有没有更好的方法从多个字符串数组中找到任何字符串

Question

在下面的示例中，有一些预定义的字符串数组，我在下面的示例中提到过，现在我的要求是如果每个数组中存在 boName，我需要调用为特定自定义字段定义的方法。有没有其他方法可以避免多个 for 循环。

public class MyClass {

    private static final String[] OTHER_CUSTOM_FIELD_RELATED_BO_NAME = { "A", "B", "C", "D", "E", "F", "G", "H" };
    private static final String[] TRANSACTION_CUSTOM_FIELD_RELATED_BO_NAME = { "A", "AB", "CD", "DE", "EG" };
    private static final String[] CRM_CUSTOM_FIELD_RELATED_BO_NAME = { "I", "J", "K", "L", "M" };
    private static final String[] ITEM_CUSTOM_FIELD_RELATED_BO_NAME = { "N", "O", "P", "Q" };
    private static final String[] ENTITY_CUSTOM_FIELD_RELATED_BO_NAME = { "AN", "ZY", "X", "P" };

    public static List<String> BusinessObjectDescribe(String boName) {
        List<String> fields = new ArrayList<String>();
        if (boName != null && !boName.isEmpty()) {
            for (String string : OTHER_CUSTOM_FIELD_RELATED_BO_NAME) {
                if (boName.equals(string)) {
                    fields.addAll(getOtherCustomFields());
                    break;
                }
            }

            for (String string : TRANSACTION_CUSTOM_FIELD_RELATED_BO_NAME) {
                if (boName.equals(string)) {
                    fields.addAll(getTransactionCustomFields());
                    break;
                }
            }

            for (String string : CRM_CUSTOM_FIELD_RELATED_BO_NAME) {
                if (boName.equals(string)) {
                    fields.addAll(getCRMCustomFields());
                    break;
                }
            }

            for (String string : ITEM_CUSTOM_FIELD_RELATED_BO_NAME) {
                if (boName.equals(string)) {
                    fields.addAll(getITEMCustomFields());
                    break;
                }
            }

            for (String string : ENTITY_CUSTOM_FIELD_RELATED_BO_NAME) {
                if (boName.equals(string)) {
                    fields.addAll(getEntityCustomFields());
                    break;
                }
            }

        }

        return fields;

    }

}

Answer 1

如果您使用的是 Java8，则可以利用函数式接口的优势。只需创建一个新函数，它将您的数组作为第一个参数，并将供应商函数作为第二个参数。在供应商中，您可以传递有关添加的逻辑。而且您需要多次调用该函数而不是重复代码。

Answer 2

要搜索，可以在 java-8 中使用以下内容：

Arrays.stream(OTHER_CUSTOM_FIELD_RELATED_BO_NAME)
    .filter(e -> e.contains("A"))
    .findFirst();

这里我们将数组转换为流，然后应用过滤器。如果找到匹配项，它将返回元素，否则将返回null。此外，.findFirst() 将在找到匹配项后立即停止搜索流。

您可以使用相同的语法并根据需要搜索其他数组。

Answer 3

我会创建一个类来封装您在循环中执行的行为。

class MyClass {
    class BoNameHandler {
        private Collection<String> applicableNames;
        private Supplier<Collection<String>> getter;
        BoNameHandler(String[] names, Supplier<Collection<String>> g) {
            applicableNames = Arrays.asList(names);
            getter = g;
        }
        void addIfApplicable(String boName, List<String> addTo) {
            if (applicableNames.stream().filter(boName::equals).findFirst().isPresent()) {
                addTo.addAll(getter.get());
            }
        }
    }
    List<BoNameHandler> handlers = new ArrayList<>();
    MyClass() {
        handlers.add(new BoNameHandler(OTHER_CUSTOM_FIELD_RELATED_BO_NAME, this::getOtherCustomFields));
        handlers.add(new BoNameHandler(TRANSACTION_CUSTOM_FIELD_RELATED_BO_NAME, this::getTransactionCustomFields));
        // and the others
    }
}

这会将您的代码减少到

public static List<String> BusinessObjectDescribe(String boName) {
    List<String> fields = new ArrayList<String>();
    for (BoNameHandler handler : handlers) {
        handler.addIfApplicable(boName, fields);
    }
    return fields;
}

我不知道您的方法是什么样的，但如果它们与 BO_NAME 一样固定，您不妨将它们放在一个 enum 中。

Answer 4

作为Set，数学集合{ a, b, c }，其实是最符合逻辑的数据结构，用那个吧。

private static final Set<String> OTHER_CUSTOM_FIELD_RELATED_BO_NAME =
    Set.of( "A", "B", "C", "D", "E", "F", "G", "H" );
private static final Set<String> TRANSACTION_CUSTOM_FIELD_RELATED_BO_NAME =
    Set.of( "A", "AB", "CD", "DE", "EG" );
private static final Set<String> CRM_CUSTOM_FIELD_RELATED_BO_NAME =
    Set.of( "I", "J", "K", "L", "M" );
private static final Set<String> ITEM_CUSTOM_FIELD_RELATED_BO_NAME =
    Set.of( "N", "O", "P", "Q" );
private static final Set<String> ENTITY_CUSTOM_FIELD_RELATED_BO_NAME =
    Set.of( "AN", "ZY", "X", "P" );


public static List<String> BusinessObjectDescribe(String boName) {
    List<String> fields = new ArrayList<>();
    if (boName != null && !boName.isEmpty()) {
        if (OTHER_CUSTOM_FIELD_RELATED_BO_NAME.contains(boName)) {
            fields.addAll(getOtherCustomFields());
        }
        if (TRANSACTION_CUSTOM_FIELD_RELATED_BO_NAME.contains(boName)) {
            fields.addAll(getTransactionCustomFields());
        }
        if (CRM_CUSTOM_FIELD_RELATED_BO_NAME.contains(boName)) {
            fields.addAll(getCRMCustomFields());
        }
        if (ITEM_CUSTOM_FIELD_RELATED_BO_NAME.contains(boName)) {
            fields.addAll(getITEMCustomFields());
        }
        if (ENTITY_CUSTOM_FIELD_RELATED_BO_NAME.contains(boName)) {
            fields.addAll(getEntityCustomFields());
        }
    }
    return fields;
}

如果你的java版本还是不知道Set.of（java 9之前）：

private static final Set<String> OTHER_CUSTOM_FIELD_RELATED_BO_NAME = new HashSet<>();
static {
    Collections.addAll(OTHER_CUSTOM_FIELD_RELATED_BO_NAME,
        "A", "B", "C", "D", "E", "F", "G", "H" );
}

Set 是最快的数据结构。那么为什么不做到最好。

Answer 5

我还会考虑将所有这些数组放到一个 Map 或 Map 中，然后再进行处理。