I am trying to call back all rows who's email starts with a character which is not AlphaNumeric.
The line I am trying to use in the statement is
REGEXP_LIKE (SUBSTR(hcp_email.email_address,1,2), '![%a-zA-Z%]')
This does not bring back the relevant lines.
I am able to bring back results with the below but this is not as practical as using a catch all range of text and numbers to ignore.
REGEXP_LIKE (SUBSTR(hcp_email.email_address, 1,2), '[":., ]')
Ideally I would like to use a NOT LIKE statement with a range a-z 0-9.
No need to embed a SUBSTR call. Just anchor the regex to the start of the string, and look at the first character. This example uses the POSIX shorthand for readability.
where regexp_like(hcp_email.email_address, '^[^[:alnum:]]');
EDIT - Added regex explanation
^ - Anchor to the start of the string
[ - Start a character class
^ - Inside of a character class this means NOT
[:alnum:] - POSIX shorthand for an alphanumeric (A-Za-z0-9)
] - End character class (the character class describes 1 character)
Related
Hi I tried using Regex_replace and it is still not working.
select CASE WHEN sbbb <> ' ' THEN regexp_replace(sbbb,'[a-zA-Z _-#]','']
ELSE sbbb
AS ABCDF
from Table where sccc=1;
This is the query which I am using to remove alphabets and specials characters from string and have only numbers. but it doesnot work. Query returns me the complete string with numbers,characters and special characters .What is wrong in the above query
I am working on a sql query. There is a column in database which contains characters,special characters and numbers. I want to only keep the numbers and remove all the special characters and alphabets. How can I do it in query of DB2. If a use PATINDEX it is not working. please help here.
The allowed regular expression patterns are listed on this page
Regular expression control characters
Outside of a set, the following must be preceded with a backslash to be treated as a literal
* ? + [ ( ) { } ^ $ | \ . /
Inside a set, the follow must be preceded with a backslash to be treated as a literal
Characters that must be quoted to be treated as literals are [ ] \
Characters that might need to be quoted, depending on the context are - &
So for you, this should work
regexp_replace(sbbb,'[a-zA-Z _\-#]','')
I'm trying to figure out the base regex to capture the middle of a google url out of a sql database.
For example, a few links:
https://www.google.com/cars/?year=2016&model=dodge+durango&id=1234
https://www.google.com/cars/?year=2014&model=jeep+cherokee+crossover&id=6789
What would be the regex to capture the text to get dodge+durango , or jeep+cherokee+crossover ? (It's alright that the + still be in there.)
My Attempts:
1)
\b[=.]\W\b\w{5}\b[+.]?\w{7}
, but this clearly does not work as this is a hard coded scenario that would only work like something for the dodge durango example. (would extract "dodge+durango)
2) Using positive lookback ,
[^+]( ?=&id )
but I am not fully sure how to use this, as this only grabs one character behind the & symbol.
How can I extract a string of (potentially) any length with any amount of + delimeters between the "model=" and "&id" boundaries?
seems like you could use regexp_replace and access match groups:
regexp_replace(input, 'model=(.*?)([&\\s]|$)', E'\\1')
from here:
The regexp_replace function provides substitution of new text for
substrings that match POSIX regular expression patterns. It has the
syntax regexp_replace(source, pattern, replacement [, flags ]). The
source string is returned unchanged if there is no match to the
pattern. If there is a match, the source string is returned with the
replacement string substituted for the matching substring. The
replacement string can contain \n, where n is 1 through 9, to indicate
that the source substring matching the n'th parenthesized
subexpression of the pattern should be inserted, and it can contain \&
to indicate that the substring matching the entire pattern should be
inserted. Write \ if you need to put a literal backslash in the
replacement text. The flags parameter is an optional text string
containing zero or more single-letter flags that change the function's
behavior. Flag i specifies case-insensitive matching, while flag g
specifies replacement of each matching substring rather than only the
first one
I may be misunderstanding, but if you want to get the model, just select everything between model= and the ampersand (&).
regexp_matches(input, 'model=([^&]*)')
model=: Match literally
([^&]*): Capture
[^&]*: Anything that isn't an ampersand
*: Unlimited times
I was testing a regular expression in Oracle SQL and found something I could not understand:
-- NO MATCH
SELECT 1 FROM DUAL WHERE REGEXP_LIKE ('Professor Frank', '(^|\s)Prof[^\s]*(\s|$)');
Above doesn't match, while the following matches:
-- MATCH
SELECT 1 FROM DUAL WHERE REGEXP_LIKE ('Professor Frank', '(^|\s)Prof\S*(\s|$)');
In other regex flavors, It will be like \bProf[^\s]*\b versus \bProf\S*\b and have similar results. Note: Oracle SQL regex does not have \b or word boundary.
