首页 \ 问答 \ 为什么要从ADO切换到ADO.NET？(Why switch from ADO to ADO.NET?)

为什么要从ADO切换到ADO.NET？(Why switch from ADO to ADO.NET?)

 我有一些朋友是“老派”的VB6数据库开发人员，我向他们讲述.NET及其功能，特别是ADO.NET。  
 在我们的谈话中，他们提出了以下原因，为什么他们宁愿坚持使用ADO而不愿转向ADO.NET：  
 
  数据集断开（如果电源故障会怎么样？）  
  仍然需要编写相同数量的代码  
  Dataset ， BindingSource和TableAdapter的新选项似乎令人困惑  
  编写相同的代码以编程方式访问数据库，所有变化都是如何布置该命令  
 
 我正在寻找答案或理由，说明为什么ADO.NET在涉及到Windows Forms应用程序时比ADO在数据访问方面“更好”。 ADO.NET提供的ADO没有提供什么？ 它比ADO做得更好吗？  
 
  注意：我想要不涉及LINQ的示例。  

I have some friends who are 'old-school' VB6 database developers and I was telling them about .NET and its features, specifically ADO.NET. 
In our conversation, they brought up the following reasons why they would rather stick with ADO than move to ADO.NET: 
 
 The Dataset is disconnected (What if power fails?) 
 The same amount of code still has to be written 
 The new options of Dataset, BindingSource and TableAdapter seem confusing 
 the same code is written programatically access the Database, all that changes is how that command is laid out 
 
I'm looking for answers or reasons why ADO.NET is 'better' than ADO with regards to data access when it comes to Windows Forms applications. What does ADO.NET provide that ADO does not? What does it do better than ADO? 
 
 NB: I'd like examples that do not involve LINQ. 

原文：https://stackoverflow.com/questions/1160639

更新时间：2023-10-29 09:10

最满意答案

 你所描述的是所谓的线性插值 。  
 对于一般情况，假设在a和b之间有一个值c ，并且需要一个介于0和1之间的值，该值基于c与a和b之间的相对位置。 x的等式如下：  
x := (c - a) / (b - a)
 
 因此，如果您的值在10到15之间（比如说11），并且您想要一个介于0和1之间的值，则可以将这些值插入上面的公式中：  
x := (11 - 10) / (15 - 10)
x := 1/5
 
 换句话说，11是从10到15的五分之一。  
 
 更为一般的情况（当你在a和b之间有一个值c ，并且你希望y和z之间有一个值x ）， x的计算方法如下：  
x := (c - a) * (z - y) / (b - a) + y
 
 在你的情况下， z = 1和y = 0 。 

What you're describing is called linear interpolation. 
For the general case, suppose you have a value c between a and b, and you want a value x between 0 and 1 which is based on c's relative position between a and b. The equation for x is as follows: 
x := (c - a) / (b - a)
 
So if you have a value between 10 and 15 (let's say 11), and you want a value between 0 and 1, you punch the values into the equation above: 
x := (11 - 10) / (15 - 10)
x := 1/5
 
In other words, 11 is one-fifth of the way from 10 to 15. 
 
The even more general case (when you have a value c between a and b and you want a value x between y and z), x is calculated as follows: 
x := (c - a) * (z - y) / (b - a) + y
 
In your case, z = 1 and y = 0.

为什么要从ADO切换到ADO.NET？(Why switch from ADO to ADO.NET?)

最满意答案

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