I am still a beginner :)
I need to get a substring ignoring the last section inside [] (including the brackets []), i.e. ignore the [something inside] section in the end.
Note - There could be other single occurances of [ in the string. And they should appear in the result.
Example
Input of the form -
1 checked arranged [1678]
Desired output -
1 checked arranged
I tried with this
var item = "1 checked arranged [1678]";
var parsed = item.match(/([a-zA-Z0-9\s]+)([(\[d+\])]+)$/);
|<-section 1 ->|<-section 2->|
alert(parsed);
I tried to mean the following -
section 1 - multiple occurrences of words (containing literals and nos.) followed by spaces
section 2 - ignore the pattern [something] in the end.
But I am getting 1678],1678,] and I am not sure which way it is going.
Thanks
OK here is the problem in your expression
([a-zA-Z0-9\s]+)([(\[d+\])]+)$
The Problem is only in the last part
([(\[d+\])]+)$
^ ^
here are you creating a character class,
what you don't want because everything inside will be matched literally.
((\[d+\])+)$
^ ^^
here you create a capturing group and repeat this at least once ==> not needed
(\[d+\])$
^
here you want to match digits but forgot to escape
That brings us to
([a-zA-Z0-9\s]+)(\[\d+\])$
See it here on Regexr, the complete string is matched, the section 1 in capturing group 1 and section 2 in group 2.
When you now replace the whole thing with the content of group 1 you are done.
You could do this
var s = "1 checked arranged [1678]";
var a = s.indexOf('[');
var b = s.substring(0,a);
alert(b);
http://jsfiddle.net/jasongennaro/ZQe6Y/1/
This s.indexOf('['); checks for where the first [ appears in the string.
This s.substring(0,a); chops the string, from the beginning to the first [.
Of course, this assumes the string is always in a similar format
var item = '1 check arranged [1678]',
matches = item.match(/(.*)(?=\[\d+\])/));
alert(matches[1]);
The regular expression I used makes use of a positive lookahead to exclude the undesired portion of the string. The bracketed number must be a part of the string for the match to succeed, but it will not be returned in the results.
Here you can find how to delete stuff inside square brackets. This will leave you with the rest. :)
Regex: delete contents of square brackets
try this if you only want to get rid of that [] in the end
var parsed = item.replace(/\s*\[[^\]]*\]$/,"")
var item = "1 checked arranged [1678]";
var parsed = item.replace(/\s\[.*/,"");
alert(parsed);
That work as desired?
Use escaped brackets and non-capturing parentheses:
var item = "1 checked arranged [1678]";
var parsed = item.match(/([\w\s]+)(?:\s+\[\d+\])$/);
alert(parsed[1]); //"1 checked arranged"
Explanation of regex:
([\w\s]+) //Match alphanumeric characters and spaces
(?: //Start of non-capturing parentheses
\s* //Match leading whitespace if present, and remove it
\[ //Bracket literal
\d+ //One or more digits
\] //Bracket literal
) //End of non-capturing parentheses
$ //End of string
Related
I have string [FBWS-1] comes first than [FBWS-2]
In this string, I want to find all occurance of [FBWS-NUMBER]
I tried this :
var term = "[FBWS-1] comes first than [FBWS-2]";
alert(/^([[A-Z]-[0-9]])$/.test(term));
I want to get all the NUMBERS where [FBWS-NUMBER] string is matched.
But no success. I m new to regular expressions.
Can anyone help me please.
Note that ^([[A-Z]-[0-9]])$ matches start of a string (^), a [ or an uppercase ASCII letter (with [[A-Z]), -, an ASCII digit and a ] char at the end of the string. So,basically, strings like [-2] or Z-3].
You may use
/\[[A-Z]+-[0-9]+]/g
See the regex demo.
NOTE If you need to "hardcode" FBWS (to only match values like FBWS-123 and not ABC-3456), use it instead of [A-Z]+ in the pattern, /\[FBWS-[0-9]+]/g.
Details
\[ - a [ char
[A-Z]+ - one or more (due to + quantifier) uppercase ASCII letters
- - a hyphen
[0-9]+ - one or more (due to + quantifier) ASCII digits
] - a ] char.
The /g modifier used with String#match() returns all found matches.
JS demo:
var term = "[FBWS-1] comes first than [FBWS-2]";
console.log(term.match(/\[[A-Z]+-[0-9]+]/g));
You can use:
[\w+-\d]
var term = "[FBWS-1] comes first than [FBWS-2]";
alert(/[\w+-\d]/.test(term));
There are several reasons why your existing regex doesn't work.
