I want to format the strings in a table column, in a specific format.
Input table:
file_paths
my-file-path/wefw/wefw/2022-03-20
my-file-path/wefw/2022-01-02
my-file-path/wef/wfe/wefw/wef/2021-02-03
my-file-path/wef/wfe/wef/
I want to remove everything after the last / sign, if the only thing after it resembles a date (i.e. YYYY-MM-dd or ####-##-##).
Output:
file_paths
my-file-path/wefw/wefw/
my-file-path/wefw/
my-file-path/wef/wfe/wefw/wef/
my-file-path/wef/wfe/wef/
I'm thinking of doing something like:
SELECT regexp_replace(file_paths, 'regex_here', '', 1, 'i')
FROM my_table
I'm unsure of how to write the RegEx for this though. I'm also open to easier methods of string manipulation, if there are any. Thanks in advance!
You can use
REGEXP_REPLACE ( file_paths, '/[0-9]{4}-[0-9]{1,2}-[0-9]{1,2}$', '/' )
See the regex demo.
The /[0-9]{4}-[0-9]{1,2}-[0-9]{1,2}$ is a POSIX ERE compliant pattern matching
/ - a slash
[0-9]{4} - four digits
- - a hyphen
[0-9]{1,2} - one or two digits
-[0-9]{1,2} - a hyphen and one or two digits
$ - end of string.
If your values can contain trailing whitespace, insert [[:space:]]* before $: /[0-9]{4}-[0-9]{1,2}-[0-9]{1,2}[[:space:]]*$.
You may use this regex:
^(.*?\/)\d{4}-\d{2}-\d{2}$
You may try this query:
select regexp_replace(file_paths, '^(.*?\/)\\d{4}-\\d{2}-\\d{2}$','$1')
Demo
Related
how would I be able to grab the number 2627995 from this string
"hellotest/2627995?hl=en"
I want to grab the number 2627995, here is my current regex but it does not work when I use regex extract from big query
(\/)\d{7,7}
SELECT
REGEXP_EXTRACT(DESC, r"(\/)\d{7,7}")
AS number
FROM
`string table`
here is the output
Thank you!!
I think you just want to match all digits coming after the last path separator, before either the start of the query parameter, or the end of the URL.
SELECT REGEXP_EXTRACT(DESC, r"/(\d+)(?:\?|$)") AS number
FROM `string table`
Demo
Try this one: r"\/(\d+)"
Your code returns the slash because you captured it (see the parentheses in (\/)\d{7,7}). REGEXP_EXTRACT only returns the captured substring.
Thus, you could just wrap the other part of your regex with the parentheses:
SELECT
REGEXP_EXTRACT(DESC, r"/(\d{7})")
AS number
FROM
`string table`
NOTE:
In BigQuery, regex is specified with string literals, not regex literals (that are usually delimited with forward slashes), that is why you do not need to escape the / char (it is not a special regex metacharacter)
{7,7} is equal to {7} limiting quantifier, meaning seven occurrences.
Also, if you are sure the number is at the end of string or is followed with a query string, you can enhance it as
REGEXP_EXTRACT(DESC, r"/(\d+)(?:[?#]|$)")
where the regex means
/ - a / char
(\d+) - Group 1 (the actual output): one or more digits
(?:[?#]|$) - either ? or # char, or end of string.
I am having following string in my query
.\ABC\ABC\2021\02\24\ABC__123_123_123_ABC123.txt
beginning with a period from which I need to extract the segment between the final \ and the file extension period, meaning following expected result
ABC__123_123_123_ABC123
Am fairly new to using REGEXP and couldn't help myself to an elegant (or workable) solution with what Q&A here or else. In all queries the pattern is the same in quantity and order but for my growth of knowledge I'd prefer to not just count and cut.
You can use REGEXP_REPLACE function such as
REGEXP_REPLACE(col,'(.*\\)(.*)\.(.*)','\2')
in order to extract the piece starting from the last slash upto the dot. Preceding slashes in \\ and \. are used as escape characters to distinguish the special characters and our intended \ and . characters.
Demo
You need just regexp_substr and simple regexp ([^\]+)\.[^.]*$
select
regexp_substr(
'.\ABC\ABC\2021\02\24\ABC__123_123_123_ABC123.txt',
'([^\]+)\.[^.]*$',
1, -- position
1, -- occurence
null, -- match_parameter
1 -- subexpr
) substring
from dual;
([^\]+)\.[^.]*$ means:
([^\]+) - find one or more(+) any characters except slash([] - set, ^ - negative, ie except) and name it as group \1(subexpression #1)
\. - then simple dot (. is a special character which means any character, so we need to "escape" it using \ which is an escape character)
[^.]* - zero or more any characters except .
