DateTime.Subtraction 运算符

定义

命名空间:

System

程序集:

System.Runtime.dll

从指定的 DateTime 减去指定的 DateTime 或 TimeSpan。

重载

Subtraction(DateTime, DateTime)	将指定的日期和时间与另一个指定的日期和时间相减，返回一个时间间隔。
Subtraction(DateTime, TimeSpan)	从指定的日期和时间减去指定的时间间隔，返回新的日期和时间。

Subtraction(DateTime, DateTime)

Source:

DateTime.cs

将指定的日期和时间与另一个指定的日期和时间相减，返回一个时间间隔。

public:
 static TimeSpan operator -(DateTime d1, DateTime d2);

public static TimeSpan operator - (DateTime d1, DateTime d2);

static member ( - ) : DateTime * DateTime -> TimeSpan

Public Shared Operator - (d1 As DateTime, d2 As DateTime) As TimeSpan

参数

d1

DateTime

要从中减去的日期和时间值（被减数）。

d2

DateTime

要减去的日期和时间值（减数）。

返回

TimeSpan

d1 和 d2 之间的时间间隔；即 d1 减去 d2。

示例

以下示例演示 Subtract 了 方法和减法运算符。

System::DateTime date1 = System::DateTime( 1996, 6, 3, 22, 15, 0 );
System::DateTime date2 = System::DateTime( 1996, 12, 6, 13, 2, 0 );
System::DateTime date3 = System::DateTime( 1996, 10, 12, 8, 42, 0 );

// diff1 gets 185 days, 14 hours, and 47 minutes.
System::TimeSpan diff1 = date2.Subtract( date1 );

// date4 gets 4/9/1996 5:55:00 PM.
System::DateTime date4 = date3.Subtract( diff1 );

// diff2 gets 55 days 4 hours and 20 minutes.
System::TimeSpan diff2 = date2 - date3;

// date5 gets 4/9/1996 5:55:00 PM.
System::DateTime date5 = date1 - diff2;

open System

let date1 = DateTime(1996, 6, 3, 22, 15, 0)
let date2 = DateTime(1996, 12, 6, 13, 2, 0)
let date3 = DateTime(1996, 10, 12, 8, 42, 0)

// diff1 gets 185 days, 14 hours, and 47 minutes.
let diff1 = date2.Subtract date1

// date4 gets 4/9/1996 5:55:00 PM.
let date4 = date3.Subtract diff1

// diff2 gets 55 days 4 hours and 20 minutes.
let diff2 = date2 - date3

// date5 gets 4/9/1996 5:55:00 PM.
let date5 = date1 - diff2

System.DateTime date1 = new System.DateTime(1996, 6, 3, 22, 15, 0);
System.DateTime date2 = new System.DateTime(1996, 12, 6, 13, 2, 0);
System.DateTime date3 = new System.DateTime(1996, 10, 12, 8, 42, 0);

// diff1 gets 185 days, 14 hours, and 47 minutes.
System.TimeSpan diff1 = date2.Subtract(date1);

// date4 gets 4/9/1996 5:55:00 PM.
System.DateTime date4 = date3.Subtract(diff1);

// diff2 gets 55 days 4 hours and 20 minutes.
System.TimeSpan diff2 = date2 - date3;

// date5 gets 4/9/1996 5:55:00 PM.
System.DateTime date5 = date1 - diff2;

Dim date1 As New System.DateTime(1996, 6, 3, 22, 15, 0)
Dim date2 As New System.DateTime(1996, 12, 6, 13, 2, 0)
Dim date3 As New System.DateTime(1996, 10, 12, 8, 42, 0)

Dim diff1 As System.TimeSpan
' diff1 gets 185 days, 14 hours, and 47 minutes.
diff1 = date2.Subtract(date1)

Dim date4 As System.DateTime
' date4 gets 4/9/1996 5:55:00 PM.
date4 = date3.Subtract(diff1)

Dim diff2 As System.TimeSpan
' diff2 gets 55 days 4 hours and 20 minutes.
diff2 = System.DateTime.op_Subtraction(date2, date3)

Dim date5 As System.DateTime
' date5 gets 4/9/1996 5:55:00 PM.
date5 = System.DateTime.op_Subtraction(date1, diff2)

注解

方法 Subtraction(DateTime, DateTime) 在执行减法时不考虑 Kind 两 DateTime 个值的 属性的值。 在减 DateTime 去对象之前，请确保对象表示同一时区中的时间。 否则，结果将包括时区之间的差异。