You trying to match the beginning and ending of your string when you
actually want everything in between, don't use ^$
Your only trying to match one alpha character [A-Z] you need to make this greedy using the +
You can shorten [A-Z] and [0-9] by using the shorthands \w and \d. The brackets are generally unnecessary.
Note your code only returns a true false value (your using test) ATM it's unclear if this is what you want. You may want to use match with a global modifier (//g) instead of test to get a collection.
Here is an example using string.match(reg) to get all matches strings:
var term = "[FBWS-1] comes first than [FBWS-2]";
var reg1 = /\[[A-Z]+-[0-9]\]/g;
var reg2 = /\[FBWS-[0-9]\]/g;
var arr1 = term.match(reg1);
var arr2 = term.match(reg2)
console.log(arr1);
console.log(arr2);
Your regular expression /^([[A-Z]-[0-9]])$/ is wrong.
Give this regex a try, /\[FBWS-\d\]/g
remove the g if you only want to find 1 match, as g will find all similar matches
Edit: Someone mentioned that you want ["any combination"-"number"], hence if that's what you're looking for then this should work /\[[A-Z]+-\d\]/
I'm trying to create a regex using javascript that will allow names like abc-def but will not allow abc-
(hyphen is also the only nonalpha character allowed)
The name has to be a minimum of 2 characters. I started with
^[a-zA-Z-]{2,}$, but it's not good enough so I'm trying something like this
^([A-Za-z]{2,})+(-[A-Za-z]+)*$.
It can have more than one - in a name but it should never start or finish with -.
It's allowing names like xx-x but not names like x-x. I'd like to achieve that x-x is also accepted but not x-.
Thanks!
Option 1
This option matches strings that begin and end with a letter and ensures two - are not consecutive so a string like a--a is invalid. To allow this case, see the Option 2.
^[a-z]+(?:-?[a-z]+)+$
^ Assert position at the start of the line
[a-z]+ Match any lowercase ASCII letter one or more times (with i flag this also matches uppercase variants)
(?:-?[a-z]+)+ Match the following one or more times
-? Optionally match -
[a-z]+ Match any ASCII letter (with i flag)
$ Assert position at the end of the line
var a = [
"aa","a-a","a-a-a","aa-aa-aa","aa-a", // valid
"aa-a-","a","a-","-a","a--a" // invalid
]
var r = /^[a-z]+(?:-?[a-z]+)+$/i
a.forEach(function(s) {
console.log(`${s}: ${r.test(s)}`)
})
Option 2
If you want to match strings like a--a then you can instead use the following regex:
^[a-z]+[a-z-]*[a-z]+$
var a = [
"aa","a-a","a-a-a","aa-aa-aa","aa-a","a--a", // valid
"aa-a-","a","a-","-a" // invalid
]
var r = /^[a-z]+[a-z-]*[a-z]+$/i
a.forEach(function(s) {
console.log(`${s}: ${r.test(s)}`)
})
You can use a negative lookahead:
/(?!.*-$)^[a-z][a-z-]+$/i
Regex101 Example
Breakdown:
// Negative lookahead so that it can't end with a -
(?!.*-$)
// The actual string must begin with a letter a-z
[a-z]
// Any following strings can be a-z or -, there must be at least 1 of these
[a-z-]+
let regex = /(?!.*-$)^[a-z][a-z-]+$/i;
let test = [
'xx-x',
'x-x',
'x-x-x',
'x-',
'x-x-x-',
'-x',
'x'
];
test.forEach(string => {
console.log(string, ':', regex.test(string));
});
The problem is that the first assertion accepts 2 or more [A-Za-z]. You will need to modify it to accept one or more character:
^[A-Za-z]+((-[A-Za-z]{1,})+)?$
Edit: solved some commented issues
/^[A-Za-z]+((-[A-Za-z]{1,})+)?$/.test('xggg-dfe'); // Logs true
/^[A-Za-z]+((-[A-Za-z]{1,})+)?$/.test('x-d'); // Logs true
/^[A-Za-z]+((-[A-Za-z]{1,})+)?$/.test('xggg-'); // Logs false
Edit 2: Edited to accept characters only
/^[A-Za-z]+((-[A-Za-z]{1,})+)?$/.test('abc'); // Logs true
Use this if you want to accept such as A---A as well :
^(?!-|.*-$)[A-Za-z-]{2,}$
https://regex101.com/r/4UYd9l/4/
If you don't want to accept such as A---A do this:
^(?!-|.*[-]{2,}.*|.*-$)[A-Za-z-]{2,}$
https://regex101.com/r/qH4Q0q/4/
So both will accept only word starting from two characters of the pattern [A-Za-z-] and not start or end (?!-|.*-$) (negative lookahead) with - .
