I'm attempting to match the first 3 letters that could be a-z followed by a specific character.
For testing I'm using a regex online tester.
I thought this should work (without success):
^[a-z]{0,3}$[z]
My test string is abcz.
Hope you can tell me what I'm doing wrong.
If you need to match a whole string abcz, use
/^[a-z]{0,3}z$/
^^
or - if the 3 letters are compulsory:
/^[a-z]{3}z$/
See the regex demo.
The $[z] in your pattern attempts to match a z after the end of string anchor, which makes the regex fail always.
Details:
^ - string start
[a-z]{0,3} - 0 to 3 lowercase ASCII letters (to require 3 letters, remove 0,)
z - a z
$ - end of string anchor.
You've got the end of line identifier too early
/^[a-z]{0,3}[z]$/m
You can see a working version here
You can do away with the [] around z. Square brackets are used to define a range or list of characters to match - as you're matching only one they're not needed here.
/^[a-z]{0,3}z$/m
Related
I'm trying this regular expressión, but I can't validate correctly the end white space and the letter:
/^\d{0,2}(\-\d{0,2})?(\-\d{0,2})?(\ ?\d[W,E]?)?$/
Examples of correct values:
33-39-10 N //OK
85-50 W //OK
-85-50 E //Wrong
What's wrong?
\d{0,2} this quantifier also matches a digit zero times so that would match the leading - in the 3rd example.
In the character class [W,E] you could omit the comma and list the characters you allow to match [ENW]
If only the third group is optional you could try including the whitespace before the end of the line $
^\d{2}(-\d{2})(-\d{2})? [ENW] $
I have used this regular expression : ^(?!\-)\d{0,2}?(\-\d{0,2}).+\s(N|E|W|S)$
Using a negative lookahead, we have excluded anything that starts with a dash (-).
(?!\-) = Starting at the current position in the expression,
ensures that the given pattern will not match
\s(N|E|W|S) matches anything with a space (\s) and one of the letters using OR operator |.
You may also use \s+(N|E|W|S).
+ = Matches between one and unlimited times, as many times as
possible, giving back as needed
I have a filename that will be something along the lines of this:
Annual-GDS-Valuation-30th-Dec-2016-082564K.docx
It will contain 5 numbers followed by a single letter, but it may be in a different position in the file name. The leading zero may or may not be there, but it is not required.
This is the code I come up with after checking examples, however SelectedFileClientID is always null
var SelectedFileClientID = files.match(/^d{5}\[a-zA-Z]{1}$/);
I'm not sure what is it I am doing wrong.
Edit:
The 0 has nothing to do with the code I am trying to extract. It may or may not be there, and it could even be a completely different character, or more than one, but has nothing to do with it at all. The client has decided they want to put additional characters there.
There are at least 3 issues with your regex: 1) the pattern is enclosed with anchors, and thus requires a full string match, 2) the d matches a letter d, not a digit, you need \d to match a digit, 3) a \[ matches a literal [, so the character class is ruined.
Use
/\d{5}[a-zA-Z]/
Details:
\d{5} - 5 digits
[a-zA-Z] - an ASCII letter
JS demo:
var s = 'Annual-GDS-Valuation-30th-Dec-2016-082564K.docx';
var m = s.match(/\d{5}[a-zA-Z]/);
console.log(m[0]);
All right, there are a few things wrong...
var matches = files.match(/\-0?(\d{5}[a-zA-Z])\.[a-z]{3,}$/);
var SelectedFileClientID = matches ? matches[1] : '';
So:
First, I get the matches on your string -- .match()
Then, your file name will not start with the digits - so drop the ^
You had forgotten the backslash for digits: \d
Do not backslash your square bracket - it's here used as a regular expression token
no need for the {1} for your letters: the square bracket content is enough as it will match one, and only one letter.
Hope this helps!
Try this pattern , \d{5}[a-zA-Z]
Try - 0?\d{5}[azA-Z]
As you mentioned 0 may or may not be there. so 0? will take that into account.
Alternatively it can be done like this. which can match any random character.
