REGEXP_REPLACE("My dog is funny and happy", r"(\S+ \S+ \S+)", r"*") This is my SQL for achieving this. My output should look something like this = My dog is funny *and happy
When I try the above query it removes the first few words. How do I work this out?
You should use a backreference:
REGEXP_REPLACE("My dog is funny and happy", r"^((?:\S+\s+){4})", r"\1*")
REGEXP_REPLACE("My dog is funny and happy", r"^(?:\S+\s+){4}", r"\0*")
See the regex demo. Details:
^ - start of string
((?:\S+\s+){4}) - Group 1 (\1 in the replacement will refer to this group value): four occurrences of one or more non-whitespaces followed with one or more whitespaces.
\0 refers to the whole match value.
See the regexp_replace reference:
REGEXP_REPLACE(value, regexp, replacement)
Returns a STRING where all substrings of value that match regular
expression regexp are replaced with replacement.
You can use backslashed-escaped digits (\1 to \9) within the
replacement argument to insert text matching the corresponding
parenthesized group in the regexp pattern. Use \0 to refer to the
entire matching text.
Related
I have written this expression for checking extra characters and I am counting the occurrence of those extra characters.
REGEXP_COUNT('Mr.John® Êlite', regexp_extract ('Mr.John® Êlite','[^\x00-\x7F]'))
It's working fine if the string has only one extra character e.g
Mr. John®
It will take out ® and give me count as 1.
But if my string has two extra characters, it will only pick the first one and ignore the second character e.g
Mr.John® Êlite
My function will extract ® and ignore Ê.
I have tried subquery as well.Not working.Need help
As noted by Wiktor Stribiżew REGEXP_COUNT needs just a source string and regexp:
db2 "values REGEXP_COUNT('Mr.John® Êlite', '[^\x00-\x7F]')"
1
-----------
2
Because you used REGEXP_EXTRACT, it does extract the first occurrence only:
The REGEXP_EXTRACT scalar function returns one occurrence of a substring of a string that matches the regular expression pattern.
and only then you do actual count.
I need to find rows where the phone number field contains unexpected characters.
Most of the values in this field look like:
123456-7890
This is expected. However, we are also seeing character values in this field such as * and #.
I want to find all rows where these unexpected character values exist.
Expected:
Numbers are expected
Hyphen with numbers is expected (hyphen alone is not)
NULL is expected
Empty is expected
Tried this:
WHERE phone_num is not like ' %[0-9,-,' ' ]%
Still getting rows where phone has numbers.
from https://regexr.com/3c53v address you can edit regex to match your needs.
I am going to use example regex for this purpose
select * from Table1
Where NOT REGEXP_LIKE(PhoneNumberColumn, '^[+]*[(]{0,1}[0-9]{1,4}[)]{0,1}[-\s\./0-9]*$')
You can use translate()
...
WHERE translate(Phone_Number,'a1234567890-', 'a') is NOT NULL
This will strip out all valid characters leaving behind the invalid ones. If all the characters are valid, the result would be NULL. This does not validate the format, for that you'd need to use REGEXP_LIKE or something similar.
You can use regexp_like().
...
WHERE regexp_like(phone_num, '[^ 0123456789-]|^-|-$')
[^ 0123456789-] matches any character that is not a space nor a digit nor a hyphen. ^- matches a hyphen at the beginning and -$ on the end of the string. The pipes are "ors" i.e. a|b matches if pattern a matches of if pattern b matches.
Oracle has REGEXP_LIKE for regex compares:
WHERE REGEXP_LIKE(phone_num,'[^0-9''\-]')
If you're unfamiliar with regular expressions, there are plenty of good sites to help you build them. I like this one
I am busy building a lookup table for specific names of merchants. I tried to make use of the following regex but it's returning less results than the standard "like" function in Netezza SQL. Please refer to below:
SQL Like function: where trim(upper(a.MRCH_NME)) like '%CNA %' -- returns 4622 matches
Regex function in Netezza SQL: where array_combine(regexp_extract_all(trim(upper(a.MRCH_NME)),'.*CNA\s','i'),'|') = 'CNA' -- returns 2226 matches
I looked at the two result sets and found that strings such as the following aren't matched:
!C CNA INT ARR
*CNA PLATZ 0400
015764 CNA CRAD
C#CNA PARK 0
I made use of the following regex expression: /.*CNA\s'/
Any idea why the above strings aren't being returned as matches?
Thank you.