Try this /([a-zA-Z]{1,}-[a-zA-Z]{1,})/g
I suggest the following :
^[a-zA-Z][a-zA-Z-]*[a-zA-Z]$
It validates :
that the matched string is at least composed of two characters (the first and last character classes are matched exactly once)
that the first and the last characters aren't dashes (the first and last character classes do not include -)
that the string can contain dashes and be greater than 2 characters (the second character class includes dashes and will consume as much characters as needed, dashes included).
Try it online.
^(?=[A-Za-z](?:-|[A-Za-z]))(?:(?:-|^)[A-Za-z]+)+$
Asserts that
the first character is a-z
the second is a-z or hyphen
If this matches
looks for groups of one or more letters prefixed by a hyphen or start of string, all the way to end of string.
You can also use the I switch to make it case insensitive.
My regular expression:
/(?!#REF!)([^!,]{1,99})!/g
My test string:
foo,#REF!,bar!,baz,qux!
It currently matches REF! but the desired outcome is for only bar! and qux! to be matched. I used the negative look-ahead (?!#REF!) to prevent that but REF! is being captured as is matches [^!,]{1,99}.
How can prevent REF! getting matched - is using a negative look-ahead the correct approach?
Since your string is a comma separated item list, you may split the string with a comma, remove all empty items (if any), get only those ending with a ! and then remove the ! from the end of the strings:
var s = "foo,#REF!,bar!,baz,qux!";
console.log(s.split(',')
.filter(Boolean) // remove empty items
.filter(function (x) {return x.charAt(x.length-1)==="!" && x!== "#REF!";} ) // ends with ! and not #REF!
.map(function(y) {return y.substr(0, y.length-1)}) // remove !
);
If for some reason you still need to use a regex, you may use
/(?:^|,)(?!#REF!)([^!,]{1,99})!/g
Access Group 1 value. See the regex demo here.
NOTE: You only have 1 capturing group here, as (?!...) is a lookahead that is a special regex construct. (?:...) is a non-capturing group, its value is not stored in any additional memory buffer as compared to a capturing group.
Details
(?:^|,) - either start of string or ,
(?!#REF!) - no #REF! is allowed to appear right after the current location
([^!,]{1,99}) - Capturing group 1: 1 to 99 chars other than ! and ,
! - a ! char
var s = "foo,#REF!,bar!,baz,qux!";
var rx = /(?:^|,)(?!#REF!)([^!,]{1,99})!/g, m, res=[];
while (m=rx.exec(s)) {
res.push(m[1]);
}
console.log(res);
You can use the following regex:
(?<=^|,)(?!#REF!)([^!,]{1,99})!
Explanations:
Adding (?<=^|,) forces the start of your regex matching to either the beginning of the line or to the previous comma. If you don't add it REF! will also be matched. The , will not be part of the result because it is in a lookbehind clause.
DEMO
If you can not use lookbehind, then you can go for a solution like the one proposed by WiktorStribizew
(?:^|,)(?!#REF!)([^!,]{1,99}!)
and by referencing to the 1st capturing group
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'm trying to match a Number in between a set of straight brackets, example:
Match the 0 in actionFields[actionFields][0][data[Report][action]]
This is what I have so far and I keep getting null.
var match, matchRegEx = /^\(?\[(\d)\]\)$/;
nameAttr = "actionFields[actionFields][0][data[Report][action]]",
match = matchRegEx.exec(nameAttr);
If you look at your regular expression, you're matching the beginning of the string, zero or one (, then a [, then a \d, then a ], then a ), then the end of the string.
You should just be able to get away with /\[(\d)\]/, unless you're expecting the [0] construct to show up elsewhere in your string.
Here's a RegexPal showing this.
Your regex should be:
\[(\d+)\]
and capture the first group.
One problem with your regex is that it is anchored at the beginning of input (^) and at the end $.
If there's only one number /\d+/
You can test only for the number