有没有一种有效地迭代这些不同索引的好方法？

var array = [1, 2, 3, 4, 5];
for (var i = 0; i < array.length; i++) {
    console.log(array[i]);
}

List<Integer> list = Arrays.asList(1, 2, 3, 4, 5);
for (Integer num : list) {
    System.out.println(num);
}

test_cases = ["test_case1", "test_case2", "test_case3"]
for test_case in test_cases:
    execute_test_case(test_case)

SELECT * FROM table_name;

for server in server_list:
    execute_command(server, command)

containers:
  - name: container1
    image: image1
  - name: container2
    image: image2

for (int i = 0; i < num_clients; i++) {
    handle_request(client[i]);
}

security_policies = ["policy1", "policy2", "policy3"]
for policy in security_policies:
    apply_policy(device, policy)

audio_files = ["file1.wav", "file2.wav", "file3.wav"]
for file in audio_files:
    process_audio(file)

image_files = ["file1.jpg", "file2.jpg", "file3.jpg"]
for file in image_files:
    process_image(file)

training_samples = ["sample1", "sample2", "sample3"]
for sample in training_samples:
    train_model(sample)

sensors = ["sensor1", "sensor2", "sensor3"]
for sensor in sensors:
    collect_data(sensor)

List<String> users = Arrays.asList("user1", "user2", "user3");
for (String user : users) {
    send_push_notification(user, message);
}

files = ["file1.txt", "file2.txt", "file3.txt"]
for file in files:
    read_file(file)

for (uint i = 0; i < num_blocks; i++) {
    process_block(blocks[i]);
}

virtual_worlds = ["world1", "world2", "world3"]
for world in virtual_worlds:
    explore_world(world)