I am stuck on this regex problem.
A 16-digit credit card number, with the first digit being a 5 and the second digit being a 1, 2, 3, 4, or 5 (the rest of the digits can be anything).
so far I have ^4[1,5]\d{14} and I know I'm missing a lot of things but I dont know what I'm missing..
please help and thanks!
Look at the start of your regex:
^4[1,5]
That says that the number must start with 4 (not 5), and that the second character must be 1, a comma, or 5.
You want this instead (followed by the rest, of course):
^5[1-5]
Note the use of - rather than , to indicate a range of characters.
The full regex you're looking for is the following
^5[1-5]\d{14}$
Demo
Your error lays in the fact that you used 1,5 as a range but this will just match 1 , or 5 as characters. To use a range, the - is needed between the enclosings
Related
I want to check if a number is 50 or more using a regular expression. This in itself is no problem but the number field has another regex checking the format of the entered number.
The number will be in the continental format: 123.456,78 (a dot between groups of three digits and always a comma with 2 digits at the end)
Examples:
100.000,00
50.000,00
50,00
34,34
etc.
I want to capture numbers which are 50 or more. So from the four examples above the first three should be matched.
I've come up with this rather complicated one and am wondering if there is an easier way to do this.
^(\d{1,3}[.]|[5-9][0-9]|\d{3}|[.]\d{1,3})*[,]\d{2}$
EDIT
I want to match continental numbers here. The numbers have this format due to internal regulations and specify a price.
Example: 1000 EUR would be written as 1.000,00 EUR
50000 as 50.000,00 and so on.
It's a matter of taste, obviously, but using a negative lookahead gives a simple solution.
^(?!([1-4]?\d),)[1-9](\d{1,2})?(\.\d{3})*,\d{2}\b
In words: starting from a boundary ignore all numbers that start with 1 digit OR 2 digits (the first being a 1,2,3 or 4), followed by a comma.
Check on regex101.com
Try:
EDIT ^(.{3,}|[5-9]\d),\d{2}$
It checks if:
there 3 chars or more before the ,
there are 2 numbers before the , and the first is between 5 and 9
and then a , and 2 numbers
Donno if it answer your question as it'll return true for:
aa50,00
1sdf,54
But this assumes that your original string is a number in the format you expect (as it was not a requirement in your question).
EDIT 3
The regex below tests if the number is valid referring to the continental format and if it's equal or greater than 50. See tests here.
Regex: ^((([1-9]\d{0,2}\.)(\d{3}\.){0,}\d{3})|([1-9]\d{2})|([5-9]\d)),\d{2}$
Explanation (d is a number):
([1-9]\d{0,2}\.): either d., dd. or ddd. one time with the first d between 1 and 9.
(\d{3}\.){0,}: ddd. zero or x time
\d{3}: ddd 3 digit
These 3 parts combined match any numbers equals or greater than 1000 like: 1.000, 22.002 or 100.000.000.
([1-9]\d{2}): any number between 100 and 999.
([5-9]\d)): a number between 5 and 9 followed by a number. Matches anything between 50 and 99.
So it's either the one of the parts above or this one.
Then ,\d{2}$ matches the comma and the two last digits.
I have named all inner groups, for better understanding what part of number is matched by each group. After you understand how it works, change all ?P<..> to ?:.
This one is for any dec number in the continental format.
^(?P<common_int>(?P<int>(?P<int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<int_end>\.\d{3})*|0)(?!,)|(?P<dec_int_having_frac>(?P<dec_int>(?P<dec_int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<dec_int_end>\.\d{3})*,)|0,|,)(?=\d))(?P<frac_from_comma>(?<=,)(?P<frac>(?P<frac_start>\d{3}\.)*(?P<frac_end>\d{1,3})))?$
test
This one is for the same with the limit number>=50
^(?P<common_int>(?P<int>(?P<int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<int_end>\.\d{3})+|(?P<int_short>[1-9]\d{2}|[5-9]\d))(?!,)|(?P<dec_int_having_frac>(?P<dec_int>(?P<dec_int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<dec_int_end>\.\d{3})+,)|(?P<dec_short_int>[1-9]\d{2}|[5-9]\d),)(?=\d))(?P<frac_from_comma>(?<=,)(?P<frac>(?P<frac_start>\d{3}\.)*(?P<frac_end>\d{1,3})))?$
tests
If you always have the integer part under 999.999 and fractal part always 2 digits, it will be a bit more simple:
^(?P<dec_int_having_frac>(?P<dec_int>(?P<dec_int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<dec_int_end>\.\d{3})?,)|(?P<dec_short_int>[1-9]\d{2}|[5-9]\d),)(?=\d)(?P<frac_from_comma>(?<=,)(?P<frac>(?P<frac_end>\d{1,2})))?$
test
If you can guarantee that the number is correctly formed -- that is, that the regex isn't expected to detect that 5,0.1 is invalid, then there are a limited number of passing cases:
ends with \d{3}
ends with [5-9]\d
contains \d{3},
contains [5-9]\d,
It's not actually necessary to do anything with \.
