I need to retrieve the bolded section of the below string . This value is in a column within my Postgres database table.
SEALS_LME_TRADES_MBL_20220919_00212.csv
I tried to utilize the functions; substring, reverse, strpos but they all have limitations. It seems like regex is the best option, however I was not able to do it.
Essentially I need to substring from beginning till the second last '_'. I do not want the date and sequence number along with the file extension at the end.
The closes regex I managed to get is: ^(([^]*){4})
https://regex101.com/
This look a little wonky but how about this?
select substring ('SEALS_LME_TRADES_MBL_20220919_00212.csv', '^(.+)_[^_]+_[^_]+')
Translation
^ from the beginning
(.+) any characters (capture and return this value), followed by
_ an underscore, followed by
[^_]+ one or more non-underscores, followed by
_ an underscore, followed by
[^_]+ one or more non-underscores
Regex greediness will cause any incidental underscores to be captured in the initial string.
Technically speaking the last portion (one or more non-underscores) can probably be omitted.
Related
I have two types of URL's which I would need to clean, they look like this:
["//xxx.com/se/something?SE_{ifmobile:MB}{ifnotmobile:DT}_A_B_C_D_E_F_G_H"]
["//www.xxx.com/se/car?p_color_car=White?SE_{ifmobile:MB}{ifnotmobile:DT}_A_B_C_D_E_F_G_H"]
The outcome I want is;
SE_{ifmobile:MB}{ifnotmobile:DT}_A_B_C_D_E_F_G_H"
I want to remove the brackets and everything up to SE, the URLS differ so I want to remove:
First URL
["//xxx.com/se/something?
Second URL:
["//www.xxx.com/se/car?p_color_car=White?
I can't get my head around it,I've tried this .*\/ . But it will still keep strings I don't want such as:
(1 url) =
something?
(2 url) car?p_color_car=White?
You can use
regexp_replace(FinalUrls, r'.*\?|"\]$', '')
See the regex demo
Details
.*\? - any zero or more chars other than line breakchars, as many as possible and then ? char
| - or
"\]$ - a "] substring at the end of the string.
Mind the regexp_replace syntax, you can't omit the replacement argument, see 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.
Have strings containing 'q_' which I want to extract everything that comes after it. Some rows contain occurrence of q_ which I want everything that occurs after it. Example values in the column are:
prod-q_cat_trait_cat_social_issue
_prod-q_body_modification_graffiti
event_tickets
dappled_grey
_prod-q_cat_tech_support
What is wrong with my regular expression as I'm trying to remove the trailing '_' after q.
REGEXP_EXTRACT(queue_id, '[^q_]+$')
Is just returning
issue
I've also tried the split method:
SPLIT(queue_id, 'q_')[OFFSET(2)]
But this returns
Array index 2 is out of bounds (overflow)
Any suggestions. Thanks! (I am using Google Cloud SQL)
Using a capturing group, you may extract all after the first q_ with:
REGEXP_EXTRACT(queue_id, 'q_(.*)')
You may extract all after the last q_ with:
REGEXP_EXTRACT(queue_id, '.*q_(.*)')
See the regex demo #1 and regex demo #2.
Here, q_ finds the first occurrence of q_ and (.*) grabs the rest of the line into Group 1, and this is the value returned by REGEXP_EXTRACT. .* matches any 0+ chars other than line break chars as many as possible, that is why the second regex will start capturing the rest of the line after the last occurrence of q_.
Google Cloud SQL uses MySQL. I think the simplest method is substring_index():
select substring_index(queue_id, '-q_', -1)
Can you try this : q_([^q_]+)$? You'll have what you want in the first group.
Edit: this one match all the cases > (?(?<=-q_).*|^((?!-q_).)*$)
Hi may i know what does the below query means?
REGEXP_REPLACE(number,'[^'' ''-/0-9:-#A-Z''[''-`a-z{-~]', 'xy') ext_number
part 1
In terms of explaining what the function function call is doing:
It is a function call to analyse an input string 'number' with a regex (2nd argument) and replace any parts of the string which match a specific string. As for the name after the parenthesis I am not sure, but the documentation for the function is here
part 2
Sorry to be writing a question within an answer here but I cannot respond in comments yet (not enough rep)
Does this regex work? Unless sql uses different syntax this would appear to be a non-functional regex. There are some red flags, e.g:
The entire regex is wrapped in square parenthesis, indicating a set of characters but seems to predominantly hold an expression
There is a range indicator between a single quote and a character (invalid range: if a dash was required in the match it should be escaped with a '\' (backslash))
One set of square brackets is never closed
After some minor tweaks this regex is valid syntax:
^'' ''\-\/0-9:-#A-Z''[''-a-z{-~]`, but does not match anything I can think of, it is important to know what string is being examined/what the context is for the program in order to identify what the regex might be attempting to do
It seems like it is meant to replaces all ASCII control characters in the column or variable number with xy.
[] encloses a class of characters. Any character in that class matches. [^] negates that, hence all characters match, that are not in the class.
- is a range operator, e.g. a-z means all characters from a to z, like abc...xyz.
It seams like characters enclosed in ' should be escaped (The second ' is to escape the ' in the string itself.) At least this would make some sense. (But for none of the DBMS I found having a regexp_replace() function (Postgres, Oracle, DB2, MariaDB, MySQL), I found something in the docs, that would indicate this escape mechanism. They all use \, but maybe I missed something? Unfortunately you didn't tag which DBMS you're actually using!)
Now if you take an ASCII table you'll see, that the ranges in the expression make up all printable characters (counting space as printable) in groups from space to /, 0 to 9, : to #, etc.. Actually it might have been shorter to express it as '' ''-~, space to ~.
Given the negation, all these don't match. The ones left are from NUL to US and DEL. These match and get replaced by xy one by one.
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 have data like
http://www.linz.at/politik_verwaltung/32386.asp
stored in a text column. I thought a non-greedy extraction with
select substring(turl from '\..*?$') as ext from tdata
would give me .asp but instead it still ?greedely results in
.linz.at/politik_verwaltung/32386.asp
How can I only match against the last occurence of dot .?
Using Postgresql 9.3
\.[^.]*$ matches . followed by any number of non-dot characters followed by end-of-string:
# select substring('http://www.linz.at/politik_verwaltung/32386.asp'
from '\.[^.]*$');
substring
-----------
.asp
(1 row)
As for why the non-greedy quantifiers do not work here is that they still start matching as soon as possible while still trying to match as short as possible from there on.
Try this:
\.[\w]*$
Here is how it works:
all the word characters (\w), any numbers of them with *, between dot (\.) and the end of the string ($), with the last . itself.
Note: updated the answer, now will capture the strings ends with ..