You probably should be using regexp_like:
SELECT *
FROM yourTable
WHERE REGEXP_LIKE(MRCH_NME, 'CNA[ ]', 'i');
This would be logically identical to the following query using LIKE:
SELECT *
FROM yourTable
WHERE MRCH_NME LIKE '%CNA ';
It seems to me the problem is more with your code rather than the regex. Look: like '%CNA %' returns all entries that contain a CNA substring followed with a literal space anywhere inside the entry. The '.*CNA\s' regex matches any 0+ chars other than newline followed with CNA and **any whitespace char*.
Acc. to this reference, \s matches "a white space character. White space is defined as [\t\n\f\r\p{Z}].
Thus, you should in fact just use
WHERE REGEXP_LIKE(MRCH_NME, 'CNA ', 'i')
or, better with a word boundary check:
WHERE REGEXP_LIKE(MRCH_NME, '\bCNA\b', 'i')
where \b marks a transition from a word to non-word and non-word to word character, thus ensuring a whole word search and justifying the regex usage.
If you do not need to match the merchant name as a whole word, use the regular LIKE with '%CNA %', it should be more efficient.
I need to extract some text from a string, but only where the text matches a string pattern. The string pattern will consist of...
2 numbers, a forward slash and 6 numbers
e.g. 12/123456
or
2 numbers, a forward slash, 6 numbers, a hyphen and 2 numbers
e.g. 12/123456-12
I know how to use INSTR to find a specific string. Is it possible to find a string that matches a specific pattern?
You'll need to use regexp_like to filter the results and regexp_substr to get the substring.
Here is roughly what it should look like:
select id, myValue, regexp_substr(myValue, '[0-9]{2}/[0-9]{6}') as myRegExMatch
from Foo
where regexp_like(myValue,'^([a-zA-Z0-9 ])*[0-9]{2}/[0-9]{6}([a-zA-Z0-9 ])*$')
with a link to a SQLFiddle that you can see in action and adjust to your taste.
The regexp_like provided in the sample above takes into consideration the alphanumerics and whitespace characters that may bound the number pattern.
Use regexp_like.
where regexp_like(col_name,'\s[0-9]{2}\/[0-9]{6}(-[0-9]{2})?\s')
\s matches a space. Include them at the start and end of pattern.
[0-9]{2}\/[0-9]{6} matches 2 numerics, a forward slash and 6 numerics
(-[0-9]{2})? is optional for a hyphen and 2 numerics following the previous pattern.
regexp_like(col_name,'^\d{2}/\d{6}($|-\d{2}$)')
or
regexp_like(col_name,'^\d{2}/\d{6}(-\d{2})?$')
I try to extract the '930' from 'EM 930' with following Regexp
REGEXP_REPLACE(info,'^[:space:]*[ABCDEFGHIJKLMNOPQRSTUVWXYZ]*[:space:]*([0-9]+)[:space:]*$','\1')
But it returns me the original string.
An idea why ?
Subsidiary question:
Why does the "\1" returned the original string when the pattern is not matched ? I expected it to return NULL, as in my other regexp experiences (eg Perl).
Who I can re-write this in a performant way so that I get of wel the matched string of well NULL ?
Your space character class was not exactly correct. If we change [:space:] to [[:space:]], your regexp_replace works as you expect:
REGEXP_REPLACE(info, '^[[:space:]]*[ABCDEFGHIJKLMNOPQRSTUVWXYZ]*[[:space:]]*([0-9]+)[[:space:]]*$','\1')
For the sake of succinctness, we could use the upper character class, [[:upper:]], for [ABCDEFGHIJKLMNOPQRSTUVWXYZ]. This changes the function invocation to:
regexp_replace(info, '^[[:space:]]*[[:upper:]]*[[:space:]]*([0-9]+)[[:space:]]*$','\1')
Or escape characters could be used in lieu of character classes:
\s space
\w word character
\d digit character
regexp_replace(info, '^\s*\w*\s*(\d+)\s*$','\1')
Explanation:
Since your malformed character class, [:space:], does not match the space that exists between 'EM' and '930', your search by parameter does not match any characters in the source parameter.
Your search by parameter, '^[[:space:]]*[ABCDEFGHIJKLMNOPQRSTUVWXYZ]*[[:space:]]*([0-9]+)[[:space:]]*$', is anchored to the beginning and end of the column, info, thus it can only match the column, info, one time at most.
In your case, there is no match and the character group, '\1' which is associated with '([0-9]*)', has no value.
Consequently, no characters are replaced and you are left with original value of the column, info, 'EM 930'.
interesting variations to better understand this function:
-If your corrected function invocation had no pattern_to_replace_by parameter, '\1', then a NULL would be returned:
regexp_replace(info, '^\s*\w*\s*(\d+)\s*$' ) FROM dual;
-Since you have a pattern_to_replace_by parameter, '\1', and now it has the matching character group, the repeating digit group is returned:
930