The easiest regex is to code for each of these individually:
(\d{3}$|[5-9]\d$|\d{3},|[5-9]\d)
You could make it more compact and efficient by merging some of the cases:
(\d{3}[$,]|[5-9]\d[$,])
If you need to also validate the format, you will need extra complexity. I would advise against attempting to do both in a single regex.
However unless you have a very good reason for having to do this with a regex, I recommend against it. Parse the string into an integer, and compare it with 50.
So I'm trying to capture numbers with a maximum length of 9, which can be "obfuscated", such as 1.2.3.4.5.6.7.8.9, it has a length of 17 but if we "deofuscate" it, it's 9 numbers, so it's a phone number.
I've tried many regex but couldn't find the right one, because there are missing captures and in other cases it captures more than wanted...
(Well this is just a test)
https://regex101.com/r/HmZo1s/1
In this one, it captures correctly the numbers above but not the one below, and it also captures a very long number which must be omitted...
https://regex101.com/r/d9kv8F/1
I hope I've explained correctly what I want to achieve...
And, the number must start with 6, 7 or 9 to be a valid phone number, always 9 digits.
EDIT:
I want to use for phone numbers, but the title is correct, as you might use any kind of number... The "obfuscation pattern" is not always the same, it can be "anything" but of course we won't focus on that, we use "simple" obfuscation (like in the examples above), with dots, slashes and stuff like that.
I think, that the point of this is to "extract" the number, then see if it starts with 6, 7 or 9 and then see if it is 9 digits length, then it is a phone number so we remove it from our text (this is done with a basic replace).
For example
6.6.6.7.7.7.8.8.8 => 666777888 => remove
5.5.5.4.4.43.3.3 => 555444333 => not a phone
this is for a comment based system, in which we want to remove all kind of phones in there, so they won't share them.
To extract telephone numbers with your desired format, you could come up with:
\b[679][-\w. ]{7,15}\d
# \b is a word boundary
# [679] is one of 6, 7 or 9
# followed by dashes, word characters, dots and spaces (7-15 times)
# and a digit in the end
See a demo on regex101.com.
Please note that you will still get a lot of false positives (ie 6-------9 which apparently is not a number), so you will very likely have to adjust them programmatically afterwards.
what about this one : ([679][^\d]?)(\d[^\d]?){7}\d
Question - what is the shortest form of regex to add a leading zero into single digit in date record?
So I want to convert 8/8/2014 8:04:34 to 08/08/2014 8:04:34 - add leading zero when only one digit is presented.
The record can have two single digit entry, one single digit entry or no single digit entry. Some records can be in forms like 25/06/2014 19:50:18 or 9/06/2014 8:27:35 - in other words, some of them could be already normalized and regex needs to fix only single digit entry.
Not a regex user by any means. Your help is appreciated.
How about:
Ctrl+H
Find what: \b(\d)(?=/)
Replace with: 0$1
Replace all
This will change 8/8/2014 8:04:34 into 08/08/2014 8:04:34
Use the following regex to find:
(\d)(\d)?/(\d)(\d)?/(.*)
Then use the following to replace:
(?{2}\1\2:0\1)/(?{4}\3\4:0\3)/\5
What we are using is called conditionals in terms of regex. Refer this answer for explanation.
Make sure you have unselected the checkbox which says ". matches newline".
First of all, let's do some test-driven development and write the test cases. We can ignore the time and concentrate on the date alone. Also, the year is not important. We have to find all the possible cases for the day and the month. For each of them, we can have:
A single digit
Two digits, the first of which is already a 0
Two digits, the first of which is not a 0
Two digits, the second of which is a 0 (probably not needed, but just in case).
The case where we have to do something is only the first one, and the last 3 could be joined into a single one, but I prefer to keep them separated. We need to test 16 combinations:
8/8/2014
8/08/2014
8/12/2014
8/10/2014
08/8/2014
08/08/2014
08/12/2014
08/10/2014
12/8/2014
12/08/2014
12/12/2014
12/10/2014
10/8/2014
10/08/2014
10/12/2014
10/10/2014
Of all of these, only 1, 2, 3, 4, 5, 9, 13 must be changed. I don't know how to do it with a single regex, but with 2 regexes it's easy:
First regex, for the day:
(?<!\d)(\d/\d{1,2}/\d+)
replace with:
0\1
It matches a date where the day has only one digit, followed by a month with either 1 or 2 days, followed by a year with any number of digits, and it simply adds a 0 at the beginning.
