Question

你好，我在客户端输入y时循环

，然后我想获取您输入的值并确保它包含在我的客户列表中但我不知道该怎么做

Scanner scanner = new Scanner( System.in );
ArrayList<Cliente> clientes = new ArrayList<Cliente>();
ArrayList<OrderService> orderService = new ArrayList<OrderService>();
System.out.println("add Cliente name")
string name = scanner.nextLine();
cliente.setName(name);
fazer um loop para se quiser adicionar mais clientes.

System.out.println("starting");

boolean value = true;
while(value)){
    System.out.println("Enter a new Client name");
    string name = scanner.nextLine();
    if(!name && name.trim().equals("")){
        Cliente cliente = new Cliente(name);
        clientes.add(cliente);
        System.out.println("Cliente "+cliente.getName+" Adicionado")
    }
    else{
        System.out.println("Nome invalido")
    }
    System.out.println("Do you want to add more client? Y for Yes, N for no");
    string response = scanner.nextLine(); 
    if(!response.equalsIgnoreCase(y)){
        value = false
    }
}

System.out.println("Enter a Client name that you want to create a work order.");
string clientName = scanner.nextLine();

Answer 1

尝试使用流。

public static Cliente findByName(Collection<Cliente> clientes, String name)
{
    return clientes.stream().filter(cliente -> name.equals(cliente.getName())).findFirst().orElse(null);
}

Answer 2

您可以使用流api轻松实现您的问题假设您的Cliente类具有name属性。根据您提供的注释，可以在第二个if语句中使用checkedCliente

if(!name && name.trim().equals("")){
    Cliente checkedCliente = clientes.stream()
                .filter(x -> name.equals(x.getName()))    // check name is there?
                .findAny()
                .orElse(null);

    if(checkedCliente == null){
        Cliente cliente = new Cliente(name);
        clientes.add(cliente);
        System.out.println("Cliente "+cliente.getName+" Adicionado")
    }
    else {
        System.out.println("Cliente " + cliente.getName() + "is already in the collection");

        // checkedCliente can be used for updates etc in here.
    }
}

Answer 3

这是检查数组列表是否包含字符串的方法

{{1}}

检查对象的列表数组中是否存在值

3 个答案: