Question

弹簧Predicate的{{1}}是静态还是非静止的？

默认实现类似如下：

Specification

但这也可以重构为：

public static Specification<Customer> byCustomerName(String name) {
    return new Specification<Customer>() {
        @Override
        public Predicate toPredicate(Root<BasePricingCache> root, CriteriaQuery<?> query, CriteriaBuilder cb) {
            return cb.equal(root.get(Customer_.name), name);
        }
    };
}

应该优先考虑什么？ private static final CUSTOMER_SPEC = return new Specification<Customer>() { @Override public Predicate toPredicate(Root<BasePricingCache> root, CriteriaQuery<?> query, CriteriaBuilder cb) { return cb.equal(root.get(Customer_.name), name); } }; public static Specification<Customer> byCustomerName(String name) { return CUSTOMER_SPEC; }线程是否安全且可以这样使用？

Answer 1

You're refactored code wouldn't even compile. There's a superfluous return, CUSTOMER_SPEC doesn't have a type, and you're referring to a name variable that doesn't exist in the scope.

I think you're overthinking the issue here and are in the process of micro-optimizing. Keeping the Specification in a static factory method should be fine.

Answer 2

线程安全不会成为实施的关注点。你根本不保留任何状态，每个参数都由调用者传递。

它可能是微优化的，但是当你的规范发生很多实例化时，你的重构会很有用。这可能会经常触发GC，从而导致响应速度明显滞后。

这种微观优化的缺点可能是你可能最终需要相当大的内存，因为实例不能用于GC。

首先通过收集数据（分析堆，......）来备份您的决定，并测试您的解决方案对应用程序的影响。

Spring规范的谓词应该是静态的吗？

2 个答案: