I have this regex pattern:
^[.]{5,}$
Which I want to return true if the tested string has 5 or more characters.
I.E it'll only return false if the string contains 4 or less characters.
At the moment it seems to return true regardless of the number of characters and I can't see why.
You want
^.{5,}$
But really - just use the built-in string length function of the language of your choice
Try this regex:
.{5,}
more chars to make up the minimum post...
I think you dont need the ^ and $. Try just:
.{5,}
Related
This works as expected:
([^\u0000-\u007F])+-हा([^\u0000-\u007F])+
Returns:
ब-हाणपूर
ब-हाणी
बनियन-हाफ
But I am looking for 1 character followed by dash. The expected output is:
ब-हाणपूर
ब-हाणी
I tried to replace + sign with character count like this...
([^\u0000-\u007F]){1}-हा([^\u0000-\u007F])+
But it returned the same 3 results. How do I return the first 2?
You need anchors:
^([^\u0000-\u007F])-हा([^\u0000-\u007F])+$
Demo
You asked 'What if I need 5 characters to the left of dash?'
The regex portion [^\u0000-\u007F] as written matches a single character that meets that criterion. If you want more or less than one, use a regex quantifier to describe how many you want.
In this case, if you want 5, you would use:
^([^\u0000-\u007F]{5})-हा([^\u0000-\u007F])+$
Probably like this:
^([^\u0000-\u007F]){1}-हा([^\u0000-\u007F])+
^([^\u0000-\u007F]{1})-हा([^\u0000-\u007F]+)
(\b[^\u0000-\u007F]{1})-हा([^\u0000-\u007F]+)
Regex demo
I want to validate a text that need have more than 3 [aA-zZ] chars, not need continous.
/^(?![_\-\s0-9])(?!.*?[_\-\s]$)(?=.*[aA-zZ]{3,})[_\-\sa-zA-Z0-9]+$/.test("aaa123") => return true;
/^(?![_\-\s0-9])(?!.*?[_\-\s]$)(?=.*[aA-zZ]{3,})[_\-\sa-zA-Z0-9]+$/.test("a1b2c3") => return false;
Can anybody help me?
How about replacing and counting?
var hasFourPlusChars = function(str) {
return str.replace(/[^a-zA-Z]+/g, '').length > 3;
};
console.log(hasFourPlusChars('testing1234'));
console.log(hasFourPlusChars('a1b2c3d4e5'));
You need to group .* and [a-zA-Z] in order to allow optional arbitrary characters between English letters:
^(?![_\-\s0-9])(?!.*?[_\-\s]$)(?=(?:.*[a-zA-Z]){3,})[_\-\sa-zA-Z0-9]+$
^^^ ^
Add this
Demo:
var re = /^(?![_\-\s0-9])(?!.*?[_\-\s]$)(?=(?:.*[aA-zZ]){3,})[_\-\sa-zA-Z0-9]+$/;
console.log(re.test("aaa123"));
console.log(re.test("a1b2c3"));
By the way, [aA-zZ] is not a correct range definition. Use [a-zA-Z] instead. See here for more details.
Correction of the regex
Your repeat condition should include the ".*". I did not check if your regex is correct for what you want to achieve, but this correction works for the following strings:
$testStrings=["aaa123","a1b2c3","a1b23d"];
foreach($testStrings as $s)
var_dump(preg_match('/^(?![_\-\s0-9])(?!.*?[_\-\s]$)(?=.*[a-zA-Z]){3,}[_\-\sa-zA-Z0-9]+$/', $s));
Other implementations
As the language seems to be JavaScript, here is an optimised implementation for what you want to achieve:
"a24be4Z".match(/[a-zA-Z]/g).length>=3
We get the list of all matches and check if there are at least 3.
That is not the "fastest" way as the result needs to be created.
)
/(?:.*?[a-zA-Z]){3}/.test("a24be4Z")
is faster. ".*?" avoids that the "test" method matches all characters up to the end of the string before testing other combinations.
As expected, the first suggestion (counting the number of matches) is the slowest.
Check https://jsperf.com/check-if-there-are-3-ascii-characters .
I was told to validate the string like this -,-,-,9,-
It was separated by , and contains 1 number(0-9), others are all -
some examples:
9,-,-,-,-
-,-,-,-,9
-,-,2,-,-
How can I match this? And what concepts should I learn in regex?
Update
I miss the times, sorry, this string can only contains 5 part,so the length can be only 9,it means a string like below should not be passed:
-,-,9,-,-,-
and of course, it should have only one number.
