I want to know if the 4th integer in the ID, is even, or if its odd.
If the 4th number is even (if the number is either 0,2,4,6,8 I want to put the ID into a new column named 'even'
IF the 4th number is odd, the column should have the name 'Odd'
select ID as 'Female'
from Users2
where ID LIKE '%[02468]'
This shows if any of the numbers are even. I want to specify the 4th number
Try this:
select *, OddOrEven = iif(substring(ID,4,1) in ('0','2','4','6','8') , 'Even', 'Odd') from Users2
This will tell you whether the 4th character is Odd or Even.
This is of course assuming that the 4th character of ID column will be numeric.
To make it permanently part of the table, you can add a computed column as shown below.
alter table Users2
add OddOrEven as iif(substring(ID,4,1) in ('0','2','4','6','8'), 'Even', 'Odd')
Substring the character you are interested in
Convert to an int
Check whether modulus 2 returns 0 (i.e. even).
select id
, case when convert(int,substring(id, 4, 1)) % 2 = 0 then 'Even' else 'Odd' end
from Users;
Example:
select id
, case when convert(int,substring(id, 4, 1)) % 2 = 0 then 'Even' else 'Odd' end
from (values ('4545-4400'), ('4546-4400')) X (id);
Returns
id
4545-4400
Odd
4546-4400
Even
Thats assuming there is always a 4th character. If not you would need to check for it.
You were close, but only need to check a single character against a set of characters:
where Substring( Id, 4, 1 ) like '[02468]'
Note that there is no wildcard (%) in the pattern.
It can be used in an expression like:
case when Substring( Id, 4, 1 ) like '[02468]' then 'Even' else 'Odd' end as Oddity
Related
How can I select records where in the column Value the 5th character is letter A?
For example the following records:
ID Value
-------------------------
1 1234A5636A6363
2 1234A4343B6363
3 1234B5353A6363
if I run
select * from table
where Value like '%A%'
this will return all records
but all I want is the first 2 where the 5th character is A, regardless if there are more A characters in the text or not
select *
from your_table
where substring(Value, 5, 1) = 'A'
The LIKE operator, in addition to %, which matches any number of any character, can use _, which matches any one single character. You may try:
SELECT *
FROM yourTable
WHERE Value LIKE '____A%'; -- 4 underscores here
use like below by using _(underscore)
LIKE '____A%'
SQL Server
select *
from YourTableName
where CHARINDEX('A', ColumnName) = 5
Note:- This finds where string 'A' starts at position 5
AND specify Your ColumnName
SQL Server 2008 R2
I have a table similar to this:
Example table:
ID Column
---------------
xxx1234
xxx12345
xxx123456
20150001
I am trying to get a conditional MAX value depending on the value of the column based on whether it meets as certain format. Using the above example, the fourth record, 20150001, represents a "good record" because it contains the current year, and the start of an increment. So, in my table, records that are considered "good" (those subject to the query I am trying to write) have the format "year + increment". The first three, that do not follow this format, should not be conditioned to the query since they don't match this format and should not be subject when computing the max value. Those are bad records. In the above example, the expected result would be "20150002".
The MAX query is simple enough to write, however I am wondering about an approach where I can sanitize the query to only include those records whom meet the particular format, and increment the last four digits (0001 to 0002).
TIA!
You can use the isdate function to filter out ID Columns that do not start with a valid year, and isnumeric to make sure the last 4 characters of the ID Column are valid increments. You also want the len to be 8, given this criteria. You can accomplish all this in the where clause:
-- load test data
declare #Example_Table table(ID_Column varchar(10))
insert into #Example_Table values
('xxx1234'),
('xxx12345'),
('xxx123456'),
('20150001')
-- return max valid ID_Column
select max(ID_Column) as max_ID_Column
from #Example_Table
where isdate(left(ID_Column,4)) = 1
and isnumeric(right(ID_Column,4)) = 1
and len(ID_Column) = 8
-- increment max valid ID_Column
update #Example_Table
set ID_Column = cast(ID_Column as int) + 1
where isdate(left(ID_Column,4)) = 1
and isnumeric(right(ID_Column,4)) = 1
and len(ID_Column) = 8
select * from #Example_Table
ID_Column
----------
xxx1234
xxx12345
xxx123456
20150002
You could use a regular expression to verify a correct year. The second half of the regular expression I taylored to your examples of 0001 and 0002, this could be opened up by adding '[0-9]' for each digit you're expecting.
