java - 这是按多个属性对项目进行分组并检查另一个属性是否为 null 或不为 null 的有效方法吗？

Question

我有以下类，它有 3 个我关心的属性:

public class Invoice {
    private String invoiceNumber;
    private String vendorCode;
    private BigDecimal vatAmount;
}

我需要按 invoiceNumber 和 vendorCode 对它们进行分组，并检查对于所有这些分组的项目，vatAmount 对于所有项目都为 null条目或不为空。如果任一条件失败，我都会抛出一个已检查异常 (RuleException)。到目前为止我有这个:

List<Invoice> invoices = ....;

Map<List<String>, List<Invoice>> uniqueInvoiceNumbers = invoices.stream()
    .collect(Collectors.groupingBy(e-> Arrays.asList(e.getInvoiceNumber(), e.getVendorCode())));

for (Map.Entry<List<String>, List<Invoice>> entrySet : uniqueInvoiceNumbers.entrySet()) {
    boolean allNull = entrySet.getValue().stream().allMatch(e-> e.getVatAmount() == null);
    boolean allFilled = entrySet.getValue().stream().allMatch(e-> e.getVatAmount() != null);
    if (!allNull && !allFilled) {
        throw new RuleException("The invoices associated with " +  entrySet.getKey().get(0) + " and vendor " + entrySet.getKey().get(1) + " have had their vat amount partially filled. They should either all be filled in or none at all");
    }
}

完成检查后，我会丢弃分组的项目uniqueInvoiceNumbers，因为它们仅用于检查。

有更好的方法吗？发票集合的大小范围为 1 到 200 项。

<小时/>

我用过this answer让代码按两个不同的键进行分组。

Answer 1

您只需要一次检查，因为“全部为空”或“全部非空”就是为了找到矛盾的元素。因此，列表至少需要两个元素才能包含矛盾的元素。当您检查 null 的第一个元素时，你知道，是否所有其他元素都必须是 null或非null .

Optional<List<String>> o = uniqueInvoiceNumbers.entrySet()
    .filter(e -> e.getValue().size() > 1)
    .filter(e -> {
        boolean isNull = e.getValue().get(0).getVatAmount() == null;
        return e.getValue().stream().map(Invoice::getVatAmount)
            .anyMatch(isNull? Objects::nonNull: Objects::isNull);
        })
    .map(Map.Entry::getKey)
    .findAny();

if(o.isPresent()) { // can't use ifPresent with checked exception
    List<String> key = o.get();
    throw new RuleException("The invoices associated with "
        +  key.get(0) + " and vendor " + key.get(1)
        + " have had their vat amount partially filled."
        + " They should either all be filled in or none at all");
}

您可以内联上面代码中使用的一些变量，这将使代码更简洁，但不一定更具可读性。

<小时/>

如果您不需要Map<List<String>, List<Invoice>> ，但仅收集它以进行此检查，您可以在分组时在一个操作中执行检查，但是，由于检查的异常，我将其实现为循环:

List<Invoice> invoices = ....;

Map<List<String>, Boolean> uniqueInvoiceNumbers = new HashMap<>();
for(Invoice i : invoices) {
    List<String> key = Arrays.asList(i.getInvoiceNumber(), i.getVendorCode());
    boolean isNull = i.getVatAmount() == null;
    Boolean existing = uniqueInvoiceNumbers.putIfAbsent(key, isNull);
    if(existing != null && existing != isNull) {
    throw new RuleException("The invoices associated with "
        +  key.get(0) + " and vendor " + key.get(1)
        + " have had their vat amount partially filled."
        + " They should either all be filled in or none at all");
    }
}

如果RuleException是一个未经检查的异常，我们可以这样做:

invoices.stream()
    .collect(Collectors.toMap(
        i -> Arrays.asList(i.getInvoiceNumber(), i.getVendorCode()),
        Function.identity(),
        (i1,i2) -> {
            if((i1.getVatAmount() == null) != (i2.getVatAmount() == null)) {
                 throw new RuleException("The invoices associated with "
                    +  i.getInvoiceNumber() + " and vendor " + i.getVendorCode()
                    + " have had their vat amount partially filled."
                    + " They should either all be filled in or none at all");
            }
            return i1;
        }));

这是相同的逻辑，尽管我们不能只映射到 Boolean value，这样，合并函数就不会知道异常消息的键。而且，如上所述，它需要取消检查异常，或者必须将其包装到另一个异常中，以便在流操作之外提取。这使得循环变体更可取，除非并行处理发挥作用，我不会考虑 200 个元素......