I'm exploring regex, but I simply could not achieve exactly what I want yet. I'm using NetBeans and I need to swap all strncpy(... , sizeof(x)) to strncpy(... , sizeof(x) -1 ), i.e, add the "-1" between the last parenthesis.
An example should be:
strncpy(data->error, t_result[ID(data->modulo)].status, sizeof(data->error)); //need below
strncpy(data->error, t_result[ID(data->modulo)].status, sizeof(data->error) - 1);
See regex in use here
(strncpy\(.*?sizeof\([^)]*\))
(strncpy\(.*?sizeof\([^)]*\)) Capture the following into capture group 1
strncpy\( Matches strncpy( literally
.*? Matches any character any number of times, but as few as possible
sizeof\( Matches sizeof( literally
[^)]* Matches any character except ) any number of times
\) Matches ) literally
Replacement $1 - 1
Result in:
strncpy(data->error, t_result[ID(data->modulo)].status, sizeof(data->error) - 1);
Related
I would like to check if a specific column in one of my tables meets the following conditions:
String must contain at least three characters
String must contain at least two different numbers [e.g. 123 would work but 111 would not]
Characters which are allowed in the string:
Numbers (0-9)
Uppercase letters
Lowercase letters
Underscores (_)]
Dashes (-)
I have some experience with Regex but am having issues with Snowflake's syntax. Whenever I try using the '?' regex character (to mark something as optional) I receive an error. Can someone help me understand a workaround and provide a solution?
What I have so far:
SELECT string,
LENGTH(string) AS length
FROM tbl
WHERE REGEXP_LIKE(string,'^[0-9]+{3,}[-+]?[A-Z]?[a-z]?$')
ORDER BY length;
Thanks!
Your regex looks a little confusing and invalid, and it doesn't look like it quite meets your needs either. I read this expression as a string that:
Must start with one or more digits, at least 3 or more times
The confusing part to me is the '+' is a quantifier, which is not quantifiable with {3,} but somehow doesn't produce an error for me
Optionally followed by either a dash or plus sign
Followed by an uppercase character zero or one times (giving back as needed)
Followed by and ending with a lowercase character zero or one times (giving back as needed)
Questions
You say that your string must contain 3 characters and at least 2 different numbers, numbers are characters but I'm not sure if you mean 3 letters...
Are you considering the numbers to be characters?
Does the order of the characters matter?
Can you provide an example of the error you are receiving?
Notes
Checking for a second digit that is not the same as the first involves the concept of a lookahead with a backreference. Snowflake does not support backreferences.
One thing about pattern matching with regular expressions is that order makes a difference. If order is not of importance to you, then you'll have multiple patterns to match against.
Example
Below is how you can test each part of your requirements individually. I've included a few regexp_substr functions to show how extraction can work to check if something exists again.
Uncomment the WHERE clause to see the dataset filtered. The filters are written as expressions so you can remove any/all of the regexp_* columns.
select randstr(36,random(123)) as r_string
,length(r_string) AS length
,regexp_like(r_string,'^[0-9]+{3,}[-+]?[A-Z]?[a-z]?$') as reg
,regexp_like(r_string,'.*[A-Za-z]{3,}.*') as has_3_consecutive_letters
,regexp_like(r_string,'.*\\d+.*\\d+.*') as has_2_digits
,regexp_substr(r_string,'(\\d)',1,1) as first_digit
,regexp_substr(r_string,'(\\d)',1,2) as second_digit
,first_digit <> second_digit as digits_1st_not_equal_2nd
,not(regexp_instr(r_string,regexp_substr(r_string,'(\\d)',1,1),1,2)) as first_digit_does_not_appear_again
,has_3_consecutive_letters and has_2_digits and first_digit_does_not_appear_again as test
from table(generator(rowcount => 10))
//where regexp_like(r_string,'.*[A-Za-z]{3,}.*') // has_3_consecutive_letters
// and regexp_like(r_string,'.*\\d+.*\\d+.*') // has_2_digits
// and not(regexp_instr(r_string,regexp_substr(r_string,'(\\d)',1,1),1,2)) // first_digit_does_not_appear_again
;
Assuming the digits need to be contiguous, you can use a javascript UDF to find the number in a string with with the largest number of distinct digits:
create or replace function f(S text)
returns float
language javascript
returns null on null input
as
$$
const m = S.match(/\d+/g)
if (!m) return 0
const lengths = m.map(m=> [...new Set (m.split(''))].length)
const max_length = lengths.reduce((a,b) => Math.max(a,b))
return max_length
$$
;
Combined with WHERE-clause, this does what you want, I believe:
select column1, f(column1) max_length
from t
where max_length>1 and length(column1)>2 and column1 rlike '[\\w\\d-]+';
Yielding:
COLUMN1 | MAX_LENGTH
------------------------+-----------
abc123def567ghi1111_123 | 3
123 | 3
111222 | 2
Assuming this input:
create or replace table t as
select * from values ('abc123def567ghi1111_123'), ('xyz111asdf'), ('123'), ('111222'), ('abc 111111111 abc'), ('12'), ('asdf'), ('123 456'), (null);
The function is even simpler if the digits don't have to be contiguous (i.e. count the distinct digits in a string). Then core logic changes to:
const m = S.match(/\d/g)
if (!m) return 0
const length = [...new Set (m)].length
return length
Hope that's helpful!