DECLARE #Sample VARCHAR(30) = '20150001';
SELECT CASE WHEN (#Sample LIKE '[12][09][0-9][0-9]000[12]') THEN 'Yes' ELSE 'No' END;
SELECT
SUBSTRING(#Sample, 1, 4),
SUBSTRING(#Sample, 5, 4),
CASE WHEN (SUBSTRING(#Sample, 1, 4) LIKE '[12][09][0-9]') THEN 'Yes' ELSE 'No' END,
CASE WHEN (SUBSTRING(#Sample, 5, 4) LIKE '[0-9][0-9][0-9][0-9]') THEN 'Yes' ELSE 'No' END;
I have a column DOOR which is a VARCHAR2 in a Table ADDRESS. I want to sort the column DOOR.
DOOR contains only two digits and no - sign
currently when I use the query
select sname, door, zip from address a order by door
I get the following result:
a
b
1
10
11
2
3
31
But I want the result to look like this:
a
b
1
2
3
10
11
31
I tried to convert DOOT into a numeric value using to_number as
select sname, to_number(door) dnr, zip from address a order by dnr
but it is giving me an error ORA-01722.
You can do this with the logic in the order by:
order by (case when regexp_like(door, '^[0-9]*$') = 0 then 1 else 0 end) desc,
(case when regexp_like(door, '^[0-9]*$') = 0 then door end),
length(door),
door
This first puts the non-numeric values first. The second clauses sorts these alphabetically. The third and fourth are for numbers. By sorting for the length before the value, you will get the numbers in order.
ORA-01722 error coming because of value 'a' ,'b',
Go for custom function which will take varchar and return number to convert , use custom function in order by clause of your query.
CREATE OR REPLACE FUNCTION tonumber (no_str varchar2)
RETURN number IS
num number := 0;
BEGIN
RETURN to_number(no_str);
EXCEPTION -- exception handlers begin A < B < 1 < 2
WHEN value_error THEN -- handles all other errors
dbms_output.put_line('in other exception catch.');
CASE
WHEN ( upper(no_str) = 'B' ) THEN return -1;
WHEN ( upper(no_str) ='A') THEN return -2;
ELSE return -999;
END CASE;
END;
Add when condition as in required. now assumed it can have only A B. for rest it will return default.
(This approach assumes that there aren't any mixed values like "234abc567".)
So, going old school...just 0-pad the strings to the maximum length of the column so that they'll sort properly as characters. But, to get the "non-numeric" values to sort first, nullify non-numeric values, put the NULLs first and padded values after that.
select door
from address
order by case when replace(translate(door, '012345679', '0000000000'), '0', '') is null
then lpad(door, 10, '0') -- value was entirely made of digits (change the 10 to the max width of the column)
else null
end nulls first
, door -- sorting within the group of NULL rows generated in the previous expression.
use the below query
SELECT PUMP_NAME
FROM MASTER.PUMPS
ORDER BY LPAD(PUMP_NAME, 10);
I have a column in my table called TO which is a comma separated list of email addresses. (1-n)
I am not concerned with a row if it ONLY contains addresses to Whatever#mycompany.com and want to flag that as 0. However, if a row contains a NON mycompany address (even if there are mycompany addresses present) I'd like to flag it as 1. Is this possible using one LIKE statement?
I've tried;
AND
[To] like '%#%[^m][^y][^c][^o][^m][^p][^a][^n][^y]%.%'
The ideal output will be:
alice#mycompany.com, bob#mycompany.com, malory#yourcompany.com 1
alice#mycompany.com, bob#mycompany.com 0
malory#yourcompany.com 1
Would it be better to write some kind of parsing function to split out addresses into a table if this isnt possible? I don't have an exhaustive list of other domains in the data.
