In fiddle : http://jsfiddle.net/rtucgv74/
I'm attempting to match the first char with 3 digits. So below code should alert f234. But instead null is returned ?
src :
var reg = /^\[a-zA-Z]\d{3}/;
alert(reg.exec(("test f234 sadfas")[1]))
How do extract char/digit/digit/digit from string ?
Almost right, but the first backslash in your regex is making the [ match an actual bracket instead of defining a character set.
And the ^ is preventing it from matching anything not at the beginning of the string.
So the correct regex is /[a-zA-Z]\d{3}/ or /[a-z]\d{3}/i
The other problem is that you try to read the second element of the result array. But you're not capturing anything in your regex, so there will be only one element ([0]) which is the whole match.
Try:
reg.exec("test f234 sadfas")[0]
In your example, you are picking out a string from a "sentence", so remove the ^. Also remove the \ that escapes the [. So you want:
[a-zA-Z]\d{3}
Test it here: http://regexr.com/3bta7
Try something like this:
var reg = /[a-zA-Z]{1}[0-9]{3}/;
alert("test f234 asdfghj".match(reg))
Check the Demo fiddle
UPDATE WITH BOUNDARY SUPPORT:
The regex will match also part of words like abc12335 so to void this behaviour you could try this regex:
var reg = /\b[a-zA-Z]{1}[0-9]{3}\b/;
alert("test aaf23412 asdfghj".match(reg)) // not match
alert("test f234 asdfghj".match(reg)) // match !
alert("test f2343 asdfghj".match(reg)) // not match
Check the demo fiddle 2
The \ in your regex escapes character [ and makes it non special. And the character ^ prevents your regex to match any part that is not in the beginning of the string.
So this would work :
var reg = /[a-zA-Z]\d{3}/;
alert(reg.exec("test f234 sadfas")[0]);
Note that there is only one element in the result because your regex don't contain any capture (characters between brackets).
This can help you find the first character occurring before 3 digits:
\w(?=\d{3}).
Related
I need a regular expression for :
<<12.txt>> <<45.txt>
I have created a regular expression :
<<.+.txt>>
But this found one match in whole string but here is 2 matches:
<<12.txt>>
<<45.txt>>
if anyone have solution for this problem please help me out there
Part of the issue is that the string you've specified wouldn't match because the second > is missing in <<45.txt>.
Also, you're using the . (dot) selector, and also trying to find a period. It works, but now how you think it is.
Here's the regex you want:
var regex = /<<\d+\.txt>>/g
\d matches only numbers
\. matches an actual period
/g means global, so it won't stop at the first match
Practice Regular Expressions
https://regexr.com/43bs4
Demo
var string = "<<12.txt>> <<45.txt>>";
var regex = /<<\d+\.txt>>/g;
var matches = string.match(regex);
console.log(matches);
P.S., if you actually want to match with 1 > or 2 >>, you can with:
var regex = /<<\d+\.txt>>?/g
? optionally matches the character right before it
/<<.+.txt>>/gm
g is for global (will search through entire source)
m is for multi line search support
I am passing codes to an API. These codes are alphanumeric, like this one: M84.534D
I just found out that the API does not use the trailing letters. In other words, the API is expecting M84.534, no letter D at the end.
The problem I am having is that the format is not the same for the codes.
I may have M84.534DAC, or M84.534.
What I need to accomplish before sending the code is to remove any non-numeric characters from the end of the code, so in the examples:
M84.534D -> I need to pass M84.534
M84.534DAC -> I also need to pass M84.534
Is there any function or regex that will do that?
Thank you in advance to all.
You can use the regex below. It will remove anything from the end of the string that is not a number
let code = 'M84.534DAC'
console.log(code.replace(/[^0-9]+?$/, ""));
[^0-9] matches anything that is not a numer
+? Will match between 1 and unlimited times
$ Will match the end of the string
So linked together, it will match any non numbers at the end of the string, and replace them with nothing.
You could use the following expression:
\D*$
As in:
var somestring = "M84.534D".replace(/\D*$/, '');
console.log(somestring);
Explanation:
\D stands for not \d, the star * means zero or more times (greedily) and the $ anchors the expression to the end of the string.
Given your limited data sample, this simple regular expression does the trick. You just replace the match with an empty string.
I've used document.write just so we can see the results. You use this whatever way you want.
var testData = [
'M84.534D',
'M84.534DAC'
]
regex = /\D+$/
testData.forEach((item) => {
var cleanValue = item.replace(regex, '')
document.write(cleanValue + '<br>')
})
RegEx breakdown:
\D = Anything that's not a digit
+ = One or more occurrences
$ = End of line/input
My string can be something like A01, B02, C03, possibly AA18 in the future as well. I thought I could use a regex to get just the letters and work on my regex since I haven't done much with it. I wrote this function:
function rowOffset(sequence) {
console.log(sequence);
var matches = /^[a-zA-Z]+$/.exec(sequence);
console.log(matches);
var letter = matches[0].toUpperCase();
return letter;
}
var x = "A01";
console.log(rowOffset(x));
My matches continue to be null. Am I doing this correctly? Looking at this post, I thought the regex was correct: Regular expression for only characters a-z, A-Z
You can use String#replace to remove all non letters from input string:
var r = 'AA18'.replace(/[^a-zA-Z]+/g, '');
//=> "AA"
Your main issue is the use of the ^ and $ characters in the regex pattern. ^ indicates the beginning of the string and $ indicates the end, so you pattern is looking for a string that is ONLY a group of one or more letters, from the beginning to the end of the string.
