Here is a regex pattern I created in Objective C:
^\n?([#]{1,2}$|[*]{1,2}$|[0-9]{1,3}.$)
I want to match:
starts with \n or empty
ends with # or * or .
if ends with . there will be 1 or 2 or 3 digits in between
If ends with # or *, there could be 1 more # or * in between
The regex I created matches '\n1#' which is not what I want.
Can anyone help me correct this? Is this fastest one? The regex will be used frequently, so I want it to be as fast as possible.
UPDATE:
Here's a sample strings for testing:
"\n#", "11*1", "1#", "a1.", "111*", "\n1#", "\n11.", "a11.", "1. ", "*1."
The 1# and 111* were matched. Not sure what went wrong.
You're matching #1 and 111# because of [0-9]{1,3}.. You haven't escaped the . and this group basically matches any sequence of 1 to 3 digits followed by any character.
What you're looking for is
^\n?(#{1,2}|\*{1,2}|[0-9]{1,3}\.)$
Properly escaped in ObjC, it would be
#"^\n?(#{1,2}|\\*{1,2}|[0-9]{1,3}\\.)$"
If this regex is used quite a lot, you might want to cache the NSRegularExpression object to avoid compiling it everytime.
Regexpal is very useful to test regular expressions.
Related
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
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.
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
What is the regex for these cases:
29000.12345678900, expected result 29000.123456789
29000.000, expected result 29000
29000.00003400, expected result 29000.000034
In short, I want to eliminate the 0 point if there is no 1-9 found again behind decimal and I also want to eliminate the dot (.) if actually the number can be considered as integer.
I use this regex
(?:.0*$|0*$)
but it gives me this result:
29123.6 from 29123.6400, 4 is gone from there.
When I tested the regex separately, it works perfectly,
.0*$ gives me 29123 from 29123.0000
0*$ gives me 29123.6423 from 29123.642300
Am I missing something with the combined regex?
If you think regex is the best way of doing it, you can just use something like this:
\.?0+$
It works for both cases:
> '12300000.000001130000000'.replace(/\.?0+$/g, '')
"12300000.00000113"
> '12300000.000000000000'.replace(/\.?0+$/g, '')
"12300000"
You can use this regex
^\d+(\.\d*[1-9])?
- -------------
| |->this would match only if the digits after . end with [1-9]
|
|->^ depicts the start of the string..it is necessary to match the pattern
that solves your problem
try it here
You simply want this:
^\d*(\.?\d*[1-9])?
^\d* that means one or more digit before the first group.
In the () that describes matching group.
\.? means single DOT(.) can be there but optional. eg. (.)
\d* there can be one or more digits. eg. (1234)
\.?\d* there can be one DOT and one or more digit eg. (.123)
[1-9] this includes only digit from 1 to 9 only excluding 0. eg. (2344)
Regex
I don't know whether Objective-C supports something like the following construct, but in Python you can do it completely without regular expressions using str.rstrip():
In [1]: def shorten_number(number):
...: return number.rstrip('0').rstrip('.')
In [2]: shorten_number('29000.12345678900')
Out[2]: '29000.123456789'
In [3]: shorten_number('29000.000')
Out[3]: '29000'
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.