I have a String of words/address ..
ID | ADDRESS
1 barangay1, City Province
2 barangay2, City Province
what i want to do is to select only the barangay without City and province
ID | ADDRESS
1 barangay1
2 barangay2
I tried using Position() but it only returns integer .. Can someone help
In addition to the POSITION() function you need the SUBSTRING() function to extract the part of the string. If the string you're intrested in is from the start of the string till the first comma then
select SUBSTRING(ADDRESS from 1 for POSITION(',' in ADDRESS)-1) from T
should work. You also might want to run the result throught TRIM() function to get rid of any leading and/or trailing whitespace.
Related
Currently, I have two tables named users and posts. Ever posts row includes a userId as reference and now I want to run query to search by userId array.
select * from posts where 'userId' = ANY(ARRAY['9', '77']);
And posts table looks like:
id userId text
1 9 text1
2 77 text2
The query returns empty while it returns full list if I remove where section. I guess the problem is that userId is a reference value but not sure how I can fix this problem.
Single quotes denote string literals. 'userId' is the userId column, but a string literal. Likewise, assuming you IDs are numbers, you should remove the quotes to get numeric literals instead of string literals:
select * from posts where userId = ANY(ARRAY[9, 77]);
-- No quotes -------------^----^-------------^--^^
Perhaps this works:
select * from posts where userId in (9,77);
I've got a results set of UK postcodes. Some are formatted with spaces, and some are not e.g. S14HG and S1 4HG
I want my select query to just return the outer part of the post code value in the results, i.e. 'S1'
I can do this in Excel using the following formula:
=IF(ISERROR(LEFT(A1,LEN(A1)-3)),””,LEFT(A1,LEN(A1)-3))
Is it possible to perform the same function in SQL through a SELECT query?
UK postcode can have one of many formats for their outward code.
However, as you can see from the possible formats in that link, there is a consistent format for the remainder of the postcode. If you are confident your postcodes are correct, you can simply remove any spaces and the last 3 characters:
declare #Postcodes table (Postcode nvarchar(10));
insert into #Postcodes values
('S1 4HG')
,('S14HG')
,('S10 4HG')
,('S104HG');
select Postcode
,replace(left(Postcode,len(Postcode)-3),' ','') as OutwardCode
from #Postcodes
Output:
Postcode OutwardCode
S1 4HG S1
S14HG S1
S10 4HG S10
S104HG S10
You can use LEFT() regardless of spaces since you only want the first two, which won't have a space.
SELECT LEFT('S1 4HG',2)
Or just get rid of the spaces...
declare #t varchar(64) = 'S 1 4 H G'
SELECT LEFT(REPLACE(#t,' ',''),2)
I have a table storing first names and surnames; some may be stored with capitalisation. Is there a query I could use to return only those rows with capitalisation?
For instance, if I have the following entries:
firstname | surname
-----------+-----------
Bob | Jones
john | bobbins
I'd only expect to be returned the record for "Bob Jones".
I'm sure it's not a difficult thing to do, but I haven't been able to find any examples anywhere.
Compare the value with the value where the first character is upper-case:
select *
from the_table
where firstname = initcap(firstname)
and surname = initcap(surname);
The function initcap() converts the first letter of each word to upper case and the rest to lower case.
I am trying to grab variable length string from a primary string.
Example:
ABC*12*1*name name****XX*123456789~
ABC*12*1*diffname diffname****XX*234567890~
ABC*12*1*diffname2 diffname2***XX*345678901~
I need to pull out the 'name name', 'diffname diffname', 'diffname2 diffname2'
etc from the string. And then replace the ' ' between the names with an asterisk - but, I cant just insert in the first space in the string, there could be multiple names, and so I would want to insert the '*' into the second, or third space, depending on the length of the name string.
SELECT
CHARINDEX('*1*',data)+3 AS startpos,
CHARINDEX('***',data) AS Endpos,
data
from #t
where data like '%ABC*12*1*%'
This gives me a start point and end point for the variable length string. So I try:
SELECT SUBSTRING(data,CHARINDEX('*1*',data)+3,CHARINDEX('***',data) -CHARINDEX('*1*',data)+3)
FROM #t
WHERE data like '%ABC*12*1*name%'
But this gives me
name n name aa*****X
as a result set, basically starting at the start point and then running well past the end point.
What am I doing wrong?
This part is the problem :
SELECT .....-CHARINDEX('*1*',data)+3
FROM .....
WHERE .....
You want to substract with Endpos so it supposed to be written in brackets like so :
-(CHARINDEX('*1*',data)+3)
and if the brackets are removed the last part should become -3 :
-CHARINDEX('*1*',data)-3
I have a table that contains a column that has data in the following format - lets call the column "title" and the table "s"
title
ab.123
ab.321
cde.456
cde.654
fghi.789
fghi.987
I am trying to get a unique list of the characters that come before the "." so that i end up with this:
ab
cde
fghi
I have tried selecting the initial column into a table then trying to do an update to create a new column that is the position of the dot using "ss".
something like this:
t: select title from s
update thedot: (title ss `.)[0] from t
i was then going to try and do a 3rd column that would be "N" number of characters from "title" where N is the value stored in "thedot" column.
All i get when i try the update is a "type" error.
Any ideas? I am very new to kdb so no doubt doing something simple in a very silly way.
the reason why you get the type error is because ss only works on string type, not symbol. Plus ss is not vector based function so you need to combine it with each '.
q)update thedot:string[title] ss' "." from t
title thedot
---------------
ab.123 2
ab.321 2
cde.456 3
cde.654 3
fghi.789 4
There are a few ways to solve your problem:
q)select distinct(`$"." vs' string title)[;0] from t
x
----
ab
cde
fghi
q)select distinct(` vs' title)[;0] from t
x
----
ab
cde
fghi
You can read here for more info: http://code.kx.com/q/ref/casting/#vs
An alternative is to make use of the 0: operator, to parse around the "." delimiter. This operator is especially useful if you have a fixed number of 'columns' like in a csv file. In this case where there is a fixed number of columns and we only want the first, a list of distinct characters before the "." can be returned with:
exec distinct raze("S ";".")0:string title from t
`ab`cde`fghi
OR:
distinct raze("S ";".")0:string t`title
`ab`cde`fghi
Where "S " defines the types of each column and "." is the record delimiter. For records with differing number of columns it would be better to use the vs operator.
A variation of WooiKent's answer using each-right (/:) :
q)exec distinct (` vs/:x)[;0] from t
`ab`cde`fghi