It's ugly but it works. Case statement compares number of occurences of # symbol with number of occurences of #mycompany.com (XXX.. is just for keeping the length of the string):
select
*
, flag = case when len(field) - len(replace(replace(field,'#mycompany.com','XXXXXXXXXXXXXX'),'#','')) > 0 then 1 else 0 end
from (
select 'alice#mycompany.com, bob#mycompany.com, malory#yourcompany.com' as field union all
select 'alice#mycompany.com, bob#mycompany.com' union all
select 'malory#yourcompany.com'
) x
I would suggest a simple counting approach. Count the number of times that "#mycompany" appears and count the number of commas. If these differ, then you have an issue:
select emails,
(case when len(emails) - len(replace(emails, ',', '')) =
len(emails) - len(replace(emails, '#mycompany.com', 'mycompany.com'))
then 0
else 1
end) as HasNonCompanyEmail
from t
To simplify the arithmetic, I replace "#mycompany.com" with "mycompany.com". This removes exactly one character.
oI have a teradata table with about 10 million records in it, that stores a numeric id field as a varchar. i need to transfer the values in this field to a bigint column in another table, but i can't simply say cast(id_field as bigint) because i get an invalid character error. looking through the values, i find that there could be a character at any position in the string, so let's say the string is varchar(18) i could filter out invalid rows like so :
where substr(id_field,1,1) not in (/*big,ugly array of non-numeric chars*/)
and substr(id_field,2,1) not in (/*big,ugly array of non-numeric chars*/)
etc, etc...
then the cast would work, but this is not feasible in the long run. it's slow and if the string has 18 possible characters, it makes the query unreadable. how can i filter out rows that have a value in this field that will not cast as a bigint without checking each character individually for an array of non-numeric characters?
example values would be
123abc464
a2.3v65
a_356087
........
000000000
BOB KNIGHT
1235468099
the values follow no specific patterns, I simply need to filter out the ones that contain ANY non-numeric data.
123456789 is okay but 123.abc_c3865 is not...
Starting with TD14 Teradata added some functions, now there are multiple ways, e.g.:
WHERE RTRIM(col, '0123456789') = ''
But the easiest way is TO_NUMBER, which returns NULL for bad data:
TO_NUMBER(col)
The best that I've ever managed is this:
where char2hexint(upper(id_field)) = char2hexint(lower(id_field))
Since upper case characters give a different hex value to lower case ones, this will ensure that you have no alphabetical characters, but will still leave you with underscores, colons and so forth. If this doesn't meet your requirements, you may need to write an UDF.
could we also try to divide the values in the field by some integer "if divided then must be a number and if not and throws some error,then must have some character...." guess this would be lot fast as has just mathematics involved...
I've faced the same issue to try to exclude alpha characters from street address house numbers. The following will work if you don't mind concatanating all the numeric numbers together......
It checks if the upper of a string equals the lower of the string, if so it's a number, if not it becomes null.
select cast(case when upper(substring('12E'from 1 for 1)) = lower(substring('12E'from 1 for 1)) then substring('12E'from 1 for 1) else null end ||
case when upper(substring('12E'from 2 for 1)) = lower(substring('12E'from 2 for 1)) then substring('12E'from 2 for 1) else null end ||
case when upper(substring('12E'from 3 for 1)) = lower(substring('12E'from 3 for 1)) then substring('12E'from 3 for 1) else null end ||
case when upper(substring('12E'from 4 for 1)) = lower(substring('12E'from 4 for 1)) then substring('12E'from 4 for 1) else null end ||
case when upper(substring('12E'from 5 for 1)) = lower(substring('12E'from 5 for 1)) then substring('12E'from 5 for 1) else null end ||
case when upper(substring('12E'from 2 for 1)) = lower(substring('12E'from 2 for 1)) then substring('12E'from 2 for 1) else null end
as integer)
Try using this code segment
WHERE id_Field NOT LIKE '%[^0-9]%'
I found lins314159 answer to be very helpful with a similar issue. It may be an old thread but for what it's worth, I used:
char2hexint(upper(id_field)) = char2hexint(lower(id_field)) AND substr(id_field,1,1) IN ('1' to '9')
to successfully cast the remaining VARCHAR results to INT
SELECT customer_id
FROM t
WHERE UPPER(customer_id)(CASESPECIFIC) <>
LOWER(customer_id)(CASESPECIFIC);
This works perfectly fine to check whether the values in a numeric field is non-numeric.
SELECT id_field
WHERE oTranslate(id_field, '0123456789','')<>'';
This works well for me! It reveals any id_field containing a non-numeric value