I have a table with medication_product_amount column where there are spaces between numbers and characteres like below:
medication_product_amount
1 UN DE 50 ML
20 UN
1 UN DE 600 G
What I want is to remove the single space ONLY between numbers and characters, something like this:
new_medication_product_amount
1UN DE 50ML
20UN
1UN DE 600G
To do this, I am looking for a regular expression to use in the function REGEXP_REPLACE. I tried using the pattern below, indicating to replace the single space after the numbers, but the output remained the same as the input:
select REGEXP_REPLACE(medication_product_amount, '(^[0-9])( )', '\1') as new_medication_product_amount
from medications
Can anyone help me come up with the right way to do this? Thanks!
Your regex is a little off. First what yours does. '(^[0-9])( )', '\1')
(^[0-9]) Start Capture (field 1) at the beginning of the string for 1 digit
followed by Start Capture (field 2) for 1 space.
Replace the string by field1.
The problems and correction:
What you want to capture does not necessary the first character of the string. So eliminate the anchor ^.
What you want to capture may be more that 1 digit in length. So replace [0-9] by [0-9]+. I.E any number of digits.
Not actually a problem but a space holds no special meaning in a regexp, it is just a space so no need to capture it unless user later. Replace ( ) with just .
END of Pattern. But there may be other occurrences. Tell Postgres to continue with the above pattern until end of string. (see flag 'g').
Resulting Expression/Query: (demo here)
select regexp_replace(medication_product_posology, '([0-9]+) ', '\1','g') as new_medication_product_posology
from medications;
Match "digit space letter", capturing and the digit and letter using '([0-9]) ([A-Z])', then put them back using back references.
select REGEXP_REPLACE(medication_product_amount, '([0-9]) ([A-Z])', '\1\2') as new_medication_product_amount
from medications
Related
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 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 have a requirement where the persons can have more than one first name and i need to convert the name into first characters from the 2 or more names with capital letters.
Examples:
Srinivas Kalyan ,Sai Kishore
if the above is one having 4 first names then i need to display as below
S K S K
The comma is also be replaced and take all the first characters
2. Srinivas-Kalyan Sai Kishore
For the above name the value should be as
S S K
since Srinivas-Kalyan is considered as one name
Also the name can be in small letters
srinivas kalyan sai kishore
For this
S K S K
This has to be in oracle
Tried the below regex_replace which is working fine in sql developer but in the application it is changing into space
replace(trim(regexp_replace(to_char(regexp_replace(initcap(regexp_replace(regexp_replace
(FIRST_NAME, '[0-9]', ''), '( *[[:punct:]])', '')), '([[:lower:]]| )')), '(.)', '\1 ')),',',null)
The missing letter in your output is caused by this regex for character removal:
( *[[:punct:]])
This will turn Kalyan ,Sai into KalyanSai, which will be treated as one word by the rest of the process, and so you will not have the S of Sai in your output.
I would suggest this shorter expression:
trim(upper(regexp_replace(
regexp_replace(first_name, '([[:alpha:]])(-|[[:alpha:]])+', '\1'),
'.*?([[:alpha:]]|$)', ' \1'
)))
Explanation
([[:alpha:]])(-|[[:alpha:]])+ -> \1
This replaces any sequence of letters (or hyphen) with the first of those. So it reduces words to their initial letter.
.*?([[:alpha:]]|$) -> (space)\1
This replaces anything that precedes the next letter, with a space. As no letter will be skipped (because .*? is non-greedy) this effectively replaces all non-letter sequences with a space. To also replace the non-letters at the very end (which do not precede a letter), the special case $ (end-of-string) is added as alternative.
After these two steps there just remains to:
Upper case everything
Remove blanks at the start and end of the result with trim
I find the advantage of this method is that it does not use any other class than [[:alpha]]. Digits, punctuation, lowercase -- or whatever else -- does not need to be identified explicitly, as it is just the negation of [[:alpha]]. The only exception that has to be made, is for the hyphen.
As some names might include some other non-letters, like a quote, you might want to add such characters as well in the first regular expression.
I have a questions regarding below data.
You clearly can see each EMP_IDENTIFIER has connected with EMP_ID.
So I need to pull only identifier which is 10 characters that will insert another column.
How would I do that?
I did some traditional way, using INSTR, SUBSTR.
I just want to know is there any other way to do it but not using INSTR, SUBSTR.
EMP_ID(VARCHAR2)EMP_IDENTIFIER(VARCHAR2)
62049 62049-2162400111
6394 6394-1368000222
64473 64473-1814702333
61598 61598-0876000444
57452 57452-0336503555
5842 5842-0000070666
75778 75778-0955501777
76021 76021-0546004888
76274 76274-0000454999
73910 73910-0574500122
I am using Oracle 11g.
