C# 十进制的最佳数据注释(18,2)
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/19811180/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Best Data annotation for a Decimal(18,2)
提问by John John
I have a column inside my sql server 2008 wih type of Decimal(18,2). But on entity framework what is the best data annotation validation I can apply to this property, inside my asp.net MVC web application ?
我的 sql server 2008 中有一个列，类型为Decimal(18,2). 但是在实体框架上，我可以在我的 asp.net MVC Web 应用程序中应用于此属性的最佳数据注释验证是什么？
回答by ediblecode
There is no explicit data annotation for a decimal so you need to use two separate ones to add constraints.
小数没有明确的数据注释，因此您需要使用两个单独的注释来添加约束。
Two Decimal Points
两个小数点
[RegularExpression(@"^\d+\.\d{0,2}$")]
This regular expression will make sure that the property has at most two decimal places.
此正则表达式将确保该属性最多有两位小数。
Max 18 digits
最多 18 位
[Range(0, 9999999999999999.99)]
Assuming you aren't accepting any negative numbers. Otherwise, replace 0with -9999999999999999.99.
假设您不接受任何负数。否则，替换0为-9999999999999999.99。
Result
结果
[RegularExpression(@"^\d+\.\d{0,2}$")]
[Range(0, 9999999999999999.99)]
public decimal Property { get; set; }
回答by Schmalls
I think @jumpingcode's answer can be combined into one RegularExpressionAttribute.
我认为@jumpingcode 的答案可以合二为一RegularExpressionAttribute。
[RegularExpression(@"^(0|-?\d{0,16}(\.\d{0,2})?)$")]
public decimal Property
{
get;
set;
}
This can be used for any precisionand scale. The 16 is replaced by precision- scaleand the 2 is replaced by the scale. The regular expression should match numbers entered like ###, 0.##, .##, 0, and ###.##as well as negative values.
这可以用于任何precision和scale。16 被替换为precision- scale，而 2 被替换为scale。正则表达式匹配应该像输入的数字###，0.##，.##，0，和###.##以及为负值。
回答by mgalpy
This seems to be the correct answer ( the above answers either restrict valid numbers that can be inserted into a data type of Decimal(18,2) or cause compile errors if you apply them to your code -- please confirm for yourself):
这似乎是正确的答案（上述答案要么限制了可以插入 Decimal(18,2) 数据类型的有效数字，要么在将它们应用于代码时导致编译错误——请自行确认）：
Use the following two constraints together:
一起使用以下两个约束：
Two Decimal Points
两个小数点
[RegularExpression(@"^\d+.?\d{0,2}$", ErrorMessage = "Invalid Target Price; Maximum Two Decimal Points.")]
Max 18 digits
最多 18 位
[Range(0, 9999999999999999.99, ErrorMessage = "Invalid Target Price; Max 18 digits")]
回答by mjyazdani
[Range(1,(double) decimal.MaxValue, ErrorMessage="value should be between{1} and {2}."]
回答by Adam Diament
For a different approach which some may consider more readable, you can override the OnModelCreating method of your DbContext to set precision, like so:
对于某些人可能认为更具可读性的不同方法，您可以覆盖 DbContext 的 OnModelCreating 方法以设置精度，如下所示：
protected override void OnModelCreating(DbModelBuilder modelBuilder)
{
modelBuilder.Entity<YourEntity>()
.Property(x => x.TheProprty)
.HasPrecision(18, 2);
}
Advantage: strongly typed vs custom regular expression
优势：强类型 vs 自定义正则表达式
Disadvantage: can't see it on the class with just a scan
缺点：仅扫描在课堂上看不到
回答by ransems
Im using almost excplusively (b/c it's simple and works)
我几乎完全使用（b/c 它很简单而且有效）
[Range(typeof(decimal), "0", "1")]
public decimal Split { get; set; }
Then if I need to convert back to double I add a conversion
然后如果我需要转换回双精度我添加一个转换
(double)model.Split
回答by Breeno
Following on from @Schmalls example (and comment re building it into an attribute) I've created a working example (uses C# 6 string interpolation):
继@Schmalls 示例（并评论将其重新构建到属性中）之后，我创建了一个工作示例（使用 C# 6 字符串插值）：
public class PrecisionAndScaleAttribute : RegularExpressionAttribute
{
public PrecisionAndScaleAttribute(int precision, int scale) : base($@"^(0|-?\d{{0,{precision - scale}}}(\.\d{{0,{scale}}})?)$")
{
}
}
Usage:
用法：
[PrecisionAndScale(6, 2, ErrorMessage = "Total Cost must not exceed 99.99")]
public decimal TotalCost { get; set; }
回答by Daniel Jacobson
If you write the 'column' annotation, will work fine
如果您编写“列”注释，则可以正常工作
[Required]
[Column(TypeName = "decimal(18, 6)")]
public decimal Foo { get; set; }
