I am trying to extract a item_subtype field from an URL.
This regex works fine in the to get the first item item_type
SELECT REGEXP_EXTRACT('info?item_type=icecream&item_subtype=chocolate/cookies%20cream,vanilla&page=1', r'item_type=(\w+)')
but what is the correct regex to get everything starting from 'chocolate' all the way to before the '&page1'
I have tried this, but can't seem to get it to work to go further
SELECT REGEXP_EXTRACT('info?item_type=icecream&item_subtype=chocolate/cookies%20cream,vanilla&page=1', r'item_subtype=(\w+[^Z])')
basically, I want to extract 'chocolate/cookies%20cream,vanilla'
In your case, \w+ only matches one or more letters, digits or underscores. Your expected values may contain other characters, too.
You may use
SELECT REGEXP_EXTRACT('info?item_type=icecream&item_subtype=chocolate/cookies%20cream,vanilla&page=1', r'item_subtype=([^&]+)')
See the regex demo.
Notes:
item_subtype= - this string is matched as a literal char sequence
([^&]+) - a Capturing group 1 that matches and captures one or more chars other than & into a separate memory buffer that is returned by REGEXP_EXTRACT function.
Related
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.
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.
In Bigquery, I am trying to find a way to extract particular segments of a string based on how many dashes come before it. The number of total dashes in the string will always be the same. For example, I could be looking for the string after the second dash and before the third dash in the following string:
abc-defgh-hij-kl-mnop
Currently, I am using the following regex to extract, which counts the dashes from the back:
([^-]+)(?:-[^-]+){2}$
The problem is that if there is nothing in between the dashes, the regex doesn't work. For example, something like this returns null:
abc-defgh-hij--mnop
Is there a way to use regex to extract a string after a certain number of dashes and cut it off before the subsequent dash?
Thank you!
Below is for BigQuery Standrd SQL
The simplest way in your case is to use SPLIT and OFFSET as in below example
SELECT SPLIT(str, '-')[OFFSET(3)]
above will return empty string for abc-defgh-hij--mnop
to prevent error in case of calling non-existing element - better to use SAFE_OFFSET
SELECT SPLIT(str, '-')[SAFE_OFFSET(3)]
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 ..