Regular expression for pattern xxxx/xxxxU - sql

I'm looking for a regular expression to match the pattern xxxx/xxxxU, where x can be 0-9 and the "U" at the end is optional.
Valid examples: 1111/1111, 1111/1111U
Invalid examples: 1111/1111Z, 111/1111
I could reach until '[0-9]{4}/[0-9]{4}', but I'm not sure how to handle the optional "U" at the end.

Always match numbers-slash-numbers; U included, if present:
[0-9]{4}/[0-9]{4}U?
or if you replace the number ranges with \d (for a digit character):
\d{4}/\d{4}U?
The ? means zero or one of the preceding character. So zero or one of U. Test it
The entire string 1111\2222U will be matched, while the match for 1111\2222Z will include the digits-slash-digits part but not the Z.
Only match if string ends in a digit or U:
If a string fragment ending in any letter other than U is not to be matched at all, try something like:
^\d{4}/\d{4}U?$
which matches if the numbers-slash-numbers plus optional U is the only content in the string (test it) or
\d{4}/\d{4}U?(\s|$)
which matches if the numbers-slash-numbers plus optional U is followed by either a white space character (included in the match) or the end of the string. (Test it.)
(Note: the test it links show the "/" between the numbers escaped with "\" [e.g. "/"]--something required by that implementation. I'm not familiar with Oracle's regex syntax, so this may not be required on that platform.)

I'd use '[0-9]{4}/[0-9]{4}U?' or '[0-9]{4}/[0-9]{4}U{0,1}'
Found at: http://docs.oracle.com/cd/B12037_01/appdev.101/b10795/adfns_re.htm

You could try this expression:
[0-9]{4}/[0-9]{4}U?
The ? means: optional (0 or 1). Have a look at this useful regex overview table.

Thanks for all your comments, but there were some cases which your answer did not cover.
'^[0-9]{4}/[0-9]{4}U?$'
The above works for all of my above cases

Related

Trino regexp_replace this character in the beginning but not in the middle Trino [duplicate]

I am a complete Reg-exp noob, so please bear with me. Tried to google this, but haven't found it yet.
What would be an appropriate way of writing a Regular expression matching files starting with a dot, such as .buildpath or .htaccess?
Thanks a lot!
In most regex languages, ^\. or ^[.] will match a leading dot.
The ^ matches the beginning of a string in most languages. This will match a leading .. You need to add your filename expression to it.
^\.
Likewise, $ will match the end of a string.
You may need to substitute the \ for the respective language escape character. However, under Powershell the Regex I use is: ^(\.)+\/
Test case:
"../NameOfFile.txt" -match '^(\\.)+\\\/'
works, while
"_./NameOfFile.txt" -match '^(\\.)+\\\/'
does not.
Naturally, you may ask, well what is happening here?
The (\\.) searches for the literal . followed by a +, which matches the previous character at least once or more times.
Finally, the \\\/ ensures that it conforms to a Window file path.
It depends a bit on the regular expression library you use, but you can do something like this:
^\.\w+
The ^ anchors the match to the beginning of the string, the \. matches a literal period (since an unescaped . in a regular expression typically matches any character), and \w+ matches 1 or more "word" characters (alphanumeric plus _).
See the perlre documentation for more info on Perl-style regular expressions and their syntax.
It depends on what characters are legal in a filename, which depends on the OS and filesystem.
For example, in Windows that would be:
^\.[^<>:"/\\\|\?\*\x00-\x1f]+$
The above expression means:
Match a string starting with the literal character .
Followed by at least one character which is not one of (whole class of invalid chars follows)
I used this as reference regarding which chars are disallowed in filenames.
To match the string starting with dot in java you will have to write a simple expression
^\\..*
^ means regular expression is to be matched from start of string
\. means it will start with string literal "."
.* means dot will be followed by 0 or more characters

Regex like telephone number on Hive without prefix (+01)

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.

regex not working correctly when the test is fine

For my database, I have a list of company numbers where some of them start with two letters. I have created a regex which should eliminate these from a query and according to my tests, it should. But when executed, the result still contains the numbers with letters.
Here is my regex, which I've tested on https://www.regexpal.com
([^A-Z+|a-z+].*)
I've tested it against numerous variations such as SC08093, ZC000191 and NI232312 which shouldn't match and don't in the tests, which is fine.
My sql query looks like;
SELECT companyNumber FROM company_data
WHERE companyNumber ~ '([^A-Z+|a-z+].*)' order by companyNumber desc
To summerise, strings like SC08093 should not match as they start with letters.
I've read through the documentation for postgres but I couldn't seem to find anything regarding this. I'm not sure what I'm missing here. Thanks.
The ~ '([^A-Z+|a-z+].*)' does not work because this is a [^A-Z+|a-z+].* regex matching operation that returns true even upon a partial match (regex matching operation does not require full string match, and thus the pattern can match anywhere in the string). [^A-Z+|a-z+].* matches a letter from A to Z, +,|or a letter fromatoz`, and then any amount of any zero or more chars, anywhere inside a string.
You may use
WHERE companyNumber NOT SIMILAR TO '[A-Za-z]{2}%'
See the online demo
Here, NOT SIMILAR TO returns the inverse result of the SIMILAR TO operation. This SIMILAR TO operator accepts patterns that are almost regex patterns, but are also like regular wildcard patterns. NOT SIMILAR TO '[A-Za-z]{2}%' means all records that start with two ASCII letters ([A-Za-z]{2}) and having anything after (%) are NOT returned and all others will be returned. Note that SIMILAR TO requires a full string match, same as LIKE.
Your pattern: [^A-Z+|a-z+].* means "a string where at least some characters are not A-Z" - to extend that to the whole string you would need to use an anchored regex as shown by S-Man (the group defined with (..) isn't really necessary btw)
I would probably use a regex that specifies want the valid pattern is and then use !~ instead.
where company !~ '^[0-9].*$'
^[0-9].*$ means "only consists of numbers" and the !~ means "does not match"
or
where not (company ~ '^[0-9].*$')
Not start with a letter could be done with
WHERE company ~ '^[^A-Za-z].*'
demo: db<>fiddle
The first ^ marks the beginning. The [^A-Za-z] says "no letter" (including small and capital letters).
Edit: Changed [A-z] into the more precise [A-Za-z] (Why is this regex allowing a caret?)

REGEXP_REPLACE explanation

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.

regex capture middle of url

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