$ - end of line
So this regexp means: find a substring which consist from: one or more any characters except slash followed by dot followed by zero or more any characters except dot and it should be in the end of string. And subexpr parameter = 1, says oracle to return first subexpression (ie first matched group in (...))
Other parameters you can find in the doc.
Here is my simple full compatible example with Oracle 11g R2, PCRE2 and some other languages.
Oracle 11g R2 using function substr (Reference documentation)
select
regexp_substr(
'.\ABC\ABC\2021\02\24\ABC__123_123_123_ABC123.txt',
'((\w)+(_){2}(((\d){3}(_)){3}){1}((\w)+(\d)+){1}){1}',
1,
1
) substring
from dual;
Pattern: ((\w)+(_){2}(((\d){3}(_)){3}){1}((\w)+(\d)+){1}){1}
Result: ABC__123_123_123_ABC123
Just as simple as it can be, regular expressions always follow a minimal standard, as you can see portability also provided, just for the case someone else is interested in going the simplest way.
Hopefully, this will help you out!
I need an oracle regex to fetch data before first pipe and after the last slash from the text before pipe.
For example, from the string:
test=file://2019/13/40/9/53/**2abc123-7test-1edf-9xyz-12345678.bin**|type
the data to be fetched is:
2abc123-7test-1edf-9xyz-12345678.bin
This works in Oracle :
select regexp_substr(col,'[^|/]+\.\w+',1,1,'i')
from (
select 'test=file://2019/13/40/9/53/2abc123-7test-1edf-9xyz-12345678.bin|type=app/href|size=1234|encoding=|locale=en_|foo.bar' as col
from dual
) q
MySql & TeraData also have such REGEXP_SUBSTR function, but haven't tested it on those.
The pattern ^.+?/([^/]+?)\| starts at the beginning of the string, skips over every character, then captures all non-slash characters, between the last slash and the first pipe.
You may use:
REGEXP_SUBSTR(column, '/([^/|]+)\|', 1, 1, NULL, 1)
Live demo here
Regex breakdown:
/ Match literally
( Start of capturing group #1
[^/|]+ Match anything except slash and pipe, at least one character
) End of CG #1
\| Match a pipe
[^\/]*?(?=\|)
[^\/]*? — matches any character that is not a backslash
(?=\|) — positive lookahead to match a vertical line
Input string: ["1189-13627273","89-13706681","118-13708388"]
Expected Output: ["14013627273","14013706681","14013708388"]
What I am trying to achieve is to replace any numbers till the '-' for each item with hard coded text like '140'
SELECT replace(value_to_replace, '-', '140')
FROM (
VALUES ('1189-13627273-77'), ('89-13706681'), ('118-13708388')
) t(value_to_replace);
check this
I found the right way to achieve that using the below regular expression.
SELECT REGEXP_REPLACE (string_to_change, '\\"[0-9]+\\-', '140')
You don't need a regexp for this, it's as easy as concatenation of 140 and the substring from - (or the second part when you split by -)
select '140'||substring('89-13706681' from position('-' in '89-13706681')+1 for 1000)
select '140'||split_part('89-13706681','-',2)
also, it's important to consider if you might have instances that don't contain - and what would be the output in this case
Use regexp_replace(text,text,text) function to do so giving the pattern to match and replacement string.
First argument is the value to be replaced, second is the POSIX regular expression and third is a replacement text.
Example
SELECT regexp_replace('1189-13627273', '.*-', '140');
Output: 14013627273
Sample data set query
SELECT regexp_replace(value_to_replace, '.*-', '140')
FROM (
VALUES ('1189-13627273'), ('89-13706681'), ('118-13708388')
) t(value_to_replace);
Caution! Pattern .*- will replace every character until it finds last occurence of - with text 140.
I have strings like 'keepme:cutme' or 'string-without-separator' which should become respectively 'keepme' and 'string-without-separator'. Can this be done in PostgreSQL? I tried:
select substring('first:last' from '.+:')
But this leaves the : in and won't work if there is no : in the string.
Use split_part():
SELECT split_part('first:last', ':', 1) AS first_part
Returns the whole string if the delimiter is not there. And it's simple to get the 2nd or 3rd part etc.
Substantially faster than functions using regular expression matching. And since we have a fixed delimiter we don't need the magic of regular expressions.
Related:
Split comma separated column data into additional columns
regexp_replace() may be overload for what you need, but it also gives the additional benefit of regex. For instance, if strings use multiple delimiters.
Example use:
select regexp_replace( 'first:last', E':.*', '');
SQL Select to pick everything after the last occurrence of a character
select right('first:last', charindex(':', reverse('first:last')) - 1)