I would very much appreciate a bit of help with the following regex riddle.
I need regex statement that would validate against the following rules:
The input can contain letters, special characters and digits.
The input can't start with "0",
The input Can have up to 7 digits
Examples of valid input:
aa1234aa2.(less than 7 digits)
asd234566 (less than 7 digits)
Examples of invalid input:
0asdfd92 (starts with 0)
asd12312311 (more than 7 digits)
What I have tried so far:
^\D[0-9]{0,7}$,
validates against d0000000, but the input may be d0d0dddd1234d
The part can't start with 0 can be removed from the requirement if it complicates a lot. The most important is to have "Can have up to 7 digits" part.
Regards,
Oleg
This is what you need!
Attempt 1: ^[1-9]\d{0,6}$
Attempt 2: ^[^0][\d\w]{0,6}$
Attempt 3: ^[^0].{0,6}$
Attempt 4: ^([\D]*\d){0,7}[\D]*$
Attempt 5: ^([\D]*[1-9]){0,7}[\D]*$|^[^0]\d{0,6}$
Attempt 6: ^([\D]*[1-9]){1,7}[\D]*$|^[^0]\d{1,6}$ <- this should work
Example here
If I understand the requirements correctly, this will work:
^(?=[^0])(\D*\d){0,7}\D*$
That will allow any string that does not start with a zero and has 7 or fewer digits. Any other characters are allowed in any quantity.
Explanation
The first part (?=[^0]) is an assertion that checks to make sure the string does not start with zero. The rest matches any number of non-digits followed by a digit, up to 7 times. Then any number of non-digits before the end of the string.
Assuming Perl (it looks like Perl regular expressions):
Check for leading zero: if (subst($pass, 0, 1) eq '0') { fail }
Check for no more than seven digits: if (($pass =~ tr /0-9/0-9/) > 7) { fail }
I'm generally against trying to cram everything into a single regular expression, especially when there are other tools available to do the job. In this case, the tr will not be executed if there is a leading zero, and a leading zero is easy to spot in the beginning of a string.
Doing it this way, it's easy to add further restrictions independently of the others. For example, "there may be more than 7 digits if they are all separated by other types of characters" (a regex for this one, probably).
You can use this regex:
^[^0](?:\D*\d){1,7}\D*$
RegEx Demo
This will perform following validations:
Must start with non-zero
Has 1 to 7 digits after first char
Verbose, but does the trick.
(^[1-9][^\d]*([\d]?[^\d]*){0,6}$|^[^\d]+([\d]?[^\d]*){0,7}$)
I found it easier to split the RegEx into two cases: when the string starts with a digit, and when it doesn't.
^((?:\D+(?:\d?\D*){0,7})|(?:[1-9]\D*(?:\d?\D*){0,6}))$
You can test it here
Related
I need a regular expression to validate a concatenated string that consists of 7 digit numbers separated by commas.
Furthermore, I must ensure that:
The string is not empty.
The chain doesn't begins or finish with commas.
The numbers do not start with 0.
Example: 1234567,2345678,3456789
My solution so far: ^\d+(,\d+)*?$
The problems I still need to resolve:
Validate that the numbers are exactly 7 digits.
Validate that the numbers do not start with 0.
Thank you.
Something like ^[1-9]\d{6}(,[1-9]\d{6})+$ should work. The [1-9] ensures the number doesn't begin with 0, and \d{6} ensures that there are 6 digits to follow.
Based on Gavin answer, here is what worked for me : ^[1-9]\d{6}(,[1-9]\d{6})*$
The minor difference is the use of the * instead of + at the end of the regular expression. There are some cases where I must validate only one 7 digits number...
Thank you for the help everyone!
I am trying to write a regex that will match Roman numerals from 0 to 39 only. There are plenty of examples which match much larger Roman numerals, but I cannot figure out how to match this specific subset.
Got it. Try this:
/^(X{1,3})(I[XV]|V?I{0,3})$|^(I[XV]|V?I{1,3})$|^V$/
Update:
Zero doesn't exist in Roman numerals. Therefore feel free to tack on your own implementation for zero.
I'm not sure how to represent 0 using Roman numerals. I assume that it has separate token N (see Wikipedia).
Assuming the regex tries to match the whole string (like in Java) and you have lookahead, you can use this regex:
(?.)(X{0,3}(IX|IV|V?I{0,3})|N)
Explanation:
(?.): ensure at least one character
X{0,3}: define the tens (0, 10, 20, 30)
(...): define the final digit
IX: 9
IV: 4
V?I{0,3}: 0-3, 5-8 (0 not as whole number, require at least one X)
N: 0 (as whole number)
If you represent 0 as empty string, the regex is simpler:
X{0,3}(IX|IV|V?I{0,3})
since the lookahead and N in the previous regex is just to prevent empty string.
Assuming you know you have valid Roman numerals and want to fetch only the ones <= 39, that is easy:
^[XVI]*$
See it in action
If that is not the case, it's a little bit trickier, but you can still take advantage of the fact that all the numbers that can be represented only with X, V and I are 1..39:
^X{0,3}(?:V?I{0,3}|I[VX])$
See it in action
X{0,3} covers 10, 20, 30
X{0,3}V?I{0,3} covers all but the ones that end with 4 or 9 (14, 29, etc)
X{0,3}I[VX] exactly the ones ending with 4 or 9
Note: these will also match an empty string, which is my interpretation of a Roman zero. If that is not the case, you can replace the * with + for the first regex and add a positive lookahead at the start of the regex for the second ((?=.)).