Question: Why don't [^\s]* and \S* work the same way in Oracle SQL?
I notice if I remove the (\s|$) at the end, the first regex will match.
In Oracle regular expressions, \s is indeed the escape sequence for a space, but NOT in a matching character set (that is, [.....], or [^....] for excluding one character). In a matching character set, only two characters have a special meaning, - for ranges and ] for closing the set enumeration. They can't be escaped; if needed in the matching set, ] must always be the first character right after the opening [ (it is the ONLY position in which a closing ] stands for itself as a character, and does not denote the end of the matching set), and - must be first or last (best to leave it always to the end of the matching set) - anywhere else it is seen as a range marker. To include (or exclude, if using the [^.....] syntax) a space, just type an actual physical space in the matching set.
Edit: What I said above is not entirely right. There is another special character in a matching set, namely ^. If it is used in the first position, it means "match any character OTHER THAN." In any other position it stands for itself. For example, '[^^]' will match any single character OTHER THAN ^ (the first ^ has special meaning, the second stands in for itself). And, a closing bracket ] stands for itself if it is the second character in brackets, if the first character is ^ (with its SPECIAL meaning). That is, to match any single character OTHER THAN ], we can use the matching pattern '[^]]'.
I am using a replace function to add some quotes around a couple of keywords.
However, this replacement doesn't work for a few cases like the one below.
See example below.
This is the query:
replace(replace(aa.SourceQuery,'sequence','"sequence"'),'timestamp','"timestamp"')
Before:
select timestamp, SparkTimeStamp
from SparkRecordCounts
After:
select "timestamp", Spark"timestamp"
from SparkRecordCounts
However, I want it to be like:
select "timestamp", Sparktimestamp
from SparkRecordCounts
EDIT I wrote this before knowing what RDBMS you were using but have left it in case it helps someone else.
I think you are looking for word boundaries in your replacement, which are generally a job for regular expressions.
Oracle has one built in, called regexp_replace, and you could use something like this:
regexp_replace(aa.SourceQuery, '(^|\s|\W)timestamp($|\s|\W)', '\1"timestamp"\2')
The regular expression looks at the start for:
^ - the start of the line OR
\s - a space character OR
\W - a non-word character
It then matches timestamp, and must end with:
$ - the end of the line OR
\s - a space character OR
\W - a non-word character
Then, and only then, does it perform the replace. \1 and \2 are used to preserve what word boundary matched at the beginning and ending of the word.
I'm not sure how other databases handle regexp_replace, it looks like mysql can via a plugin like this but there may not be a native method.
SQL Server has a solution to something similar here
From within an Oracle 11g database, using SQL, I need to remove the following sequence of special characters from a string, i.e.
~!##$%^&*()_+=\{}[]:”;’<,>./?
If any of these characters exist within a string, except for these two characters, which I DO NOT want removed, i.e.: "|" and "-" then I would like them completely removed.
For example:
From: 'ABC(D E+FGH?/IJK LMN~OP' To: 'ABCD EFGHIJK LMNOP' after removal of special characters.
I have tried this small test which works for this sample, i.e:
select regexp_replace('abc+de)fg','\+|\)') from dual
but is there a better means of using my sequence of special characters above without doing this string pattern of '\+|\)' for every special character using Oracle SQL?
You can replace anything other than letters and space with empty string
[^a-zA-Z ]
here is online demo
As per below comments
I still need to keep the following two special characters within my string, i.e. "|" and "-".
Just exclude more
[^a-zA-Z|-]
Note: hyphen - should be in the starting or ending or escaped like \- because it has special meaning in the Character class to define a range.
For more info read about Character Classes or Character Sets
Consider using this regex replacement instead:
REGEXP_REPLACE('abc+de)fg', '[~!##$%^&*()_+=\\{}[\]:”;’<,>.\/?]', '')
The replacement will match any character from your list.
Here is a regex demo!
The regex to match your sequence of special characters is:
[]~!##$%^&*()_+=\{}[:”;’<,>./?]+
I feel you still missed to escape all regex-special characters.
To achieve that, go iteratively:
build a test-tring and start to build up your regex-string character by character to see if it removes what you expect to be removed.
If the latest character does not work you have to escape it.
That should do the trick.
SELECT TRANSLATE('~!##$%sdv^&*()_+=\dsv{}[]:”;’<,>dsvsdd./?', '~!##$%^&*()_+=\{}[]:”;’<,>./?',' ')
FROM dual;
result:
TRANSLATE
-------------
sdvdsvdsvsdd
SQL> select translate('abc+de#fg-hq!m', 'a+-#!', etc.) from dual;
TRANSLATE(
----------
abcdefghqm