When I execute this
select PATINDEX('%[0 ]%', '03/SI/00807/18-19')
I am getting 1.
By using ^ like this:
select PATINDEX('%[^0 ]%', '03/SI/00807/18-19')
I am getting 2.
[^] Allows you to match on any character not in the [^] brackets (for example, [^abc] would match on any character that is not a, b, or c characters) Whereas
[ ] Allows you to match on any character in the [ ] brackets (for example, [abc] would match on a, b, or c characters)
_ Allows you to match on a single character
% Allows you to match any string of any length (including zero length)
[^abcd] means: any one character EXCEPT a,b,c or d
select PATINDEX('%[0 ]%', '03/SI/00807/18-19')
The first character in your string which is (0 or space) is the 0 in the first place, so patindex returns 1.
select PATINDEX('%[^0 ]%', '03/SI/00807/18-19')
The first character in your string which is (neither 0 nor space) is the 3 in the second place, so patindex returns 2.
I am trying for a regex to
reject if input is all numbers
accept alpha-neumeric
reject colon ':'
I tried ,
ng-pattern="/[^0-9]/" and
ng-pattern="/[^0-9] [^:]*$/"
for example ,
"Block1 Grand-street USA" must be accepted
"111132322" must be rejected
"Block 1 grand : " must be rejected
You may use
ng-pattern="/^(?!\d+$)[^:]+$/"
See the regex demo.
To only forbid a : at the end of the string, use
ng-pattern="/^(?!\d+$)(?:.*[^:])?$/"
See another regex demo
The pattern matches
^ - start of string
(?!\d+$) - no 1+ digits to the end of the string
[^:]+ - one or more chars other than :
(?:.*[^:])? - an optional non-capturing group that matches 1 or 0 occurrences of
.* - any 0+ chars other than line break chars, as many as possible
[^:] - any char other than : (if you do not want to match an empty string, replace the (?: and )?)
$ - end of string.
According to comments, you want to match any character but colon.
This should do the job:
ng-pattern="/^(?!\d+$)[^:]+$/"
I'm looking to extract all the text up until a '\' (backslash).
The substring is required to remove all proceeding characters (17 in total) and so I would like to return all after the 17th until it comes across a backslash.
I've tried using charindex but it doesn't seem to stop at the \ it returns characters afterward. My code is as follows
SELECT path, substring(path,17, CHARINDEX('\',Path)+ LEN(Path)) As Data
FROM [Table].[dbo].[Projects]
WHERE Path like '\ENQ%\' AND
Deleted = '0'
Example
The below screen shot shows the basic query and result i.e the whole string
I then use substring to remove the first X characters as there will always be the same amount of proceeding characters
But what Im actually after is (based on the above result) the "Testing 1" "Testing 2" and "Testing ABC" section
The substring is required to remove all proceeding characters (17 in total) and so I would like to return all after the 17th until it comes across a backslash.
select
substring(path,17,CHARINDEX('\',Path)-17)
from
table
To overcome Invalid length parameter passed to the LEFT or SUBSTRING function error, you can use CASE
select
substring(path,17,
CASE when CHARINDEX('\',Path,17)>0
Then CHARINDEX('\',Path)-17)
else VA end
)
from
table
I have this string:
{"name": "Fancy HaXXor123Name","profession": 1,"race": 2,"map_id": 1052,"world_id": 268435461,"team_color_id": 0,"commander": false,"fov": 0.768}
I want to get an array back which includes the following information (from left to right from the string):
Fancy HaXXor123Name
1
2
1052
268435461
0
false
0.768
I tried to mess with RegExBuddy and got a promissing pattern which looks like this
(\d{1,}).(\d{1,})|(\d{1,})|(?i)"(.*?)"
This is what I got back
name
Fancy HaXXor123Name
profession
1
race
2
map_id
10
2
world_id
2684354
1
team_color_id
0
commander
fov
0
768
So there are large spaces between the informations, torn numbers and the false is missing. I can't fix this problem and I'm completely new to StringRegExp.
I'm using AutoIT which uses the PCRE RegExp-Engine (this is what think).
You may use a regex like the following:
"\s*:\s*(?:"\K[^"]*|\K[^][\s,{}]+)
See the regex demo
Details:
"\s*:\s* - a literal ", 0+ whitespaces, :, 0+ whitespaces
(?:"\K[^"]*|\K[^][\s,{}]+) - A non-capturing group matching 2 alternatives:
"\K[^"]* - a ", then \K zeros the text matched so far, and then matches 0+ chars other than " with [^"]*
\K[^][\s,{}]+ - \K drops the text matched so far, and [^][\s,{}]+ matches 1+ chars other than ], [, whitespace, ,, { and }.