E.g. match 353812345678 So far I have ^3538{1}[\d]{8} which works but does not restrict the length. How do I make sure the length is only a maximum of 12 digits?
If you want the number to be the only thing in the string: ^3538\d{8}$
If you just want the number in a string: \b3538\d{8}\b
^ is the start-of-string anchor, while $ is the end-of-string anchor, so the first one restricts the number to be the only thing on the line.
In the other one, \b means a word boundary, so it just means no other letters or digits may come directly before or after the number.
Also, note, in your original regex, the {1} is redundant, and [\d] means the same as \d.
^3538{1}[\d]{8}[^\d] will ensure you have 3538 followed by 8 digits and something that is NOT a digit -- thus limiting the length.
Add a dollar sign ($) at the end of the regex:
^3538{1}[\d]{8}$
Related
I am trying to write a regex to max a sequence of numbers that is 5 digits long or over, but I ignore any spaces, dashes, parens, or hashes when doing that analysis. Here's what I have so far.
(\d|\(|\)|\s|#|-){5,}
The problem with this is that this will match any sequence of 5 characters including those characters I want to ignore, so something like "#123 " would match. While I do want to ignore the # and space character, I still need the number itself to be 5 digits or more in order to qualify at a match.
To be clear, these would match:
1-2-3-4-5
123 45
2(134) 5
Bonus points if the matching begins and ends with a number rather than with one of those "special characters" I am excluding.
Any tips for doing this kind of matching?
If I understood requirements right you can use:
^\d(?:[()\s#-]*\d){4,}$
RegEx Demo
It always matches a digit at start. Then it is followed by 4 or more of a non-capturing group i.e. (?:[()\s#-]*\d) which means 0 or more of any listed special character followed by a digit.
So just repeat a digit, followed by any other sequence of allowed characters 5 or more times:
^(\d[()\s#-]*){5,}$
You can ensure it ends on a digit if you subtract one of the repetitions and add an explicit digit at the end:
^(\d[()\s#-]*){4,}\d$
You can suggest non-digits with \D so et would be something like:
(\d\D*){5,}
Here is a guide.
New to Regular Expressions. Thanks in advance!
Need to validate field is 1-10 mixed-case alphanumeric and spaces are allowed. First character must be alphanumeric, not space.
Good Examples:
"Larry King"
"L King1"
"1larryking"
"L"
Bad Example:
" LarryKing"
This is what I have and it does work as long as the data is exactly 10 characters. The problem is that it does not allow less than 10 characters.
[0-9a-zA-Z][0-9a-zA-Z ][0-9a-zA-Z ][0-9a-zA-Z ][0-9a-zA-Z ][0-9a-zA-Z ][0-9a-zA-Z ][0-9a-zA-Z ][0-9a-zA-Z ][0-9a-zA-Z ]
I've read and tried many different things but am just not getting it.
Thank you,
Justin
I don't know what environment you are using and what engine. So I assume PCRE (typically for PHP)
this small regex does exact what you want: ^(?i)(?!\s)[a-z\d ]{1,10}$
What's going on?!
the ^ marks the start of the string (delete it, if the expression must not match the whole string)
the (?i) tells the engine to be case insensitive, so there's no need to write all letter lower and upper case in the expression later
the (?!\s) ensures the following char won't be a white space (\s) (it's a so called negative lookahead)
the [a-z\d ]{1,10} matches any letter (a-z), any digit (\d) and spaces () in a row with min 1 and max 10 occurances ({1,10})
the $ at the end marks the end of the string (delete it, if the expression must not match the whole string)
Here's also a small visualization for better understanding.
Debuggex Demo
Try this: [0-9a-zA-Z][0-9a-zA-Z ]{0,9}
The {x,y} syntax means between x and y times inclusive. {x,} means at least x times.
You want something like this.
[a-zA-Z0-9][a-zA-Z0-9 ]{0,9}
This first part ensures that it is alphanumeric. The second part gets your alphanumeric with a space. the {0,9} allows from anywhere from 0 to 9 occurrences of the second part. This will give your 1-10
Try this: ^[(^\s)a-zA-Z0-9][a-z0-9A-Z ]*
Not a space and alphanumeric for the first character, and then zero or more alphanumeric characters. It won't cap at 10 characters but it will work for any set of 1-10 characters.
The below is probably most semantically correct:
(?=^[0-9a-zA-Z])(?=.*[0-9a-zA-Z]$)^[0-9a-zA-Z ]{1,10}$
It asserts that the first and last characters are alphanumeric and that the entire string is 1 to 10 characters in length (including spaces).
I assume that the space is not allowed at the end too.
^[a-zA-Z0-9](?:[a-zA-Z0-9 ]{0,8}[a-zA-Z0-9])?$
or with posix character classes:
^[[:alnum:]](?:[[:alnum:] ]{0,8}[[:alnum:]])?$
i think the simplest way is to go with \w[\s\w]{0,9}
Note that \w is for [A-Za-z0-9_] so replace it by [A-Za-z0-9] if you don't want _
Note that \s is for any white char so replace it by if you don't want the others
I need to find the text of all the one-digit number.
My code:
$string = 'text 4 78 text 558 my.name#gmail.com 5 text 78998 text';
$pattern = '/ [\d]{1} /';
(result: 4 and 5)
Everything works perfectly, just wanted to ask it is correct to use spaces?
Maybe there is some other way to distinguish one-digit number.
Thanks
First of all, [\d]{1} is equivalent to \d.
As for your question, it would be better to use a zero width assertion like a lookbehind/lookahead or word boundary (\b). Otherwise you will not match consecutive single digits because the leading space of the second digit will be matched as the trailing space of the first digit (and overlapping matches won't be found).
