I have string like below
BANKNIFTY-13-FEB-2020-31200-ce
I want to convert the string to 13-FEB-31200-ce
so I tried below code
str.match(/(.*)-(?:.*)-(?:.*)-(.*)-(?:.*)-(?:.*)/g)
But its returning whole string
Two capture groups is probably the way to go. Now you have two options to use it. One is match which requires you to put the two pieces together
var str = 'BANKNIFTY-13-FEB-2020-31200-ce'
var match = str.match(/[^-]+-(\d{2}-[A-Z]{3}-)\d{4}-(.*)/)
// just reference the two groups
console.log(`${match[1]}${match[2]}`)
// or you can remove the match and join the remaining
match.shift()
console.log(match.join(''))
Or just string replace which you do the concatenation of the two capture groups in one line.
var str = 'BANKNIFTY-13-FEB-2020-31200-ce'
var match = str.replace(/[^-]+-(\d{2}-[A-Z]{3}-)\d{4}-(.*)/, '$1$2')
console.log(match)
Regex doesn't seem to be the most appropriate tool here. Why not use simple .split?
let str = 'BANKNIFTY-13-FEB-2020-31200-ce';
let splits = str.split('-');
let out = [splits[1], splits[2], splits[4], splits[5]].join('-');
console.log(out);
If you really want to use regexp,
let str = 'BANKNIFTY-13-FEB-2020-31200-ce';
let splits = str.match(/[^-]+/g);
let out = [splits[1], splits[2], splits[4], splits[5]].join('-');
console.log(out);
I would not use Regex at all if you know exact positions. Using regex is expensive and should be done differently if there is way. (https://blog.codinghorror.com/regular-expressions-now-you-have-two-problems/)
const strArr = "BANKNIFTY-13-FEB-2020-31200-ce".split("-"); // creates array
strArr.splice(0,1); // remove first item
strArr.splice(2,1); // remove 2020
const finalStr = strArr.join("-");
If the pattern doesn't need to be too specific.
Then just keep it simple and only capture what's needed.
Then glue the captured groups together.
let str = 'BANKNIFTY-13-FEB-2020-31200-ce';
let m = str.match(/^\w+-(\d{1,2}-[A-Z]{3})-\d+-(.*)$/)
let result = m ? m[1]+'-'+m[2] : undefined;
console.log(result);
In this regex, ^ is the start of the string and $ the end of the string.
You can have something like this by capturing groups with regex:
const regex = /(\d{2}\-\w{3})(\-\d{4})(\-\d{5}\-\w{2})/
const text = "BANKNIFTY-13-FEB-2020-31200-ce"
const [, a, b, c] = text.match(regex);
console.log(`${a}${c}`)
Related
I'm trying to come up with a regex that will do following.
I have a string
var input_string = "E100T10P200E3000T3S10";
var output=input_string.split(**Trying to find this**);
this should give an array with all the letters in order with repetitions
output = ["E","T","P","E","T","S"]
See below. \d+ means one or more digits; filter (x => x) removes empty strings that can appear in the beginning or the end of the array if the input string begins or ends with digits.
var input_string = "E100T10P200E3000T3S10";
var output = input_string.split (/\d+/).filter (x => x);
console.log (output);
We can try just matching for capital letters here:
var input_string = "E100T10P200E3000T3S10";
var output = input_string.match(/[A-Z]/g);
console.log(output);
Another approach is spread the string to array and use isNaN as filter callback
var input_string = "E100T10P200E3000T3S10";
var output = [...input_string].filter(isNaN);
console.log(output);
You can use regex replace method. First replace all the digits with empty string and then split the resultant string.
const input_string = 'E100T10P200E3000T3S10';
const ret = input_string.replace(/\d/g, '').split('');
console.log(ret);
I have the following string
"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049,
I need to get string after "ssu":" the Result should be 89c4eef0-3a0d-47ae-a97f-42adafa7cf8f. How do I do it in Javascript but very simple? I am thinking to collect 36 character after "ssu":".
You could build a valid JSON string and parse it and get the wanted property ssu.
var string = '"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049,',
object = JSON.parse(`{${string.slice(0, -1)}}`), // slice for removing the last comma
ssu = object.ssu;
console.log(ssu);
One solution would be to use the following regular expression:
/\"ssu\":\"([\w-]+)\"/
This pattern basically means:
\"ssu\":\" , start searching from the first instance of "ssu":"
([\w-]+) , collect a "group" of one or more alphanumeric characters \w and hypens -
\", look for a " at the end of the group
Using a group allows you to extract a portion of the matched pattern via the String#match method that is of interest to you which in your case is the guid that corresponds to ([\w-]+)
A working example of this would be:
const str = `"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049,`
const value = str.match(/\"ssu\":\"([\w-]+)\"/)[1]
console.log(value);
Update: Extract multiple groupings that occour in string
To extract values for multiple occurances of the "ssu" key in your input string, you could use the String#matchAll() method to achieve that as shown:
const str = `"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049,"ssu":"value-of-second-ssu","ssu":"value-of-third-ssu"`;
const values =
/* Obtain array of matches for pattern */
[...str.matchAll(/\"ssu\":\"([\w-]+)\"/g)]
/* Extract only the value from pattern group */
.map(([,value]) => value);
console.log(values);
Note that for this to work as expected, the /g flag must be added to the end of the original pattern. Hope that helps!
