I have text in a column like RAPP 01. RAPP 02 upto RAPP 45 and RAPP 99.I included all these values manually in my IN statement in my WHERE clause but it slows the query as the data set in huge. I tried WHERE SUBSTR(REMARK_TXT,1,7) LIKE 'RIPA [01-45,99]' and it did not return any data. Can you please help?
Thanks!
You could use REGEXP functionality here:
WHERE REGEXP_SIMILAR(REMARK_TXT, '^RAPP [0-9]{2}$') = 1;
That regex matches with a string that starts with RAPP followed by a space then followed by 2 numbers and the end of the string.
Updating to deal with two number ranges (01-49) and (99). This isn't the best thing to do with regex, but it's still possible:
WHERE REGEXP_SIMILAR(REMARK_TXT, '^RAPP ([0-4][0-9]|99)$') = 1;
This is saying a string that starts with RAPP and then ends in either a two digit number that starts with 0 through 4 OR the number 99
You could use the following:
where column like ('RAPP %')
Which would return anything beginning with the string RAPP and a whitespace. Notice the '%' sign, this will be your wildcard.
Careful on using like, especially not like and putting the wildcard at the beginning of your condition, it would have much bigger performance issues.
Related
how do I write a SQL where statement that checks if a string contains some substring and a number. For example:
string: macsea01
where string like 'macsea' plus a number
Regex is the most obvious solution to this question. Without more detail about the specific format of the string, I can suggest the following, which will match a sequence of a letter in the alphabet followed immediately by a digit:
where column_name like '%[a-zA-Z][0-9]%'
If you're literally looking for macsea at the beginning of the string followed by a digit, it would be:
where column_name like 'macsea[0-9]%'
Regex seem to bee a little slippery here, depending on your needs you can for instance divide the string into several parts, first the text part, and take the rest of the string, try to convert it into a number.
Somthing like this (but I think this perticular code is broken
where substring(column_name, 1, 6) = 'macsea' and cast(substring(column_name, 7, 1000) as int) > 0
I am using Snowflake SQL. I would like to remove characters from a string after a special character ~. How can I do that?
here is the whole scenario. Let me explain. I do have a string like 'CK#123456~fndkjfgdjkg'. Now, i want only the number after #.And not anything after ~. This is number length varies for that field value. It might be 1 or 5 or 3. And i want to add the condition in where class where this number is equal to check_num from other table after joining. I am trying REGEXP_SUBSTR(A.SRC_TXT, '(?<=CK#)(.+?\b)') = C.CHK_NUM in the where condition. I am getting the error as 'No repititive argument after ?'
You can use a regex for this
-- To remove just the character after a ~
select regexp_replace('fo~o bar','~.', '');
-- returns 'fo bar'
--If you want to keep the ~
select regexp_replace('fo~o bar','~.', '~');
-- returns 'fo~ bar'
--If you want to remove everything after the ~
select regexp_replace('fo~o bar','~.*', '');
--returns 'fo'
If you need to remove other specific character sets after a ~, you can probably do this with a slightly more complicated regex, but I'd need examples of your desired input/output to help with that.
EDIT for updated question
This regex replace should get what you need.
select regexp_replace('CK#123456~fndkjfgdjkg','CK#(\\d*)~.*', '\\1');
-- returns 123456
(\\d*) gets ANY number of digits in a row, and the \\1 causes it to replace the match with what was in the first set of parenthesis, which is your list of digits. the CK# and ~.* are there to make sure the whole string gets matched and replaced.
If the CK# can vary as well, you can use .*? like this.
select regexp_replace('ABCD123HI#123456~fndkjfgdjkg','.*?#(\\d*)~.*', '\\1')
-- returns 123456
I'd probably do something like the following, easy enough but not as cool as RegEx type of functions.
set my_string='fooo~12345';
set search_for_me = '~';
SELECT SUBSTR($my_string, 1, DECODE(position($search_for_me, $my_string), 0, length($my_string), position($search_for_me, $my_string)));
I hope this helps...Rich
It looks like lookahead and lookbehinds are not supported in REGEXP functions, they seem to work in the PATTERN clause of a LIST command. Snowflake documentation makes no mention either way of lookahead or lookbehinds.
In your example:
It seems that the query engine is looking for that repeating argument, where you are attempting a lookbehind
You have not specified what you wanted extracted. You have two capture groups, but in this scenario everything would be returned
Since you are looking to remove everything after ~ you have a delimiter, why not use it in your REGEXP_SUBSTR function?
