I want to retrieve file names from urls in sql.
for example:
Input:
url:
https://www.google.co.in/root/subdir/file.extension?p1=v1&p2=v2
https://www.abxdhcak.com/sitemap-companies.xml
then Output should be:
file.extension
sitemap-companies.xml
To match your expected output you can use REGEXP_REPLACE
REGEXP_REPLACE(txt, '^.*/|\?.*$') as rg
This does 2 things:
'^.*/'
This removes all characters up to and including the last forward-slash in the string.
'\?.*$'
This removes all characters after and including a question mark.
This may not work for all cases, but it works for the examples provided.
Related
We have a problem with a regular expression on hive.
We need to exclude the numbers with +37 or 0037 at the beginning of the record (it could be a false result on the regex like) and without letters or space.
We're trying with this one:
regexp_like(tel_number,'^\+37|^0037+[a-zA-ZÀÈÌÒÙ ]')
but it doesn't work.
Edit: we want it to come out from the select as true (correct number) or false.
To exclude numbers which start with +01 0r +001 or +0001 and having only digits without spaces or letters:
... WHERE tel_number NOT rlike '^\\+0{1,3}1\\d+$'
Special characters like + and character classes like \d in Hive should be escaped using double-slash: \\+ and \\d.
The general question is, if you want to describe a malformed telephone number in your regex and exclude everything that matches the pattern or if you want to describe a well-formed telephone number and include everything that matches the pattern.
Which way to go, depends on your scenario. From what I understand of your requirements, adding "not starting with 0037 or +37" as a condition to a well-formed telephone number could be a good approach.
The pattern would be like this:
Your number can start with either + or 00: ^(\+|00)
It cannot be followed by a 37 which in regex can be expressed by the following set of alternatives:
a. It is followed first by a 3 then by anything but 7: 3[0-689]
b. It is followed first by anything but 3 then by any number: [0-24-9]\d
After that there is a sequence of numbers of undefined length (at least one) until the end of the string: \d+$
Putting everything together:
^(\+|00)(3[0-689]|[0-24-9]\d)\d+$
You can play with this regex here and see if this fits your needs: https://regex101.com/r/KK5rjE/3
Note: as leftjoin has pointed out: To use this regex in hive you might need to additionally escape the backslashes \ in the pattern.
You can use
regexp_like(tel_number,'^(?!\\+37|0037)\\+?\\d+$')
See the regex demo. Details:
^ - start of string
(?!\+37|0037) - a negative lookahead that fails the match if there is +37 or 0037 immediately to the right of the current location
\+? - an optional + sign
\d+ - one or more digits
$ - end of string.
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.
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.
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)]
If I have a column, say:
paths
And paths holds the absolute path to a file.
If I wanted, to remove n characters from the beginning and end of the records, for the entire path column - would this be possible?
Edit - example:
Lets say there are records in paths like so:
C:\Users\Alex\Documents\Files\File-1.txt
C:\Users\Alex\Documents\Files\File-2.txt
C:\Users\Alex\Documents\Files\File-3.txt
C:\Users\Alex\Documents\Files\File-4.txt
I would like to update the records to be:
Users\Alex\Documents\Files\File-1
Users\Alex\Documents\Files\File-2
Users\Alex\Documents\Files\File-3
Users\Alex\Documents\Files\File-4
So essentially removing n characters from the beginning and end of an entire column.
A general regex solution would be:
^(?:.{3})(.*)(?:.{4})$
# that is match exactly three chars in the beginning (non-capturing group)
# match everything up to the end of the string
# but as .* is sweet-tempered...
# it gives back the four characters in the end ($)
# your match is in the first group
See a demo on regex101.com
use this pattern and replace with nothing
^.{3}|.{4}$
Demo
As you tagged SQL which implies Standard SQL:
Answers to questions tagged with SQL should use ANSI SQL.
There's no need for anything more complicated than a SUBSTRING, you just need to calculate the number of characters you want to strip:
SUBSTRING(path FROM 4 FOR CHARACTER_LENGTH(path)-7)
This is also working as-is in PostgreSQL :)