Here is how I would write this:
(?<!\S)\d(?!\S)
This means "match a digit only if there is not a non-whitespace character before it, and there is not a non-whitespace character after it".
I used the double negative like (?!\S) instead of (?=\s) so that you will also match single digits that are at the beginning or end of the string.
I prefer this over \b\d\b for your example because it looks like you really only want to match when the digit is surrounded by spaces, and \b\d\b would match the 4 and the 5 in a string like 192.168.4.5
To allow punctuation at the end, you could use the following:
(?<!\S)\d(?![^\s.,?!])
Add any additional punctuation characters that you want to allow after the digit to the character class (inside of the square brackets, but make sure it is after the ^).
Use word boundaries. Note that the range quantifier {1} (a single \d will only match one digit) and the character class [] is redundant because it only consists of one character.
\b\d\b
Search around word boundaries:
\b\d\b
As explained by the others, this will extract single digits meaning that some special characters might not be respected like "." in an ip address. To address that, see F.J and Mike Brant's answer(s).
It really depends on where the numbers can appear and whether you care if they are adjacent to other characters (like . at the end of a sentence). At the very least, I would use word boundaries so that you can get numbers at the beginning and end of the input string:
$pattern = '/\b\d\b/';
But you might consider punctuation at the end like:
$pattern = '/\b\d(\b|\.|\?|\!)/';
If one-digit numbers can be preceded or followed by characters other than digits (e.g., "a1 cat" or "Call agent 7, pronto!") use
(?<!\d)\d(?!\d)
Demo
The regular expression reads, match a digit (\d) that is neither preceded nor followed by digit, (?<!\d) being a negative lookbehind and (?!\d) being a negative lookahead.
I have a barcode of the format 123456########. That is, the first 6 digits are always the same followed by 8 digits.
How would I check that a variable matches that format?
You haven't specified a language, but regexp. syntax is relatively uniform across implementations, so something like the following should work: 123456\d{8}
\d Indicates numeric characters and is typically equivalent to the set [0-9].
{8} indicates repetition of the preceding character set precisely eight times.
Depending on how the input is coming in, you may want to anchor the regexp. thusly:
^123456\d{8}$
Where ^ matches the beginning of the line or string and $ matches the end. Alternatively, you may wish to use word boundaries, to ensure that your bar-code strings are properly separated:
\b123456\d{8}\b
Where \b matches the empty string but only at the edges of a word (normally defined as a sequence consisting exclusively of alphanumeric characters plus the underscore, but this can be locale-dependent).
123456\d{8}
123456 # Literals
\d # Match a digit
{8} # 8 times
You can change the {8} to any number of digits depending on how many are after your static ones.
Regexr will let you try out the regex.
123456\d{8}
should do it. This breaks down to:
123456 - the fixed bit, obviously substitute this for what you're fixed bit is, remember to escape and regex special characters in here, although with just numbers you should be fine
\d - a digit
{8} - the number of times the previous element must be repeated, 8 in this case.
the {8} can take 2 digits if you have a minimum or maximum number in the range so you could do {6,8} if the previous element had to be repeated between 6 and 8 times.
The way you describe it, it's just
^123456[0-9]{8}$
...where you'd replace 123456 with your 6 known digits. I'm using [0-9] instead of \d because I don't know what flavor of regex you're using, and \d allows non-Arabic numerals in some flavors (if that concerns you).
I am trying to create a regex to have a string only contain 0-9 as the characters and it must be at least 1 char in length and no more than 45. so example would be 00303039 would be a match, and 039330a29 would not.
So far this is what I have but I am not sure that it is correct
[0-9]{1,45}
I have also tried
^[0-9]{45}*$
but that does not seem to work either. I am not very familiar with regex so any help would be great. Thanks!
You are almost there, all you need is start anchor (^) and end anchor ($):
^[0-9]{1,45}$
\d is short for the character class [0-9]. You can use that as:
^\d{1,45}$
The anchors force the pattern to match entire input, not just a part of it.
Your regex [0-9]{1,45} looks for 1 to 45 digits, so string like foo1 also get matched as it contains 1.
^[0-9]{1,45} looks for 1 to 45 digits but these digits must be at the beginning of the input. It matches 123 but also 123foo
[0-9]{1,45}$ looks for 1 to 45 digits but these digits must be at the end of the input. It matches 123 but also foo123
^[0-9]{1,45}$ looks for 1 to 45 digits but these digits must be both at the start and at the end of the input, effectively it should be entire input.
The first matches any number of digits within your string (allows other characters too, i.e.: "039330a29"). The second allows only 45 digits (and not less). So just take the better from both:
^\d{1,45}$
where \d is the same like [0-9].
Use this regular expression if you don't want to start with zero:
^[1-9]([0-9]{1,45}$)
If you don't mind starting with zero, use:
^[0-9]{1,45}$
codaddict has provided the right answer. As for what you've tried, I'll explain why they don't make the cut:
[0-9]{1,45} is almost there, however it matches a 1-to-45-digit string even if it occurs within another longer string containing other characters. Hence you need ^ and $ to restrict it to an exact match.
^[0-9]{45}*$ matches an exactly-45-digit string, repeated 0 or any number of times (*). That means the length of the string can only be 0 or a multiple of 45 (90, 135, 180...).
A combination of both attempts is probably what you need:
^[0-9]{1,45}$
^[0-9]{1,45}$ is correct.
Rails doesnt like the using of ^ and $ for some security reasons , probably its better to use \A and \z to set the beginning and the end of the string
For this case word boundary (\b) can also be used instead of start anchor (^) and end anchor ($):
\b\d{1,45}\b
\b is a position between \w and \W (non-word char), or at the beginning or end of a string.