If you want the second part of the identifier and it is always 10 characters:
select t.*, substr(emp_identifier, -10) as secondpart
from t;
Here is one way:
REGEXP_SUBSTR (EMP_IDENTIFIER, '-(.{10})',1,1,null,1)
That will give the 1st 10 character string that follows a dash ("-") in your string. Thanks to mathguy for the improvement.
Beyond that, you'll have to provide more details on the exact logic for picking out the identifier you want.
Since apparently this is for learning purposes... let's say the assignment was more complicated. Let's say you had a longer input string, and it had several groups separated by -, and the groups could include letters and digits. You know there are at least two groups that are "digits only" and you need to grab the second such "purely numeric" group. Then something like this will work (and there will not be an instr/substr solution):
select regexp_substr(input_str, '(-|^)(\d+)(-|$)', 1, 2, null, 2) from ....
This searches the input string for one or more digits ( \d means any digit, + means one or more occurrences) between a - or the beginning of the string (^ means beginning of the string; (a|b) means match a OR b) and a - or the end of the string ($ means end of the string). It starts searching at the first character (the second argument of the function is 1); it looks for the second occurrence (the argument 2); it doesn't do any special matching such as ignore case (the argument "null" to the function), and when the match is found, return the fragment of the match pattern included in the second set of parentheses (the last argument, 2, to the regexp function). The second fragment is the \d+ - the sequence of digits, without the leading and/or trailing dash -.
This solution will work in your example too, it's just overkill. It will find the right "digits-only" group in something like AS23302-ATX-20032-33900293-CWV20-3499-RA; it will return the second numeric group, 33900293.
I have a select query and it fetches a field with complex data. I need to parse that data in specified format. please help with your expertise:
selected string = complexType|ChannelCode=PB - Phone In A Box|IncludeExcludeIndicator=I
expected output - PB|I
Please help me in writing a sql regular expression to accomplish this output.
The first step in figuring out the regular expression is to be able to describe it plain language. Based on what we know (and as others have said, more info is really needed) from your post, some assumptions have to be made.
I'd take a stab at it by describing it like this, which is based on the sample data you provided: I want the sets of one or more characters that follow the equal signs but not including the following space or end of the line. The output should be these sets of characters, separated by a pipe, in the order they are encountered in the string when reading from left to right. My assumptions are based on your test data: only 2 equal signs exist in the string and the last data element is not followed by a space but by the end of the line. A regular expression can be built using that info, but you also need to consider other facts which would change the regex.
Could there be more than 2 equal signs?
Could there be an empty data element after the equal sign?
Could the data set after the equal sign contain one or more spaces?
All these affect how the regex needs to be designed. All that said, and based on the data provided and the assumptions as stated, next I would build a regex that describes the string (really translating from the plain language to the regex language), grouping around the data sets we want to preserve, then replace the string with those data sets separated by a pipe.
SQL> with tbl(str) as (
2 select 'complexType|ChannelCode=PB - Phone In A Box|IncludeExcludeIndicator=I' from dual
3 )
4 select regexp_replace(str, '^.*=([^ ]+).*=([^ ]+)$', '\1|\2') result from tbl;
RESU
----
PB|I
The match regex explained:
^ Match the beginning of the line
. followed by any character
* followed by 0 or more 'any characters' (refers to the previous character class)
= followed by an equal sign
( start remembered group 1
[^ ]+ which is a set of one or more characters that are not a space
) end remembered group one
.*= followed by any number of any characters but ending in an equal sign
([^ ]+) followed by the second remembered group of non-space characters
$ followed by the end of the line
The replace string explained:
\1 The first remembered group
| a pipe character
\2 the second remember group
Keep in mind this answer is for your exact sample data as shown, and may not work in all cases. You need to analyse the data you will be working with. At any rate, these steps should get you started on breaking down the problem when faced with a challenging regex. The important thing is to consider all types of data and patterns (or NULLs) that could be present and allow for all cases in the regex so you return accurate data.
Edit: Check this out, it parses all the values right after the equal signs and allows for nulls:
SQL> with tbl(str) as (
2 select 'a=zz|complexType|ChannelCode=PB - Phone In A Box|IncludeExcludeIndicator=I - testing|test1=|test2=test2 - testing' from dual
3 )
4 select regexp_substr(str, '=([^ |]*)( |||$)', 1, level, null, 1) output, level
5 from tbl
6 connect by level <= regexp_count(str, '=')
7 ORDER BY level;
OUTPUT LEVEL
-------------------- ----------
zz 1
PB 2
I 3
4
test2 5
SQL>