注意

方法 DateTimeOffset.Subtraction(DateTimeOffset, DateTimeOffset) 在执行减法时会考虑时区之间的差异。

此运算符的等效方法是 DateTime.Subtract(DateTime)

另请参阅

• Subtract(DateTime)
• Ticks
• 使用日期和时间执行算术运算

Subtraction(DateTime, TimeSpan)

Source:

DateTime.cs

从指定的日期和时间减去指定的时间间隔，返回新的日期和时间。

public:
 static DateTime operator -(DateTime d, TimeSpan t);

public static DateTime operator - (DateTime d, TimeSpan t);

static member ( - ) : DateTime * TimeSpan -> DateTime

Public Shared Operator - (d As DateTime, t As TimeSpan) As DateTime

参数

d

DateTime

要从其中减去的日期和时间值。

t

TimeSpan

待减去的时间间隔。

返回

DateTime

一个对象，其值为 d 的值减去 t 的值。

例外

ArgumentOutOfRangeException

生成的 DateTime 小于 DateTime.MinValue 或大于 DateTime.MaxValue。

示例

以下示例演示 Subtract 了 方法和减法运算符。

System::DateTime date1 = System::DateTime( 1996, 6, 3, 22, 15, 0 );
System::DateTime date2 = System::DateTime( 1996, 12, 6, 13, 2, 0 );
System::DateTime date3 = System::DateTime( 1996, 10, 12, 8, 42, 0 );

// diff1 gets 185 days, 14 hours, and 47 minutes.
System::TimeSpan diff1 = date2.Subtract( date1 );

// date4 gets 4/9/1996 5:55:00 PM.
System::DateTime date4 = date3.Subtract( diff1 );

// diff2 gets 55 days 4 hours and 20 minutes.
System::TimeSpan diff2 = date2 - date3;

// date5 gets 4/9/1996 5:55:00 PM.
System::DateTime date5 = date1 - diff2;

open System

let date1 = DateTime(1996, 6, 3, 22, 15, 0)
let date2 = DateTime(1996, 12, 6, 13, 2, 0)
let date3 = DateTime(1996, 10, 12, 8, 42, 0)

// diff1 gets 185 days, 14 hours, and 47 minutes.
let diff1 = date2.Subtract date1

// date4 gets 4/9/1996 5:55:00 PM.
let date4 = date3.Subtract diff1

// diff2 gets 55 days 4 hours and 20 minutes.
let diff2 = date2 - date3

// date5 gets 4/9/1996 5:55:00 PM.
let date5 = date1 - diff2

System.DateTime date1 = new System.DateTime(1996, 6, 3, 22, 15, 0);
System.DateTime date2 = new System.DateTime(1996, 12, 6, 13, 2, 0);
System.DateTime date3 = new System.DateTime(1996, 10, 12, 8, 42, 0);

// diff1 gets 185 days, 14 hours, and 47 minutes.
System.TimeSpan diff1 = date2.Subtract(date1);

// date4 gets 4/9/1996 5:55:00 PM.
System.DateTime date4 = date3.Subtract(diff1);

// diff2 gets 55 days 4 hours and 20 minutes.
System.TimeSpan diff2 = date2 - date3;

// date5 gets 4/9/1996 5:55:00 PM.
System.DateTime date5 = date1 - diff2;

Dim date1 As New System.DateTime(1996, 6, 3, 22, 15, 0)
Dim date2 As New System.DateTime(1996, 12, 6, 13, 2, 0)
Dim date3 As New System.DateTime(1996, 10, 12, 8, 42, 0)

Dim diff1 As System.TimeSpan
' diff1 gets 185 days, 14 hours, and 47 minutes.
diff1 = date2.Subtract(date1)

Dim date4 As System.DateTime
' date4 gets 4/9/1996 5:55:00 PM.
date4 = date3.Subtract(diff1)

Dim diff2 As System.TimeSpan
' diff2 gets 55 days 4 hours and 20 minutes.
diff2 = System.DateTime.op_Subtraction(date2, date3)

Dim date5 As System.DateTime
' date5 gets 4/9/1996 5:55:00 PM.
date5 = System.DateTime.op_Subtraction(date1, diff2)

注解

此方法从 的刻度值 t 中减去 的 d刻度值。

此运算符的等效方法是 DateTime.Subtract(DateTime)

另请参阅

• Subtract(TimeSpan)
• Ticks
• TimeSpan
• 使用日期和时间执行算术运算