I'm attempting to isolate eight digits from a cell that contains other numbers as well as text and no rhyme or reason to where it is placed. An example return would look something like this:
will deliver 11/07 in USA at 12:30 with conf# 12345678
I need the conf# only, but it could be at the end, beginning, middle of the string and I don't know how to isolate it. I'm working in DB2 so I can't use functions such as PATINDEX or CHARINDEX, so what are my other option for pulling out only "12345678" regardless of where it is located?
While DB2 doesn't have PATINDEX or CHARINDEX, it does have LOCATE.
If your DB2 version supportx pureXML, you can use the regular expression support in XQuery, something like:
select xmlcast(
xmlquery(
' if (fn:matches( $YOURCOLUMN, "(^|.*[^\d])(\d{8})([^\d].*$|$)")) then fn:replace( $YOURCOLUMN,"(^|.*[^\d])(\d{8})([^\d].*$|$)","$2") else "" '
)
as varchar(20)
)
from YOURTABLE
This assumes that 8-digit sequence appears only once in the column. You may need to tweak the regex to support some border cases.
Related
I am trying to split a field by delimiter in LookML. This field either follows the format of:
Managers (AE)
Managers (AE - MM)
I was able to split to first case using this
sql: case
when rlike (${user_role_name}, '^.*[\\(\\)].*$') then split_part(${user_role_name}, ' ', -1)
However, I haven't been able to get the 2nd case to do the same. It's in a case statement so I am going to add another when statement, but am not able to figure out the regex for parentheses that contains spaces.
Thanks in advance for the help!
By "split" the string, I think you mean you want to extract the part in parentheses, right?
I would do this using a regex substring method. You didn't mention what warehouse you're using, and the syntax will vary a little, but on snowflake that would look like:
regexp_substr(${user_role_name}, '\\([^)]*\\)')
So, for example, with the inputs you gave:
select regexp_substr('Managers (AE)', '\\([^)]*\\)')
union all
select regexp_substr('Managers (AE - MM)', '\\([^)]*\\)')
result
(AE)
(AE - MM)
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($1,'\\w+#(.+?)~',1,1,'is',1)
FROM VALUES
('CK#123456~fndkjfgdjkg')
,('QH#128fklj924~fndkjfgdjkg')
;
This looks for:
One or more word characters
Followed by #
Capturing one or more characters upto and not including ~
Returns the characters within the capture group
You can change the .+? to \\d+? to make sure the pattern is only digits. Backslashes must be escaped with a backslash.
The descriptions for each argument of the function can be found here:
https://docs.snowflake.net/manuals/sql-reference/functions/regexp_substr.html
You could check this!!
select substr('CK#123456~fndkjfgdjkg',4,6) from dual;
OUTPUT
123456
https://docs.snowflake.net/manuals/sql-reference/functions/substr.html
I am working with an Oracle database and would like to write a REGEXP_LIKE expression that finds any number where all digits are the same, such as '999999999' or '777777777' without specifying the length of the field. Also, I would like it to be able to identify characters as well, such as 'aaaaa'.
I was able to get it working when specifying the field length, by using this:
select * from table1
where regexp_like (field1, '^([0-9a-z])\1\1\1\1\1\1\1\1');
But I would like it to be able to do this for any field length.
If a field contains '7777771', for example, I would not want to see it in the results.
Try this:
^([0-9a-z])\1+$
Live demo
You're almost there. You just need to anchor the end of the regex.
^([0-9a-z])\1+$
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.
Does anyone have a LIKE pattern that matches whole words only?
It needs to account for spaces, punctuation, and start/end of string as word boundaries.
I am not using SQL Full Text Search as that is not available. I don't think it would be necessary for a simple keyword search when LIKE should be able to do the trick. However if anyone has tested performance of Full Text Search against LIKE patterns, I would be interested to hear.
Edit:
I got it to this stage, but it does not match start/end of string as a word boundary.
where DealTitle like '%[^a-zA-Z]pit[^a-zA-Z]%'
I want this to match "pit" but not "spit" in a sentence or as a single word.
E.g. DealTitle might contain "a pit of despair" or "pit your wits" or "a pit" or "a pit." or "pit!" or just "pit".
Full text indexes is the answer.
The poor cousin alternative is
'.' + column + '.' LIKE '%[^a-z]pit[^a-z]%'
FYI unless you are using _CS collation, there is no need for a-zA-Z
you can just use below condition for whitespace delimiters:
(' '+YOUR_FIELD_NAME+' ') like '% doc %'
it works faster and better than other solutions. so in your case it works fine with "a pit of despair" or "pit your wits" or "a pit" or "a pit." or just "pit", but not works for "pit!".
I think the recommended patterns exclude words with do not have any character at the beginning or at the end. I would use the following additional criteria.
where DealTitle like '%[^a-z]pit[^a-z]%' OR
DealTitle like 'pit[^a-z]%' OR
DealTitle like '%[^a-z]pit'
I hope it helps you guys!
Surround your string with spaces and create a test column like this:
SELECT t.DealTitle
FROM yourtable t
CROSS APPLY (SELECT testDeal = ' ' + ISNULL(t.DealTitle,'') + ' ') fx1
WHERE fx1.testDeal LIKE '%[^a-z]pit[^a-z]%'
If you can use regexp operator in your SQL query..
For finding any combination of spaces, punctuation and start/end of string as word boundaries:
where DealTitle regexp '(^|[[:punct:]]|[[:space:]])pit([[:space:]]|[[:punct:]]|$)'
Another simple alternative:
WHERE DealTitle like '%[^a-z]pit[^a-z]%' OR
DealTitle like '[^a-z]pit[^a-z]%' OR
DealTitle like '%[^a-z]pit[^a-z]'
This is a good topic and I want to complement this to someone how needs to find some word in some string passing this as element of a query.
SELECT
ST.WORD, ND.TEXT_STRING
FROM
[ST_TABLE] ST
LEFT JOIN
[ND_TABLE] ND ON ND.TEXT_STRING LIKE '%[^a-z]' + ST.WORD + '[^a-z]%'
WHERE
ST.WORD = 'STACK_OVERFLOW' -- OPTIONAL
With this you can list all the incidences of the ST.WORD in the ND.TEXT_STRING and you can use the WHERE clausule to filter this using some word.
You could search for the entire string in SQL:
select * from YourTable where col1 like '%TheWord%'
Then you could filter the returned rows client site, adding the extra condition that it must be a whole word. For example, if it matches the regex:
\bTheWord\b
Another option is to use a CLR function, available in SQL Server 2005 and higher. That would allow you to search for the regex server-side. This MSDN artcile has the details of how to set up a dbo.RegexMatch function.
Try using charindex to find the match:
Select *
from table
where charindex( 'Whole word to be searched', columnname) > 0