(\w+|\W+|\d+)?\d{5}[azA-Z]
I have this regex
/^[a-z]{1,}( (?=[a-z])){0,}(_(?=[a-z])){0,}[a-z]{0,}$/
I want to match
ag_b_cf_ajk
or
zva b c de
or
hh_b opxop a_b
so any character tokens separated by a single space or underscore.
(In the regex above, we have a literal space, which is legal, and we have look-aheads that ensure that a space or underscore is followed by a character).
The problem is, my above regex is only matching the first space or underscore, like so:
axz_be
axz be
but these fail
axz_be_j
axz be j
I believe I missing some concept with regexes in order to solve this as I have been trying for the last few hours!
It seems you can just use
^[a-z]+(?:[_ ][a-z]+)*$
See the regex demo
The regex matches
^ - start of string
[a-z]+ - one or more lowercase ASCII letters
(?:[_ ][a-z]+)* - zero or more sequences of:
[_ ] - a space or an underscore
[a-z]+ - one or more lowercase ASCII letters
$ - end of string
If the space or underscore must appear at least once, use the + quantifier instead of *:
^[a-z]+(?:[_ ][a-z]+)+$
^
To add a multicharacter alternative to the underscore and hyphen, you need to introduce another non-capturing group:
^[a-z]+(?:(?:[_ ]|\[])[a-z]+)+$
See another regex demo
I am trying to use regexp to match some specific key words.
For those codes as below, I'd like to only match those IFs at first and second line, which have no prefix and postfix. The regexp I am using now is \b(IF|ELSE)\b, and it will give me all the IFs back.
IF A > B THEN STOP
IF B < C THEN STOP
LOL.IF
IF.LOL
IF.ELSE
Thanks for any help in advance.
And I am using http://regexr.com/ for test.
Need to work with JS.
I'm guessing this is what you're looking for, assuming you've added the m flag for multiline:
(?:^|\s)(IF|ELSE)(?:$|\s)
It's comprised of three groups:
(?:^|\s) - Matches either the beginning of the line, or a single space character
(IF|ELSE) - Matches one of your keywords
(?:$|\s) - Matches either the end of the line, or a single space character.
Regexr
you can do it with lookaround (lookahead + lookbehind). this is what you really want as it explicitly matches what you are searching. you don't want to check for other characters like string start or whitespaces around the match but exactly match "IF or ELSE not surrounded by dots"
/(?<!\.)(IF|ELSE)(?!\.)/g
explanation:
use the g-flag to find all occurrences
(?<!X)Y is a negative lookbehind which matches a Y not preceeded by an X
Y(?!X) is a negative lookahead which matches a Y not followed by an X
working example: https://regex101.com/r/oS2dZ6/1
PS: if you don't have to write regex for JS better use a tool which supports the posix standard like regex101.com
I have a url like:
image/media-group/rugby-league-programme-covers-3436?sort=title
or
image/media-group/rugby-league-programme-covers-3436
I need to get everything after media-group and not including ? or anything after.
So in both instances rugby-league-programme-covers-3436 is what I need to return
I used the regular expression /media-group/(.*)\? which works for the instance where there is a query string but not in the instance where there is no query string.
I am using the below code
var patt=new RegExp('/media-group/(.*)\?');
return patt.exec(url)[1];
Your help on this would be most appreciated
I believe the best pattern would be:
/^[^\#\?]+\/media-group\/([^\?]+).*$/
which breaks out as:
^ - start of string
[^\#\?]+ - one or more non-hash, non-question-marks
\/ - literal char
media-group - literal chars
\/ - literal char
( - start capture group
[^\?]+ - one or more chars non-question-marks
) - end of capture group
.* - zero or more chars
$ - end of string
The reason this works is because [^\?]+ is "greedy" in that it will attempt the longest possible match, which encompasses either a question-mark followed by arbitrary chars, or nothing, since all chars to the end of the string have already been captured in the non-question-mark capture group.
So, using
var RE=new RegExp(/^[^\#\?]+\/media-group\/([^\?]+).*$/),
url="image/media-group/rugby-league-programme-covers-3436?sort=title";
console.log(url.match(RE)[1])
prints: rugby-league-programme-covers-3436 and changing url to image/media-group/rugby-league-programme-covers-3436, produces the same result.
Update
Modified the pattern re David Foerster's comment.