Additionally, if you want to get each individual instance of the letters, you want to include the "global" indicator (g) at the end of your regex pattern: /[a-zA-Z]+/g. Leaving that out means that it will only find the first instance of the pattern and then stop searching . . . adding it will match all instances.
Those two updates should get you going.
EDIT:
Also, you may want to use match() rather than exec(). If you have a string of multiple values (e.g., "A01, B02, C03, AA18"), match() will return them all in an array, whereas, exec() will only match the first one. If it is only ever one value, then exec() will be fine (and you also wouldn't need the "global" flag).
If you want to use match(), you need to change your code order just a bit to:
var matches = sequence.match(/[a-zA-Z]+/g);
To return an array of separate letters remove +:
var matches = sequence.match(/[a-zA-Z]/g);
You're confused about what the goal of the other question was: he wanted to check that there were only letters in his string.
You need to remove the anchors ^$, who match respectively the beginning and end of the string:
[a-zA-Z]+
This will match the first of letters in your input string.
If there might be more (ie you want multiple matches in your single string), use
sequence.match(/[a-zA-Z]+/g)
This /[^a-z]/g solves the problem. Look at the example below.
function pangram(str) {
let regExp = /[^a-z]/g;
let letters = str.toLowerCase().replace(regExp, '');
document.getElementById('letters').innerHTML = letters;
}
pangram('GHV 2## %hfr efg uor7 489(*&^% knt lhtkjj ngnm!##$%^&*()_');
<h4 id="letters"></h4>
You can do this:
var r = 'AA18'.replace(/[\W\d_]/g, ''); // AA
Also can be done by String.prototype.split(regex).
'AA12BB34'.split(/(\d+)/); // ["AA", "12", "BB", "34", ""]
'AA12BB34'.split(/(\d+)/)[0]; // "AA"
Here regex divides the giving string by digits (\d+)
I want to match some keywords in the url
var parentURL = document.referrer;
var greenPictures = /redwoods-are-big.jpg|greendwoods-are-small.jpg/;
var existsGreen = greenPictures.test(parentURL);
var existsGreen turns true when it finds greendwoods-are-small.jpg but also when it finds small.jpg
What can i do that it only turns true if there is exactly greendwoods-are-small.jpg?
You can use ^ to match the beginning of a string and $ to match the end:
var greenPictures = /^(redwoods-are-big.jpg|greendwoods-are-small.jpg)$/;
var existsGreen = greenPictures.test(parentURL);
But of cause the document.referrer is not equal ether redwoods-are-big.jpg or greendwoods-are-small.jpg so i would match /something.png[END]:
var greenPictures = /\/(redwoods-are-big\.jpg|greendwoods-are-small\.jpg)$/; // <-- See how I escaped the / and the . there? (\/ and \.)
var existsGreen = greenPictures.test(parentURL);
Try this regex:
/(redwoods-are-big|greendwoods-are-small)\.jpg/i
I used the i flag for ignoring the character cases in parentURL variable.
Description
Demo
http://regex101.com/r/aI4yJ6
Dashes does not have any special meaning outside character sets, e.g.:
[a-f], [^x-z] etc.
The characters with special meaning in your regexp is | and .
/redwoods-are-big.jpg|greendwoods-are-small.jpg/
| denotes either or.
. matches any character except the newline characters \n \r \u2028 or \u2029.
In other words: There is something else iffy going on in your code.
More on RegExp.
Pages like these can be rather helpful if you struggle with writing regexp's:
regex101 (with sample)
RegexPlanet
RegExr
Debuggex
etc.
I am using regex to verify that a string only contains alphabets and spaces. Regex is defined as
var regex = /^[A-Za-z ]/;
but even if I am testing it with a string "X," , it is giving a true result.
What is the error here?
^[A-Za-z ] only matches one character. and ^ means the start of the string. To accomplish what you want, use:
+ - This means match one or more. Alternatively you can use:
* - Which means match zero or more.
But I think you're better off with the first one (+). Another thing is, match for the whole string. This means you have to search from the first to the last character.
$ - This means match the end.
Your code should be like this:
var regex = /^[A-Za-z ]+$/;
Your regex matches the first letter of your input and therefore true is returned.. You need to add $ to make sure you only match a complete string from the beginning (^) to the end ($).
var regex = /^[A-Za-z ]*$/;
Try using this:
/^[a-zA-Z\ ]+$/
This is the correct way:
/^[A-Za-z ]*$/
Regex Demo