Starting a new question as my other question solved a different issue with the regex.
Here's my regex:
(?i)\\d{1,4}(?<!v(?:ol)?\\.?\\s?)(?![^\\(]*\\))
Regex split up for clarity:
(?i) - case insensitive
\\d{1,4} - a number with 1-4 digits
(?<!v(?:ol)?\\.?\\s?) the number cannot be preceded by 'v', 'v.', 'vol', 'vol.', with or without a space on the end.
(?![^\\(]*\\)) - Number cannot be inside parentheses.
It all works except for the 'vol.' bit.:
#"Words words 342 words (2342) (words 2 words) (words).ext" result 342 - correct.
#"Words - words words (2010) (words 2 words) (words).ext" result nil - correct.
#"words words v34 35.ext" result 34 - incorrect.
#"Words vol.342 343 (1234) (3 words) (desc).ext" result 342 - incorrect.
What am I doing wrong with my 'vol.' section?
You need to put the lookbehind before the number. Also, you need to add digits as illegal characters inside the lookbehind, or the 4 in v.34 will match. Try
(?i)(?<!v(?:ol)?\\.?\\s*\\d*)\\d{1,4}(?![^(]*\\))
This is expecting (edit: wrongly, as it turns out) that regexkitlite supports infinite repetition inside lookbehind which not many regex flavors do.
A look into the docs shows that it does support finite (but variable) repetition inside lookbehind, and if you are aware that the following will only work if there is at most one space between vol. and the number, then you could try
(?i)(?<!v(?:ol)?\\.?\\s?)(?<!\\d)\\d{1,4}(?![^(]*\\))
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 a quick question.
I'm studying some SQL exercises, and one of them, I need to remove the number from a street name.
For example:
5026 S CRENSHAW BLVD - Should be CRENSHAW
2635 WHITTIER BLVD - Should be WHITTIER
308 WESTWOOD PLZ # 1390L - Should Be WESTWOOD
1111 WILSHIRE BLVD - Should be WILSHIRE
Then, the answer to treat the address above was like this:
substring(facility_address FROM '[\d]+\s?\w?\s([\w]+)\s?')
I would like to understand how the substring works, what does it means de [\D] etc.
Could someone explain?
Thank you very much indeed! :)
You should read up on regular expressions.
'[\d]+\s?\w?\s([\w]+)\s?' is a regular expression.
I'll try to break it down:
First of all, these are quantifiers:
+ means one or more
? means one or none
And now for the regular expression:
[\d]+ matches one or more digits. I think the square brackets are actually not necessary here.
\s? optionally matches a whitespace character (optionally meaning it may or may not be there)
\w? optionally matches a word character
\s matches a whitespace character, but this time it's not optional since there is no ? at the end
([\w]+) matches one or more word characters. Notice the parenthesis which denote a so called 'capture group`. Everything within the parenthesis is actually returned by the substring function.
SELECT
ocr_text,
bucket,
REGEXP_EXTRACT('-?[0-9]+(\.[0-9]+)?', ocr_text)
FROM temp
I am trying to extract amounts from a string that will not have currency present. Any number that does not have decimals should not match. Commas should be allowed assuming they follow the correct rules (at hundreds marker)
56 no (missing decimals)
56.45 yes
120 no (missing decimals)
120.00 yes
1200.00 yes
1,200.00 yes
1,200 no (missing decimals)
1200 no (missing decimals)
134.5 no (decimal not followed by 2 digits)
23,00.00 no (invalid comma location)
I'm a noob to REGEX so I know my above statement already does not meet the criteria i've listed. However, i'm already stuck getting the error (INVALID_FUNCTION_ARGUMENT) premature end of char-class on my REGEX_EXTRACT line
Can someone point me in the right direction? How can I resolve my current issue? How can I modify to correctly incorporate the other criteria listed?
