如何優化周期？

Question

有一種方法將 String [] 類型的兩個參數作為輸入，並根據接收到的內容進行過濾。

方法

 @Override
    public List<Product> filter(String[] brands, String[] cpus) {
        List<Product> products;
        List<Product> toReturn = new ArrayList<>();
        int i = 0;
        if(brands != null && cpus != null){
            for(String brand: brands){
                for(String cpu: cpus){
                    products = productRepo.findAllByBrandAndCpu(brand, cpu);
                    toReturn.addAll(products);
                    System.out.println(i++);
                }
            }
        }else{
            if (brands != null){
                for (String brand: brands) {
                    products = productRepo.findAllByBrand(brand);
                    toReturn.addAll(products);
                }
            }else if(cpus != null){
                for (String cpu: cpus) {
                    products = productRepo.findAllByCpu(cpu);
                    toReturn.addAll(products);
                }
            }else{
                return productRepo.findAll();
            }
        }
        return toReturn;
    }

如何更改此代碼，以便此處沒有那么多迭代：

這里

if(brands != null && cpus != null){
            for(String brand: brands){
                for(String cpu: cpus){
                    products = productRepo.findAllByBrandAndCpu(brand, cpu);
                    toReturn.addAll(products);
                    System.out.println(i++);
                }
            }
        } 

方法 findAllByBrandAndCpu

@Query(value = "SELECT p FROM Product p where p.brand.name = ?1 and p.cpu.name = ?2")
List<Product> findAllByBrandAndCpu(String brandName, String cpuName);

Answer 1

您好@Kzz，您可以通過使用IN SQL 運算符更新存儲庫查詢來刪除兩個嵌套循環，如下所示：

@Query(value = "SELECT p FROM Product p where p.brand.name IN ?1 and p.cpu.name IN ?2")
List<Product> findAllByBrandAndCpu(List<String> brands, List<String> cpus);