PropertyBuilder.HasPrecision Method
Definition
Important
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Overloads
| HasPrecision(Int32) |
Configures the precision of the property. |
| HasPrecision(Int32, Int32) |
Configures the precision and scale of the property. |
HasPrecision(Int32)
Configures the precision of the property.
public virtual Microsoft.EntityFrameworkCore.Metadata.Builders.PropertyBuilder HasPrecision (int precision);abstract member HasPrecision : int -> Microsoft.EntityFrameworkCore.Metadata.Builders.PropertyBuilder
override this.HasPrecision : int -> Microsoft.EntityFrameworkCore.Metadata.Builders.PropertyBuilderPublic Overridable Function HasPrecision (precision As Integer) As PropertyBuilderParameters
- precision
- Int32
The precision of the property.
Returns
The same builder instance so that multiple configuration calls can be chained.
Applies to
HasPrecision(Int32, Int32)
Configures the precision and scale of the property.
public virtual Microsoft.EntityFrameworkCore.Metadata.Builders.PropertyBuilder HasPrecision (int precision, int scale);abstract member HasPrecision : int * int -> Microsoft.EntityFrameworkCore.Metadata.Builders.PropertyBuilder
override this.HasPrecision : int * int -> Microsoft.EntityFrameworkCore.Metadata.Builders.PropertyBuilderPublic Overridable Function HasPrecision (precision As Integer, scale As Integer) As PropertyBuilderParameters
- precision
- Int32
The precision of the property.
- scale
- Int32
The scale of the property.
Returns
The same builder instance so that multiple configuration calls can be chained.
Applies to
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Entity Framework