Here is a general regex pattern for a positive/negative number with two decimal places and optional thousands comma separators:
(?<!\S)(?:-?[0-9]{1,3}(,[0-9]{3})*(\.[0-9]{2})|-?[0-9]+(\.[0-9]{2}))(?!\S)
Demo
Your updated query:
SELECT
ocr_text,
bucket,
REGEXP_EXTRACT(ocr_text, '(?<!\S)(?:-?[0-9]{1,3}(,[0-9]{3})*(\.[0-9]{2})|-?[0-9]+(\.[0-9]{2}))(?!\S)')
FROM temp;
From the Presto docs I read, it supposedly supports Java's regex syntax. In the event that lookarounds are not working, you may try this version:
SELECT
ocr_text,
bucket,
REGEXP_EXTRACT(ocr_text, '(\s|^)(?:-?[0-9]{1,3}(,[0-9]{3})*(\.[0-9]{2})|-?[0-9]+(\.[0-9]{2}))(\s|$)')
FROM temp;
REGEXP_EXTRACT('^[-]?(\d*.\d*)', ocr_text)
Pattern: ^[-]?(\d*\.\d*)
Explanation:
^ - Start of line
[-]? - With or without negative dash (-)
\d* - 0 or more digits
\. - a decimal (escaped, because in regex decimals are considered special characters)
\d* - 0 or more digits (the decimal part);
$ - End of the line.
Bonus tip: There are helpful tools online to test your regex!
The Below code works to extract the value like all numbers but it catches all, only specific to certain alphabets its not working well. Anyone, please suggest well.
-?\d+\.?\d*
I have done work on NLP using Regex.
In Teradata, I'm looking for one regular expression pattern that would allow me to find a pattern of some numbers, then a space or maybe no space, and then 'SF'. It should return 7 in both cases below:
SELECT
REGEXP_INSTR('12345 1000SF', pattern),
REGEXP_INSTR('12345 1000 SF', pattern)
Or, my actual goal is to extract the 1000 in both cases if there's an easier way, probably using REGEXP_SUBSTR. More details are below if you need them.
I have a column that contains free text and I would like to extract the square footage. But, in some cases, there is a space between the number and 'SF' and in some cases there is not:
'other stuff 1000 SF'
'other stuff 1000SF'
I am trying to use the REGEXP_INSTR function to find the starting position. Through google, I have found the pattern for the first to be
'([0-9])+ SF'
When I try the pattern for the second, I try
'([0-9])+SF'
and I get the error
SELECT Failed. [2662] SUBSTR: string subscript out of bounds
I've also found an answer to a similar questions, but they don't work for Teradata. For example, I don't think you can use ? in Teradata.
The error message indicates you're using SUBSTR, not REGEXP_SUBSTR.
Try this:
RegExp_Substr(col, '[0-9]*(?= {0,1}SF)')
Find multiple digits followed by a single optional blank followed by SF and extract those digits.
I would pattern it like this:
\b(\d+)\s*[Ss][Ff]\b
\b # word boundary
(\d+) # 1 or more digits (captured)
\s* # 0 or more white-space characters
[Ss] # character class
[Ff] # character class
\b # word boundary
Demo
I'm trying to create a Regex to check for 6-12 characters, one being a digit, the rest being any characters, no spaces. Can Regex do this? I'm trying to do this in objective-c and I'm not familiar with Regex at all. I've been reading a couple tutorials, but most are for matching simple cases of a number, or a set of numbers, but not exactly what i'm looking for. I can do it with methods, but I was wondering if it that would be too slow and I figured I could try learning something new.
asdfg1 == ok
asdfg 1 != ok
asdfgh != ok
123456 != ok
asdfasgdasgdasdfasdf != ok
use this regex ^(?=.*\d)(?=.*[a-zA-Z])[^ ]{6,12}$
It seems that you mean "letter" when you say "character", right? And (thanks to burning_LEGION for pointing that out) there may be only one digit?
In that case, use
^(?=\D*\d\D*$)[^\W_]{6,12}$
Explanation:
^ # Start of string
(?=\D*\d\D*$) # Assert that there is exactly one digit in the string
[^\W_] # Match a letter or digit (explanation below)
{6,12} # 6-12 times
$ # End of string
[^\W_] might look a little odd. How does it work? Well, \w matches any letter, digit or underscore. \W matches anything that \w doesn't match. So [^\W] (meaning "match any character that is not not alphanumeric/underscore") is essentially the same as \w, but by adding _ to this character class, we can remove the underscore from the list of allowed characters.
i didn't try though, but i think here is the answer
(^[^\d\x20]*\d[^\d\x20]*$){6,12}
This is for one digit: ^[^\d\x20]{0,11}\d{1}[^\d\x20]{0,11}$ but I can`t get limited to 6-12 length, you can use other function to check length first and if it from 6 to 12 check with this regex witch I wrote.