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具有多个后端和不同数据库设计的数据访问层(Data Access Layer with Multiple Backends and Different Database Designs)

我正在开始一个项目,并且正在为我们的数据访问层而苦苦挣扎。 基本上它需要与不同数据库设计的多个后端接口。

我想要一个通用的DAL,然后在任何后端执行一个通用功能。 后端具有用于插入,更新等的唯一代码。因此,在1个后端中添加一个Employee将在另一个代码中具有不同的代码。

我尝试了Repository模式,但这并不适用于这种情况。 我已经结束了一个工厂模式的方法,但我最终会为每个对象创建一个工厂。 我可能只能创建1个工厂,但后端对象将具有数百个函数,如“SaveEmployee”,“SavePlan”等。

现在我有以下几点:

DAL
    --> DAL.Backend1
        --> Employee.Save(employee)
        --> Plan.Save(plan)
    --> DAL.Backend2
        --> Employee.Save(employee)
        --> Plan.Save(plan)

在DAL项目中,我为每个对象,员工,计划都有一个工厂模式,以决定返回并执行哪个DAL对象。

我很确定这不是最好的架构,所以我想知道是否有更好的模式来解决我的问题。


I am starting a project and am struggling with the architecture for our data access layer. Basically it will need to interface with multiple backends with different database designs.

I would like a common DAL, which then executes a common function in any backend. The backends have unique code for inserting, updating, etc. So adding an Employee in 1 backend will have different code in another.

I tried the Repository pattern but that just doesn't apply to the situation. I've ended up with just a Factory pattern method, but I will end up creating a Factory for each object. I could maybe only create 1 factory but then the Backend object would have hundred of functions like "SaveEmployee", "SavePlan", etc.

Right now I have the following:

DAL
    --> DAL.Backend1
        --> Employee.Save(employee)
        --> Plan.Save(plan)
    --> DAL.Backend2
        --> Employee.Save(employee)
        --> Plan.Save(plan)

In the DAL project I have a Factory pattern for each Object, Employee, Plan, to decide which DAL's Object to return and execute against.

I am pretty sure this is not the best architecture for this, so I am wondering if there's a better pattern to use to solve my problem.


原文:https://stackoverflow.com/questions/14508077
更新时间:2022-04-17 06:04

最满意答案

你可以使用这样的循环:

char northl2[100] = {0};

northl2[0]='1';
northl2[1]='1';
int count_one = 0;
for (int i = 0;i<100;++i)
{
    if (northl2[i] == '1')
    {
        ++count_one;
    }
}
std::cout << count_one;

在这种情况下,这打印2,因为有2 1 。 代码遍历数组的每个元素,检查它的值,并递增其计数。 char northl2[100] = {0}; 默认情况下将每个元素设置为0.如果需要不同的值,请使用循环:

char northl2[100];
int main()
{
    int count_one = 0;
    for (int i = 0; i< 100;++i)
    {
         northl2[i] = 'C'; //Or whatever char other than '1'
    }
    northl2[0]='1';
    northl2[1]='1';
    for (int i = 0;i<100;++i)
    {
        if (northl2[i] == '1')
        {
            ++count_one;
        }
    }
}

另外,不要忘记在循环为所有元素赋值后分配1,否则它们将被覆盖


You could use a loop like this:

char northl2[100] = {0};

northl2[0]='1';
northl2[1]='1';
int count_one = 0;
for (int i = 0;i<100;++i)
{
    if (northl2[i] == '1')
    {
        ++count_one;
    }
}
std::cout << count_one;

This prints 2 in this case, because there are 2 1's. The code iterates through each element of the arrays, checks it for a value, and increments its count. The char northl2[100] = {0}; sets by-default each element to 0. If you need a different value, use a loop:

char northl2[100];
int main()
{
    int count_one = 0;
    for (int i = 0; i< 100;++i)
    {
         northl2[i] = 'C'; //Or whatever char other than '1'
    }
    northl2[0]='1';
    northl2[1]='1';
    for (int i = 0;i<100;++i)
    {
        if (northl2[i] == '1')
        {
            ++count_one;
        }
    }
}

Also, don't forget to assign the 1's after the loop assigns values to all of the elements, or else, they will be overwritten.

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