sql查询大于DATE（年）A但小于DATE（年）B

Question

I have a list of items like so:

Field: Name & Field: DOB
- David - 1962
- Robert - 1955
- Tracy - 1980
- Bono - 1964
- Betty - 1968

I need a simple SQL query that will return all name whose birth years ate in the 1960s (1960-1969). something along the lines of this logic:

$sql = "SELECT * FROM employees WHERE DOB => '1960' AND DOB =< '1969' AND ORDER BY name";

FYI: Currently the DOB field in the employee database is a VARCHAR field with 4 character designation.

Can someone point me to the correct syntax on such a query?

Answer 1

You should not be storing date/time information as varchar , but since you have one way you can screen for 1960s birthdays is to just check the first three characters of the DOB field:

SELECT *
FROM employees
WHERE SUBSTRING(DOB, 1, 3) = '196'
ORDER BY name

You could also use your original approach, with the following WHERE clause:

WHERE DOB >= '1960' AND DOB <= '1969'

However, I don't like treating strings as numbers, because you get weird things happening if the strings should ever change length. I would advise you to use a datetime to store your DOB information.

Answer 2

Change the query from:

"SELECT * FROM employees WHERE DOB => '1960' AND DOB =< '1969' AND ORDER BY name";

to

"SELECT * FROM employees WHERE DOB >= '1960' AND DOB <= '1969' AND ORDER BY name";
// Its >= or <= not => or =<

and try again

Answer 3

Assuming your varchar DOB column contains only numbers:

SELECT * 
FROM employees
WHERE CONVERT(DOB, INTEGER) >= 1960
   AND CONVERT(DOB, INTEGER) <= 1969

Answer 4

Did You try this ?

$sql = "SELECT * FROM employees WHERE DOB Like '%YYY%'AND ORDER BY name";

Consider yyy is 196X so all dob starting with 196 will result.

But this can't be specific enough

Answer 5

select * from tt1 where year >='1960' and year <='1969' order by name

希望这会有所帮助。