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 $.
If there's only one number /\d+/
You can test only for the number
Related
I have the regular expression:
const regex = /^\d*\.?\d{0,2}$/
and its inverse (I believe) of
const inverse = /^(?!\d*\.?\d{0,2}$)/
The first regex is validating the string fits any positive number, allowing a decimal and two decimal digits (e.g. 150, 14., 7.4, 12.68). The second regex is the inverse of the first, and doing some testing I'm fairly confident it's giving the expected result, as it only validates when the string is anything but a number that may have a decimal and two digits after (e.g. 12..05, a5, 54.357).
My goal is to remove any characters from the string that do not fit the first regex. I thought I could do that this way:
let myString = '123M.45';
let fixed = myString.replace(inverse, '');
But this does not work as intended. To debug, I tried having the replace character changed to something I would be able to see:
let fixed = myString.replace(inverse, 'ZZZ');
When I do this, fixed becomes: ZZZ123M.45
Any help would be greatly appreciated.
I think I understand your logic here trying to find a regex that is the inverse of the regex that matches your valid string, in the hopes that it will allow you to remove any characters that make your string invalid and leave only the valid string. However, I don't think replace() will allow you to solve your problem in this way. From the MDN docs:
The replace() method returns a new string with some or all matches of a pattern replaced by a replacement.
In your inverse pattern you are using a negative lookahead. If we take a simple example of X(?!Y) we can think of this as "match X if not followed by Y". In your pattern your "X" is ^ and your "Y" is \d*\.?\d{0,2}$. From my understanding, the reason you are getting ZZZ123M.45 is that it is finding the first ^ (i.e, the start of the string) that is not followed by your pattern \d*\.?\d{0,2}$, and since 123M.45 doesn't match your "Y" pattern, your negative lookahead is satisfied and the beginning of your string is matched and "replaced" with ZZZ.
That (I think) is an explanation of what you are seeing.
I would propose an alternative solution to your problem that better fits with how I understand the .replace() method. Instead of your inverse pattern, try this one:
const invalidChars = /[^\d\.]|\.(?=\.)|(?<=\.\d\d)\d*/g
const myString = '123M..456444';
const fixed = myString.replace(invalidChars, '');
Here I am using a pattern that I think will match the individual characters that you want to remove. Let's break down what this one is doing:
[^\d\.]: match characters that are not digits
\.(?=\.): match . character if it is followed by another . character.
(?<=\.\d\d)\d*: match digits that are preceded by a decimal and 2 digits
Then I join all these with ORs (|) so it will match any one of the above patterns, and I use the g flag so that it will replace all the matches, not just the first one.
I am not sure if this will cover all your use cases, but I thought I would give it a shot. Here's a link to a breakdown that might be more helpful than mine, and you can use this tool to tweak the pattern if necessary.
I don't think you can do this
remove any characters from the string that do not fit the first regex
Because regex matching is meant for the entire string, and replace is used to replace just a PART inside that string. So the Regex inside replace must be a Regex to match unwanted characters only, not inverted Regex.
What you could do is to validate the string with your original regex, then if it's not valid, replace and validate again.
//if (notValid), replace unwanted character
// replace everything that's not a dot or digit
const replaceRegex = /[^\d.]/g; // notice g flag here to match every occurrence
const myString = '123M.45';
const fixed = myString.replace(replaceRegex, '');
console.log(fixed)
// validate again
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 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 am having a difficult time getting a seemingly simple Regexp. I am trying to grab the last occurrences of word characters between square brackets in a string. My code:
pattern = /\[(\w+)\]/g;
var text = "item[gemstones_attributes][0][shape]";
if (pattern.test(text)) {
alert(RegExp.lastMatch);
}
The above code is outputting "gemstones_attributes", when I want it to output "shape". Why is this regexp not working, or is there something wrong with my approach to getting the last match? I'm sure that I am making an obvious mistake - regular expressions have never been my string suit.
Edit:
There are cases in which the string will not terminate with a right-bracket.
You can greedily match as much as possible before your pattern which will result in your group matching only the last match:
pattern = /.*\[(\w+)\]/g;
var text = "item[gemstones_attributes][0][shape]";
var match = pattern.exec(text);
if (match != null) alert(match[1]);
RegExp.lastMatch gives the match of the last regular expression. It isn't the last match in the text.
Regular expressions parse left to right and are greedy. So your regexp matches the first '[' it sees and grabs the words between it. When you call lastMatch it gives you the last pattern matched. What you need is to match everything you can first .* and then your pattern.
i think your problem is in your regex not in your src line .lastMatch.
Your regex returns just the first match of your square brackets and not all matches. You can try to add some groups to your regular expression - and normally you should get all matches.
krikit
Use match() instead of test()
if (text.match(pattern))
test() checks for a match inside a string. This is successfull after the first occurence, so there is no need for further parsing.
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