DateTime.Subtraction 運算子

定義

命名空間:

System

組件:

System.Runtime.dll

從指定的 DateTime 減去指定的 DateTime 或 TimeSpan。

多載

Subtraction(DateTime, DateTime)	將另一個指定的日期和時間減去指定的日期和時間，並傳回時間間隔。
Subtraction(DateTime, TimeSpan)	將指定的日期和時間減去指定的時間間隔，並傳回新的日期和時間。

Subtraction(DateTime, DateTime)

Source:

DateTime.cs

將另一個指定的日期和時間減去指定的日期和時間，並傳回時間間隔。

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

參數

d1

DateTime

位於減號左邊的日期和時間值 (被減數)。

d2

DateTime

位於減號右邊的日期和時間值 (減數)。

傳回

TimeSpan

d1 和 d2 之間的時間間隔，也就是 d1 減 d2。

範例

下列範例示範 Subtract 方法和減法運算子。

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)

備註

執行減法時，方法 Subtraction(DateTime, DateTime) 不會考慮兩 DateTime 個值的 Kind 屬性值。 在減去 DateTime 物件之前，請確定物件代表相同時區的時間。 否則，結果會包含時區之間的差異。

注意

方法 DateTimeOffset.Subtraction(DateTimeOffset, DateTimeOffset) 會在執行減法時，考慮時區之間的差異。

這個運算子的對等方法為 DateTime.Subtract(DateTime)

另請參閱

• Subtract(DateTime)
• Ticks
• 使用日期和時間執行算數運算

Subtraction(DateTime, TimeSpan)

Source:

DateTime.cs

將指定的日期和時間減去指定的時間間隔，並傳回新的日期和時間。

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

參數

d

DateTime

位於減號左邊的日期和時間值。

t

TimeSpan

要減去的時間間隔。

傳回

DateTime

物件，其值為 d 值減掉 t 值的差異值。

例外狀況

ArgumentOutOfRangeException

產生的 DateTime 小於 DateTime.MinValue 或大於 DateTime.MaxValue。

範例

下列範例示範 Subtract 方法和減法運算子。

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)

備註

這個方法會從 的刻度值 t 減去 的 d 刻度值。

這個運算子的對等方法為 DateTime.Subtract(DateTime)

另請參閱

• Subtract(TimeSpan)
• Ticks
• TimeSpan
• 使用日期和時間執行算數運算