Note 2: If they are not on separate lines (or in separate strings), you can replace ^ and $ with word boundaries (\b).
Hi I am working on RegEx. Correct response should NOT allow for number to the tenths only, as in RESPONSE = "925.0", nor should it allow for trailing zeros after the hundredths place as in RESPONSE = "925.000". Only correct responses: 925, 0925, 0925., 925., 925.00, 00925
I worked on it and finally came up with this
"^-?(0)*(\d*(\.(00))?\d+.|(\d){1,3}(,(\d){3})*(\.(00))?)$"
It works for three digit numbers but if i want it for 38400.00 it doesn't allow it
I am not quite certain whether the decimal places can be any digit or if they have to be zero. If the former, then this should do the trick:
^-?\d{1,3}(,?\d{3})*(\.(\d{2})?)?$
If the latter, then this:
^-?\d{1,3}(,?\d{3})*(\.(00)?)?$
The entire match starting with the decimal point is optional, and the two decimal places in that match are optional as well.
UPDATE I just realized that it appears you need to accept commas in the response as well - I assume for thousands, millions, etc.
UPDATE #2 per OP's comment
^-?(\d+|\d{1,3}(,\d{3})*)(\.(00)?)?$
UPDATE #3 Added link to regex101 for explanation of this regular expression.
Have a try with:
^-?\d{1,3}(?:,?\d{3})*(?:\.(?:00)?)?$
I think your problem is that you're trying to match it in chunks of three, with commas separating, but 38400.00 doesn't have commas.
Try this:
^-?\d+(\.?(\d{2})?)$
The - indicates the character, -. With the ? after, it says that it may or may not apply. This allows negative numbers, so if you only want positive numbers matched, delete the first two characters.
\d represents every digit. The + after says that there can be as many as you want, as long as there's at least one.
Then there's a \., which is just a dot in the number. The ? does the same as before.. Since you seem to allow trailing periods, I assumed you wanted it to be considered separately from the following digits.
The () encloses the next group, which is the period (\.) followed by two characters that match \d -- two digits -- and which may be repeated 0 or 1 times, as dictated by the ?. This allows people to either have no digits after the period or two, but nothing else.
The ^ at the beginning specifies it has to be at the beginning of the line, and the $ at the end specifies it has to end at the end of the line. Remember to enable the multiline (m) flag so it works properly.
Disclaimer: I've not done much regex work before, so I could well be totally off. If it doesn't work, let me know.
Couldn't you do this without the ?'s
^[0-9,]+(\.){0,1}(\d{2}){0,1}$
improved: ^\d+[0-9,]*(\.){0,1}(\d{2}){0,1}$
Edit:
Broken down a bit as requested
Old one:
[0-9,]+
1 or more digits/commas (would have accepted ',' as true) so improved version:
\d+
for starts with 1 or more digits
[0-9,]*
0 or more digits/commas
followed by
(\.){0,1}
0 or 1 decimal
Followed by
(\d{2}){0,1}
0 or 1 of (exactly 2 digits)
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 "$EXPR" file.txt will output in stdout the lines that are valid.
This worked well for me. Reads logically too according to the way you described the matching
/[^PCL]+PCL\s?*[6-9]\d*/
I have this regex to allow for only alphanumeric characters.
How can I check that the string at least contains 3 alphabet characters as well.
My current regex,
if(!/^[a-zA-Z0-9]+$/.test(val))
I want to enforce the string to make sure there is at least 3 consecutive alphabet characters as well so;
111 // false
aaa1 // true
11a // false
bbc // true
1a1aa // false
+ means "1 or more occurrences."
{3} means "3 occurrences."
{3,} means "3 or more occurrences."
+ can also be written as {1,}.
* can also be written as {0,}.
To enforce three alphabet characters anywhere,
/(.*[a-z]){3}/i
should be sufficient.
Edit. Ah, you'ved edited your question to say the three alphabet characters must be consecutive. I also see that you may want to enforce that all characters should match one of your "accepted" characters. Then, a lookahead may be the cleanest solution:
/^(?.*[a-z]{3})[a-z0-9]+$/i
Note that I am using the case-insensitive modifier /i in order to avoid having to write a-zA-Z.
Alternative. You can read more about lookaround assertions here. But it may be a little bit over your head at this stage. Here's an alternative that you may find easier to break down in terms of what you already know:
/^([a-z0-9]*[a-z]){3}[a-z0-9]*$/i
This should do the work:
^([0-9]*[a-zA-Z]){3,}[0-9]*$
It checks for at least 3 "Zero-or-more numerics + 1 Alpha" sequences + Zero-or-more numerics.
You want to match zero or more digits then 3 consecutive letters then any other number of digits?
/\d*(?:[a-zA-Z]){3,}\d*/
This is vanilla JS you guys can use. My problem is solved using this.
const str = "abcdggfhf";
const pattern = "fhf";
if(pattern.length>2) {
console.log(str.search(pattern));
}