^(?=\D*\d\D*$)[0-9-](?:,[0-9-]){4}$
You can try this.See demo.
https://regex101.com/r/nM7nT5/5
This ensures that the string must hav atleast one comma and exactly one digit.
^(?:\d(?:,-)+|-(?:,-)*,\d(?:,-)*)$
DEMO
OR
^(?=\D*(?:^|,)\d(?:,|$)\D*$)[\d-](?:,[\d-])+$
DEMO
I am using Powershell 2.0. I have file names like my_file_name_01012013_111546.xls. I am trying to get my_file_name.xls. I have tried:
.*(?=_.{8}_.{6})
which returns my_file_name. However, when I try
.*(?=_.{8}_.{6}).{3}
it returns my_file_name_01.
I can't figure out how to get the extension (which can be any 3 characters. The time/date part will always be _ 8 characters _ 6 characters.
I've looked at a ton of examples and tried a bunch of things, but no luck.
If you just want to find the name and extension, you probably want something like this: ^(.*)_[0-9]{8}_[0-9]{6}(\..{3})$
my_file_name will be in backreference 1 and .xls in backreference 2.
If you want to remove everything else and return the answer, you want to substitute the "numbers" with nothing: 'my_file_name_01012013_111546.xls' -replace '_[0-9]{8}_[0-9]{6}' ''. You can't simply pull two bits (name and extension) of the string out as one match - regex patterns match contiguous chunks only.
try this ( not tested), but it should works for any 'my_file_name' lenght , any lenght of digit and any kind of extension.
"my_file_name_01012013_111546.xls" -replace '(?<=[\D_]*)(_[\d_]*)(\..*)','$2'
non regex solution:
$a = "my_file_name_01012013_111546.xls"
$a.replace( ($a.substring( ($a.LastIndexOf('.') - 16 ) , 16 )),"")
The original regex you specified returns the maximum match that has 14 characters after it (you can change to (?=.{14}) who is the same).
Once you've changed it, it returns the maximum match that has 14 characters after it + the next 3 characters. This is why you're getting this result.
The approach described by Inductiveload is probably better in case you can use backreferences. I'd use the following regex: (.*)[_\d]{16}\.(.*) Otherwise, I'd do it in two separate stages
get the initial part
get the extension
The reason you get my_filename_01 when you add that is because lookaheads are zero-width. This means that they do not consume characters in the string.
As you stated, .*(?=_.{8}_.{6}) matches my_file_name because that string is is followed by something matching _.{8}_.{6}, however once that match is found, you've only consumed my_file_name, so the addition of .{3} will then consume the next 3 characters, namely _01.
As for a regex that would fit your needs, others have posted viable alternatives.
I'd like know if there are a way to test if a word have a letter repeated in any position?
I'm currently using this regex to test it, but not work, becouse if I add more then 2 's' the test returns true.
/s{0,2}/.test('süuaãpérbrôséê'); //expected true
/s{0,2}/.test('ssüuaãpérbrôéê'); //expected true
/s{0,2}/.test('süuaãpérbrôéê'); //expected true
/s{0,2}/.test('süuaãpérbrôséês'); //expected fail
Thanks.
/s{2,}/
or generally for any character:
/(.)\1/
/(\w)\1/ finds two alphanumeric characters next to each other
This will find and replace the duplicates:
s/(\w)\1/$1/
The only way that I found to resolve this problem is using php preg_match_all, on this way I can count how much times the character repeat.
$s = 'süuaãpérbrôséê';
preg_match_all('/s/i', $s, $m);
echo count($m[0]); //outputs 2
My initial idea was pass a regex and use preg_match to verify the match exists in a determined number of times, but I think that it's not possible, so I'll create a method that receive the word and the regex that I need match and it will return the number of matches.
Thanks.
using lookahead you can achieve something like that:
^(?=.*(\w)(.*\1){1}.*$)((?!\1).)*\1(((?!\1).)*\1){1}((?!\1).)*$
Where {1} is number of repeatings minus 1, so for finding if there are three repeations this would look like:
^(?=.*(\w)(.*\1){2}.*$)((?!\1).)*\1(((?!\1).)*\1){2}((?!\1).)*$
And for two or three:
^(?=.*(\w)(.*\1){1,2}.*$)((?!\1).)*\1(((?!\1).)*\1){1,2}((?!\1).)*$
etc.
The lookaheads with backreferences can be very powerful :)