Hi all I am trying to write sql for selecting string between two special characters.
example: in the table, field value like 7185878969-129981041-000000 . how can I select only middle portion 129981041 without hard coding. What will be the best way to go about this?.Please provide sample code. Thanks
Impala has split_part():
select split_part(col, '-', 2)
Try this for MySQL:
SELECT REVERSE(SUBSTRING_INDEX(REVERSE(SUBSTRING_INDEX(Column,'-',2)),'-',1))
FROM table_name;
Result:
129981041
Related
The environment I am currently working in is Snowflake.
As a matter of data sensitivty, I will be using pseudonyms for my following question.
I have a specific field in one of my tables called FIELD_1. The data in this field is structured as such:
I am trying to figure out how to automatically extract from my FIELD_1 the output I have in FIELD_2.
Does anyone have any idea what kind of query I would need to achieve this? Any help would be GREATLYappreciated! I am really quite stuck on this problem.
Thank you!
You seem to want everything up to the first four numbers. Then to replace the underscores with spaces. If so:
select replace(regexp_substr(field_1, '^[^0-9]*[0-9]{4}'), '_', ' ')
Or alternatively, if you want the first three components separated by underscores:
select replace(regexp_substr(field_1, '^[^_]+_[^_]+_[0-9]{4}'), '_', ' ')
If the data is as simplistic in reality as you've described here, you can use a variable-length LEFT() function in conjunction with REPLACE() to get the desired output:
SELECT FIELD_1, REPLACE(LEFT(FIELD_1, LEN(FIELD_1)-10),'_',' ') AS FIELD_2
FROM table_name
See also:
SELECT - Snowflake Documentation
LEFT - Snowflake Documentation
REPLACE - Snowflake Documentation
LENGTH, LEN - Snowflake Documentation
How can I extract a text before or after specific character in SQL.
Friday|10:30-21:00
Friday|6:00-21:00
Monday|10:30-21:00
Saturday|10:00-2:00
I want to extract after | character.
Thanks,
Use string manipulation functions. Consider:
SELECT *, Mid(fieldname, InStr(fieldname, "|")+1) AS NewData FROM tablename
This us easy with charindex and stuff, depending on your database, syntax might vary.
select Stuff(column,1,CharIndex('|',column),'')
from Yourtable
I have column store_name (varchar). In that column I have entries like prime sport, best buy... with a space. But when user typed concatenated string like primesport without space I need to show result prime sport. how can I achieve this? Please help me
SELECT *
FROM TABLE
WHERE replace(store_name, ' ', '') LIKE '%'+#SEARCH+'%' OR STORE_NAME LIKE '%'+#SEARCH +'%'
Well, I don't have much idea, and even I am searching for it. But may be what I know works for you, You can achieve this by performing different type of string operations:
Mike can be Myke or Myce or Mikke or so on.
Cat an be Kat or katt or catt or so on.
For this you should write a function to generate number of possible strings and then form a SQL Query using all these, and query the database.
A similar kind of search in known as Soundex Search from Oracle and Soundex Search from Microsoft. Have a look of it. this may work.
And overall make use of functions like upper and lower.
Have you tried using replace()
You can replace the white space in the query then use like
SELECT * FROM table WHERE replace(store_name, ' ', '') LIKE '%primesport%'
It will work for entries like 'prime soft' querying with 'primesoft'
Or you can use regex.
I am using Oracle 11 G and have the following set of data:
12.0
4.2
Version.1
7.9
abc.72
I want to return all string characters before the period. What sort of query would I run in order to achieve this? Any help would be greatly appreciated, thanks!
You can try a combination of instr and substr.
Something like this:
select substr(field, 1, instr(field, '.') - 1)
from your_table;
Assuming field always contains a . character on it.
You can also deal with strings without a . by using case, if or any other similar valid conditional function on Oracle's SQL language implementation.
Of course, you can always put this on a function to make it look nicer on your query.
I require a select query that adds a space to the data based on the placement of the capital letters i.e. 'HelpMe' using this query would be displayed as 'Help Me' . Note i cannot use a stored function to do this the it must be done in the query itself. The Data is of variable length and query must be in SQL. Any Help will be appreciated.
Thanks
You need to use user defined function for this until MS give us support for regular expressions. Solution would be something like:
SELECT col1, dbo.RegExReplace(col1, '([A-Z])',' \1') FROM Table
Aldo this would produce leading space that you can remove with TRIM.
Replace regular expresion function:
http://connect.microsoft.com/SQLServer/feedback/details/378520
About dbo.RegexReplace you can read at:
TSQL Replace all non a-z/A-Z characters with an empty string
Assume if you are using Oracle RDBMS, you use the following,
REGEX_REPLACE
SELECT REGEXP_REPLACE('ILikeToWatchCSIMiami',
'([A-Z.])', ' \1')
AS RX_REPLACE
FROM dual
;
Managed to get this output: * SQLFIDDLE
But as you see it doesn't treat well on words such as CSI though.