Try the following:
SELECT $1,REGEXP_SUBSTR($1,'\\w+#(.+?)~',1,1,'is',1)
FROM VALUES
('CK#123456~fndkjfgdjkg')
,('QH#128fklj924~fndkjfgdjkg')
;
This looks for:
One or more word characters
Followed by #
Capturing one or more characters upto and not including ~
Returns the characters within the capture group
You can change the .+? to \\d+? to make sure the pattern is only digits. Backslashes must be escaped with a backslash.
The descriptions for each argument of the function can be found here:
https://docs.snowflake.net/manuals/sql-reference/functions/regexp_substr.html
You could check this!!
select substr('CK#123456~fndkjfgdjkg',4,6) from dual;
OUTPUT
123456
https://docs.snowflake.net/manuals/sql-reference/functions/substr.html
I have a string like the following in a sqlite 1 column:
1a 2B 3c 354 AfS 151 31s2fef 1fs31 3F1e2s 84f64e 45fs
space separated, x amount of characters 0-9, a-z, A-Z, there might be punctuation I'm not sure, but it is definitely space separated.
I'm trying to make a regular expression so I can query the database by number of words. basically if I wanted to get the 6th "word" in the example I'd be looking for:
151
so I tried to make a regular expression that says if the Nth word = 151, return me that row.
Here's what I've got so far.
SELECT * FROM table1 WHERE column1 REGEXP ^((?:\S+\s+){1}){6}
That unfortunately gives me the first through sixth words, but I really want to pinpoint the 6th word like the example above.
Also, I was thinking so save room in the database I could get rid of the white space, I'd just need to know how to count a specific number of characters in to the string which I couldn't figure out either.
Thanks for the help, never written a regular expression before.
If all you need is a match, there is no need for the non-capturing group. Just match a non-space + space group 5 times and follow with a 151.
^(\S+\s+){5}151
Use following regex
^([^\s]+\s){6}(.*?)(\s|$)
this must return 151, you can change {6} with your number to match the string.
CASE
WHEN <in_data> LIKE '[0-9][0-9][0-9][0-9][0-9][0-9]' THEN SUBSTR(<in_data>,1,3)
ELSE '000'
END
We're doing a migration project from Sybase to Teradata, and having a problem figuring this one out :) I'm still new to Teradata.
I would like to ask the equivalent TD code for this -
LIKE '[0-9][0-9][0-9][0-9][0-9][0-9]' to Teradata
Basically, it just checks whether the digits are numeric value.
Can someone give me a hint on this
You can also use REGEXP_SUBSTR to directly extract the three digits:
COALESCE(REGEXP_SUBSTR(in_data,'^[0-9]{3}(?=[0-9]{3}$)'), '000')
This looks for the first three digits and then does a lookahead for three following digits without adding them to the overall match.
^ indicates the begin of the string, '$' the end, so there are no other characters before or after the six digits. (?=...) is a so-called "lookahead", i.e. those three digits are checked, but ignored.
If there's no match the regex returns NULL which is changed to '000'.
You need to use regexp instead of like, since [0-9][0-9][0-9][0-9][0-9][0-9] is a regular expression.
To do an exact match, you need to add anchors. ie, to match the string which contains an exact 6 digit chars.
regexp '^[0-9]{6}$'
or
regexp '^[[:digit:]]{6}$'
I have a nvarchar field in my table, which contains all sorts of strings.
In case there are strings which contain a number following a non-number sign, I want to insert a space before that number.
That is - if a certain entry in that field is abc123, it should be turned into abc 123, or ab12.34 should become ab 12. 34.I want this to be done throughout the entire table.
What's the best way to achieve it?
You can try something like that:
select left(col,PATINDEX('%[0-9]%',col)-1 )+space(1)+
case
when PATINDEX('%[.]%',col)<>0
then substring(col,PATINDEX('%[0-9]%',col),len(col)+1-PATINDEX('%[.]%',col))
+space(1)+
substring(col,PATINDEX('%[.]%',col)+1,len(col)+1-PATINDEX('%[.]%',col))
else substring(col,PATINDEX('%[0-9]%',col),len(col)+1-PATINDEX('%[0-9]%',col))
end
from tab
It's not simply, but I hope it will help you.
SQL Fiddle
I used functions (link to MSDN):
LEFT, PATINDEX, SPACE, SUBSTRING, LEN
and regular expression.