Second regex, for the month:
(\d{2}/)(\d/\d+)
replace with:
\10\2
This one assumes that the first one has already been run, and thus the day has 2 digits. It finds dates where the month has a single digit, and adds a 0 before it. Please note that \10\2 means: the first group that matched, followed by a 0, followed by the second group. It doesn't mean: the tenth group, followed by the second. So the digits 1 and 0 are logically separated.
Run the first one, then the second one, and it gives the correct result:
08/08/2014
08/08/2014
08/12/2014
08/10/2014
08/08/2014
08/08/2014
08/12/2014
08/10/2014
12/08/2014
12/08/2014
12/12/2014
12/10/2014
10/08/2014
10/08/2014
10/12/2014
10/10/2014
Thanks to this recent answer I finally can give you an (hopefully) correct answer ;)
Replace
\b(?:(\d\d)|(\d))/(?:(\d\d)|(\d))/(\d\d)
with
(?{1}\1:0$2)/(?{3}\3:0\4)/\5
It uses Notepad++ conditionals (which I didn't know of until I stumbled over the mention question) to handle when only one or the other is single digit.
The regex matches a word boundary \b followed by two digits, captured in group 1, or one digit, captured in group 2, followed by a /. Then the same logic is repeated for day, which is captured in group 3 (2 digit) or 4 (1 digit). Then finally it checks that a year follows (at least two digits).
The conditional replace is explained in the linked answer. But simply put the (?{1} test if a match to group 1 was made it replaces with the expression before the :, otherwise the one after.
Hope this helps.
Regards
If you had a date like (ISO format)
2017-9-5
This
replace(/(\D)(\d)(?!\d)/g, '$10$2')
will turn it into
2017-09-05
and will preserve two digits in dates like
2017-11-11 or 2017-9-05
a general approach is to search for (in this case 5 digit numbers):
(\d)??(\d)??(\d)??(\d)??(\d)
Replace with
(?1\1:0)(?2\2:0)(?3\3:0)(?4\4:0)\5
You can use /^\d\/|(?<=\/)\d\/\d/g to select text, then add 0 before selected text, it should work for all your conditions.
Am playing around with regexp's but this is my headache. I have a dynamic number which needs a suffix. The suffix is always 0 to 9, 99 or 999.
Example:
I have the number 461200 and now I want to create an regexp that will match 461200 to 461209. What I've learned it should be ^46120[0-9]$? Is this correct or somewhere to the left of hell?
Ok, let us assume it is correct and I now want to match 461200 - 461299? This is where I get lost.
^4612[0-9]{2}?
It cannot be. I am yet to figure this out.
Any help appreciated.
For 1 digit at the end you need:
^4612[0-9]$
2 digits at the end:
^4612[0-9]{2}$
3 digits at the end:
^4612[0-9]{3}$
The number in braces {} means the number of time the preceding character or set has to be repeated.
Ok, let us assume it is correct and I now want to match 461200 -
461299?
You can either repeat the desired character class by saying [0-9][0-9] or use quantifiers [0-9]{2}.
It can be either:
^4612[0-9][0-9]$
or
^4612[0-9]{2}$
Both would work.
maybe try this regex:
^4612\d{2}$
I need a little help with Regex.
I want the regex to validate the following sentences:
fdsufgdsugfugh PCL 6
dfdagf PCL 11
fdsfds PCL6
fsfs PCL13
kl;klkPCL6
fdsgfdsPCL13
some chars, than PCL and than 6 or a greater number.
How this can be done?
I'd go with something like this:
^(.*)(PCL *)([6-9][0-9]*|[1-5][0-9]+)$
Meaning:
(.*) = some chars
(PCL *) = then PCL with optional whitespaces afterwards
([6-9][0-9]*|[1-5][0-9]+) then 6 or a greater number
This one should suit your needs:
^.*PCL\s*(?:[6-9]|\d{2,})$
Visualization by Debuggex
In bash:
EXPR=^[a-zA-Z]\+ *PCL *\([6-9]\|[0-9]\{2,\}\)
Translated:
Line begins with at least 1 occurence of a character (ignore caps)
Any amount of spaces, PCL, any amount of spaces
Either a number between 6 or 9, or a number with at least 2 digits
This expression used with something like grep "$EXPR" file.txt will output in stdout the lines that are valid.
This worked well for me. Reads logically too according to the way you described the matching
/[^PCL]+PCL\s?*[6-9]\d*/