Questions about inserting time in SQL database

Insertion time problem in sql database

CONVERT(varchar(10), getDate(),120)

Convert the time format into character format (10) Note that this is very important, and then insert it

CONVERT(nvarchar(10),count_time,121): CONVERT is a date conversion function, usually in the time type

The function used to convert between (datetime, smalldatetime) and string types (nchar, nvarchar, char, varchar) has three parameters. The first parameter refers to the size after conversion, the second parameter refers to the field or function that needs to convert the date, and the third parameter refers to the format of the conversion.

0 | 0 or 100 | mon dd yyyy hh:miAM (or PM)

1 | 101 | mm/dd/yy

2 | 102 | yy-mm-dd

3 | 103 | dd/mm/yy

4 | 104 | dd-mm-yy

5 | 105 | dd-mm-yy

6 | 106 | dd mon yy

7 | 107 | mon dd,yy

8 | 108 | hh:mm:ss

9 | 9 or 109 | mon dd yyyy hh:mi:ss:mmmmAM (or PM)

10 | 110 | mm-dd-yy

11 | 111 | yy/mm/dd

12 | 112 | yymmdd

11| 13 or 113 | dd mon yyyy hh:mi:ss:mmm(24-hour format)

14 | 114 | hh:mi:ss:mmm(24-hour format)

15 | 20 or 120 | yyyy-mm-dd hh:mi:ss(24-hour format)

16 | 21 or 121 | yyyy-mm-dd hh:mi:ss:mmm(24-hour format)

How to calculate the execution time of sql statement

The datediff function is generally used in SQL to represent the time difference.

Basic syntax:

DATEDIFF(datepart,startdate,enddate)

illustrate:

The startdate and enddate parameters are legal date expressions.

The datepart parameter can be the following values:

example:

Example 1

Use the following SELECT statement:

SELECT DATEDIFF(day,'2008-12-29','2008-12-30') AS DiffDate

result:

DiffDate

1

Example 2

Use the following SELECT statement:

SELECT DATEDIFF(day,'2008-12-30','2008-12-29') AS DiffDate

result:

DiffDate

-1

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