Solve two sub problems for the database 10

Updated on technology 2024-03-04
3 answers
  1. Anonymous users2024-02-06

    41.where statement.

    42.drop statement: drop table s

    43.alter statement: alter table table1 add(columnname, varchar(32)).

    44.insert statement: insert into s values('Zhang Er','Chemistry',80)

    45. drop view viewname

    46. sname(in s) -cname(in c)

    48. where = 'Male' and = "Databases"

    49.update statement: update s set grade = 85 where sn ="Wang II" and cn = "Chemistry"

    50.insert statement: insert into s values('Zhang Er','Chemistry',80)

    51.alter statement: alter table table1 add(cn, varchar(32).

    52.The number of rows that meet the requirements is counted.

    54.Second Paradigm.

    55.Department number.

    56.First Paradigm.

    57.Third Paradigm.

    58.Completely dependent on either one.

    59.First Paradigm.

  2. Anonymous users2024-02-05

    The length of the bytes that you can only recognize is 8 bytes, and since you don't have anything to store, you can't see the remaining storage space.

  3. Anonymous users2024-02-04

    char(8) is 8 bytes, each byte holds one letter, then abc occupies three bytes, then the rest is 5 bytes.

    Because varchar(8) is a variable length, the length occupied by abc is the entire length, so varchar has no space left. It's 3 bytes + 1 byte (this byte is the length byte (sometimes this byte is not counted because this byte is not within 8)).

    nvarchar(8) is a variable Unicode character. Each character in nvarchar is a Unicode character, so each character occupies two bytes, which means that it occupies 6 bytes, and the remaining space is 0

    nchar(8) is basically the same as nvarchar(8), it is also a unicode character, but it is not mutable, so it still occupies 6 bytes, and the remaining space is 16-6=10 bytes.

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