Question

public class Address{
 private TypeAddress typeAddress;
 private String street;
 //Getters and Setters
 ...
}

public class TypeAddress{
  private String type;
  //Getters and Setters
  ...
}

我有一个地址集

Collection<Address> addresses = db.getAddresses();

我需要检查客户是否会选择任何类型的地址。它只是假装一种情况所以你们会理解。

我想要的是，我想获得一种类型的地址，即客户选择的那种类型。如果他选择了一个不存在的类型，我会选择另一个地址。所以，我基本上需要做2个循环来检查它是否有任何类型，如果有，我将在下一个循环中选择它。如果不是，我会在下一个循环中选择另一个。有没有办法使用Collection的contains方法来检查地址的类型是否是我想要的？

类似的东西：

if(addresses.contains("someType")){//remind that I'm trying to check not the object Address, but the object TypeAddress
 //get this address
}else{
 //get another address
}

Answer 1

如果您被迫使用集合（而不是像建议的Map），为什么不能只迭代一次并将最“有希望”的实例存储为一些地址引用：

Address address = null;
for (Address addr : addresses) {
    if ( isBetterThanAlreadyFound(addr, address) ) {
        address = addr;
    }
}


private static boolean isBetterThanAlreadyFound(Address addr, Address alreadyFound) {
    return alreadyFound == null || (addr.isOfDesiredType() && alreadyFound.isNot());
}

Answer 2

使用Map<TypeAddress, Address>，以便您可以验证是否存在必要的地址，并使其相对快速和轻松。

Map<TypeAddress, Address> addresses = new HashMap<TypeAddress, Address>();

//put something and then
TypeAddress type = getType();
Address address = addresses.get(type);
if (address != null) {
    return address;
} else {
   //do smth else
}

不要忘记以适当的方式覆盖类equals的方法hashCode和TypeAddress。

Answer 3

甚至包含内部循环逻辑，使用Map代替

Answer 4

我认为您可能正在寻找一种方法来摆脱循环，或者在循环开始时继续使用的方法。方便的是，它们被称为break和continue

for (Address address : addresses) {
    if (!address.getType.equals("someType")) continue;
    myAddress = address;
}

OR

for (Address address : addresses) {
    if (address.getType.equals("someType")) {
        myAddress = address;
        break;
    }
}

Answer 5

您可以使用Apache CollectionUtils API：

Predicate p = new Predicate()
{
 public boolean evaluate(Object o)
 {
   return ((Adress) o).getTypeAddress().getType().equals(TYPE_INPUTED_BY_USER);
 }
}

return CollectionUtils.exists(p);

Answer 6

Error:MySQL shutdown unexpectedly.
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检查Collection是否包含Object内的Object

6 个答案: