Say I have a column in postgres database called pid that looks like this:
set1/2019-10-17/ASR/20190416832-ASR.pdf
set1/2019-03-15/DEED/20190087121-DEED.pdf
set1/2021-06-22/DT/20210376486-DT.pdf
I want to remove everything after the last dash "-" including the dash itself. So expected results:
set1/2019-10-17/ASR/20190416832.pdf
set1/2019-03-15/DEED/20190087121.pdf
set1/2021-06-22/DT/20210376486.pdf
I've looked into replace() and split_part() functions but still can't figure out how
to do this. Please advise.
We can use a regex replacement here:
SELECT col, REGEXP_REPLACE(col, '^(.*)-[^-]+(\.\w+)$', '\1\2') AS col_out
FROM yourTable;
The regex used above captures the column value before the last dash in \1, and the extension in \2. It then builds the output using \1\2 as the replacement.
Here is a working regex demo.
Related
I am fairly new to teradata, but I was trying to understand how to use REGEXP_SUBSTR
For example I have the following cell value = ABCD^1234567890^1
How can I extract 1234567890
What I attempted to do is the following:
REGEXP_SUBSTR(x, '(?<=^).*?(?=^)')
But this didnt seem to work.
Can anyone help?
It might (or might not) be possible to use REGEXP_SUBSTR() to handle this, but you would need to use a capture group. An alternative here would be to do a regex replacement instead:
SELECT x, REGEXP_REPLACE(x, '^.*?\^|\^.*$', '') AS output
FROM yourTable;
The regex pattern used here matches:
^.*?\^ everything from the start to the first ^
| OR
\^.*$ everything from the second ^ to the end
We then replace with empty string to remove the content being matched.
I am having a deep dilemma in hive. My data set in Hive looks like this:
##214628##564#7576#7876
#12771#242###256823
###3264###7236473####3
In each instance, I want to print only the first string after the #. So the output should be something like this:
214628
12771
3264
I tried using the reg_extract function, but alas I am getting only NULL values. Since hive doesn't support reg_substr, the following synatax doesn't work:
to_number(trim(regexp_substr(col_name,'[^#]+',1,1)))
Any suggestions are wecome!
You can use regexp_replace and then substr combination.
First remove all multiple occurrences of # from the string using regexp_replace().
regexp_replace(col,'#+','#') -- for data '#####123##' this will produce '#123#'
Then remove first # using substr. And then use instr to fetch everything starting from first till #.
substr(substr(str,2),1, instr(substr(str,2),'#')-1) this will produce '123'
You can see whole sql below.
select substr(substr(str,2),1, instr(substr(str,2),'#')-1) as result
from (
SELECT regexp_replace('#####123##','#+','#') as str) a
I assumed you always have # in the beginning. if you just add if left(str,1)='#'... and handle according to the data.
I have a string containing \s\ keyword. Now, I want to replace it with NULL.
select string,REGEXP_REPLACE(string,'\\\s\\','') from test
But unable to replace with the above statement in spark sql
input: \s\help
output: help
want to use regexp_replace
To replace one \ in the actual string you need to use \\\\ (4 backslashes) in the pattern of the regexep_replace. Please do look at https://stackoverflow.com/a/4025508/9042433 to understand why 4 backslashes are needed to replace just one backslash
So, the required statement would become like below
select name, regexp_replace(name, '\\\\s\\\\', '') from test
Below screenshot has examples for better understanding
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./?', '~!##$%^&*()_+=\{}[]:”;’<,>./?',' ')
FROM dual;
result:
TRANSLATE
-------------
sdvdsvdsvsdd
SQL> select translate('abc+de#fg-hq!m', 'a+-#!', etc.) from dual;
TRANSLATE(
----------
abcdefghqm
I am new to Regular Expressions and any help is highly appreciated.
Pattern like W00000,W00001,W00002,W00004
Must begin with W
Each string before comma must be six characters
String can only be repeated four times
Comma in between
Must not begin or end with comma
I tried below pattern and some others, like (^[W]{1}\d{5}){1,4}'), and none of them work correctly:
Select 'X' from dual Where REGEXP_LIKE ('W12342','(^[W]{1}\d{5})(?<!,)$')
My understanding is that the OP is saying the match should fail if the string begins or ends with a comma, not just that the preceding or trailing commas shouldn't match, so anchors are needed. Also, based on the regex he attempted, I infer that a single group, such as W00000, should match. So, I think the regex should be this, if the characters following the W must always be digits:
^W[:digit:]{5}(,W[:digit:]{5}){0,3}$
Or this, if they can be something other than digits:
^W[^,]{5}(,W[^,]{5}){0,3}$
UPDATE:
The OP posted the following comment:
I am on Oracle 11g and [:digit:] doesn't work. When I replace it with [0-9] it then works fine.
According to the documentation, Oracle 11g conforms to the POSIX regex standard and should be able to use POSIX character classes such as [:digit:]. However, I noticed in the docs that Oracle 11g does support Perl-style backslash character class abbreviations, which I didn't think was the case when I originally wrote this answer. In that case, the following should work:
^W\d{5}(,W\d{5}){0,3}$
Well in that case, you can do this:
(W[^,]{5},){3}W[^,]{5}
If I understood correctly, this should do it!
^W[0-9]{5}(,W[0-9]{5}){0,3}$
One W12345 pattern, maybe followed by one to 3 ,W12345 blocks.
Edit1: Adding ^$ to fail if there is a comma
Edit2: Fix class, since it fails on Oracle 11g