你可以使用这个方法将List转换为String数组:

 Object[] stringlist=list.toArray();

完整的例子:

ArrayList<String> list=new ArrayList<>();
    list.add("Abc");
    list.add("xyz");

    Object[] stringlist=list.toArray();

    for(int i = 0; i < stringlist.length ; i++)
    {
          Log.wtf("list data:",(String)stringlist[i]);
    }

从Java-11开始，可以使用API集合。toArray(IntFunction<T[]> generator)实现与:

List<String> list = List.of("x","y","z");
String[] arrayBeforeJDK11 = list.toArray(new String[0]);
String[] arrayAfterJDK11 = list.toArray(String[]::new); // similar to Stream.toArray

将任何List<Type>转换为String[]的泛型解决方案:

public static  <T> String[] listToArray(List<T> list) {
    String [] array = new String[list.size()];
    for (int i = 0; i < array.length; i++)
        array[i] = list.get(i).toString();
    return array;
}

注意:必须重写toString()方法。

class Car {
  private String name;
  public Car(String name) {
    this.name = name;
  }
  public String toString() {
    return name;
  }
}
final List<Car> carList = new ArrayList<Car>();
carList.add(new Car("BMW"))
carList.add(new Car("Mercedes"))
carList.add(new Car("Skoda"))
final String[] carArray = listToArray(carList);

private String[] prepareDeliveryArray(List<DeliveryServiceModel> deliveryServices) {
    String[] delivery = new String[deliveryServices.size()];
    for (int i = 0; i < deliveryServices.size(); i++) {
        delivery[i] = deliveryServices.get(i).getName();
    }
    return delivery;
}

Java 8中的替代方案:

String[] strings = list.stream().toArray(String[]::new);

Java 11 +:

String[] strings = list.toArray(String[]::new);

你可以使用Iterator<String>来迭代数组列表<String>中的元素:

ArrayList<String> list = new ArrayList<>();
String[] array = new String[list.size()];
int i = 0;
for (Iterator<String> iterator = list.iterator(); iterator.hasNext(); i++) {
    array[i] = iterator.next();
}

现在您可以使用任何循环从String[]中检索元素。