnimals = new HashMap<>();

        // fill a map with a bunch of animals
        zooAnimals.put("Lion", new Animal("Lion", "Yellow"));
        zooAnimals.put("Tiger", new Animal("Tiger", "Orange"));
        zooAnimals.put("Bear", new Animal("Bear", "Brown"));
        zooAnimals.put("Penguin", new Animal("Penguin", "Black"));
        zooAnimals.put("Panda", new Animal("Panda", "Black and White"));
        zooAnimals.put("Polar Bear", new Animal("Polar Bear", "White"));
        zooAnimals.put("Giraffe", new Animal("Giraffe", "Yellow"));
        zooAnimals.put("Elephant", new Animal("Elephant", "Gray"));
        zooAnimals.put("Horse", new Animal("Zebra", "Brown"));
        zooAnimals.put("Monkey", new Animal("Monkey", "Brown"));
        zooAnimals.put("Kangaroo", new Animal("Kangaroo", "Brown"));
        zooAnimals.put("Koala", new Animal("Koala", "Gray"));
        zooAnimals.put("Crocodile", new Animal("Crocodile", "Green"));
        zooAnimals.put("Snake", new Animal("Snake", "Green"));
        zooAnimals.put("Turtle", new Animal("Turtle", "Green"));

        Animal[] brownAnimals = getAnimalsByColor(zooAnimals, "Brown");

        System.out.println(Arrays.toString(brownAnimals));
    }

    public static Animal[] getAnimalsByColor(Map<String, Animal> zooAnimals, String color) {
        Animal[] animals = zooAnimals.values().stream()
                .filter(animal -> animal.getColor().equalsIgnoreCase(color))
                .toArray(Animal[]::new);
        return animals;
    }
}

class Animal {
    private String name;
    private String color;

    public Animal(String name, String color) {
        this.name = name;
        this.color = color;
    }

    // getters
    public String getName() {
        return name;
    }

    public String getColor() {
        return color;
    }

    // tostring
    @Override
    public String toString() {
        return "Name: " + name + ", Color: " + color;
    }

}

我假设你在动物类中没问题

 // a List to store animals
    List<Animal> animalList = new ArrayList<>();   

 // Sort HashMap to find animals with the specified color
    for (Animal animal : zooAnimals.keyset()) {
        if (animal.getColor().equals("Brown")) {
            animalList.add(animal);
        }
    }

//then if you want convert list to array
Animal[] animalArray = animalList.toArray(new Animal[animalList.size()]);

让我知道它是否有效

"...我有一个动物的HashMap。我必须创建一个单独的数组，并从 HashMap 中找到某种颜色的动物。所有匹配项都将放入数组中。..."

请改用 Map#values 方法。

Animal[] animalArray = new Animal[0];

for (Animal a : zooAnimals.values())
    if (a.getColor().equals("Brown")) {
        animalArray = Arrays.copyOf(animalArray, animalArray.length + 1);
        animalArray[animalArray.length - 1] = a;
    }

（可选）使用流。

Animal[] animalArray
    = zooAnimals.values()
                .stream()
                .filter(x -> x.getColor().equals("Brown"))
                .toArray(Animal[]::new);