Use this regExp: /(?!"ssu":")(\w+-)+\w+/
const str = '"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049,';
const re = /(?!"ssu":")(\w+-)+\w+/;
const res = str.match(re)[0];
console.log(res);
You can use regular expressions.
var str = '"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049,'
var minhaRE = new RegExp("[a-z|0-9]*-[a-z|0-9|-]*");
minhaRE.exec(str)
OutPut: Array [ "89c4eef0-3a0d-47ae-a97f-42adafa7cf8f" ]
Looks almost like a JSON string.
So with a small change it can be parsed to an object.
var str = '"sis":4,"sct":15,"ssu":"89c4eef0-3a0d-47ae-a97f-42adafa7cf8f","ssv":384,"siw":96554,"scx":1049, ';
var obj = JSON.parse('{'+str.replace(/[, ]+$/,'')+'}');
console.log(obj.ssu)
I have an expression.
var expression = "Q101='You will have an answer here like a string for instance.'"
I have a regular expression that searches the expression.
var regEx = new regExp(/=|<>|like/)
I want to split the expression using the regular expression.
var result = expression.split(regExp)
This will return the following:
["Q101", "'You will have an answer here ", " a string for instance'"]
This is not what I want.
I should have:
["Q101", "'You will have an answer here like a string for instance'"]
How do I use the regular expression above to split only on the first match?
Since you only want to grab the two parts either side of the first delimiter it might be easier to use String.match and discard the whole match:
var expression = "Q101='You will have an answer here like a string for instance.'";
var parts = expression.match(/^(.*?)(?:=|<>|like)(.*)$/);
parts.shift();
console.log(parts);
expression = "Q101like'This answer uses like twice'";
parts = expression.match(/^(.*?)(?:=|<>|like)(.*)$/);
parts.shift();
console.log(parts);
JavaScript's split method won't quite do what you want, because it will either split on all matches, or stop after N matches. You need an extra step to find the first match, then split once by the first match using a custom function:
function splitMatch(string, match) {
var splitString = match[0];
var result = [
expression.slice(0, match.index),
expression.slice(match.index + splitString.length)
];
return result;
}
var expression = "Q101='You will have an answer here like a string for instance.'"
var regEx = new RegExp(/=|<>|like/)
var match = regEx.exec(expression)
if (match) {
var result = splitMatch(expression, match);
console.log(result);
}
While JavaScript's split method does have an optional limit parameter, it simply discards the parts of the result that make it too long (unlike, e.g. Python's split). To do this in JS, you'll need to split it manually, considering the length of the match —
const exp = "Q101='You will have an answer here like a string for instance.'"
const splitRxp = /=|<>|like/
const splitPos = exp.search(splitRxp)
const splitStr = exp.match(splitRxp)[0]
const result = splitPos != -1 ? (
[
exp.substring(0, splitPos),
exp.substring(splitPos + splitStr.length),
]
) : (
null
);
console.log(result)
I am trying to capture all data before the first _. What I have so far is
const regex = /(.*)(?=_)/g;
var s = "Mike_Jones_Jr";
console.log(s.match(regex));
The output is an array Array ["Mike_Jones","" ]
What I was expecting was Mike
Use /^[^_]*/
^ looks from the beginning of the string
[^_] negates the _
* gives any number of characters
const regex = /^[^_]*/;
var s = "Mike_Jones_Jr";
console.log(s.match(regex));
var s = "Mike_Jones_Jr";
console.log(s.split('_')[0]);
Create a capture group ((something between parentheses)) that starts at the beginning of the line (^) and is lazy (.*?), then grab the second item in the matching array.
const regex = /(^.*?)_/s
console.log('Mike_Jones_Jr'.match(regex)[1] || '')
console.log(`Mike
_Jones_Jr`.match(regex)[1] || '')
You can simply use split,
Note:- Second parameter is to limit the number of elements in final outptut
var s = "Mike_Jones_Jr";
console.log( s.split('_', 1) );
If you want to do using regex, you can drop the g flag
const regex = /^[^_]*(?=_)/;
var s = "Mike_Jones_Jr";
console.log(s.match(regex));
console.log("_ melpomene is awesome".match(regex));
In my Javascript code, I get one very long line as a string.
This one line only has around 65'000 letters. Example:
config=123&url=http://localhost/example&path_of_code=blablaba&link=kjslfdjs...
What I have to do is replace all & with an break (\n) first and then pick only the line which starts with "path_of_code=". This line I have to write in a variable.
The part with replace & with an break (\n) I already get it, but the second task I didn't.
var obj = document.getElementById('div_content');
var contentJS= obj.value;
var splittedResult;
splittedResult = contentJS.replace(/&/g, '\n');
What is the fastest way to do it? Please note, the list is usually very long.
It sounds like you want to extract the text after &path_of_code= up until either the end of the string or the next &. That's easily done with a regular expression using a capture group, then using the value of that capture group:
var rex = /&path_of_code=([^&]+)/;
var match = rex.exec(theString);
if (match) {
var text = match[1];
}
Live Example:
var theString = "config=123&url=http://localhost/example&path_of_code=blablaba&link=kjslfdjs...";
var rex = /&path_of_code=([^&]+)/;
var match = rex.exec(theString);
if (match) {
var text = match[1];
console.log(text);
}
Use combination of String.indexOf() and String.substr()
var contentJS= "123&url=http://localhost/example&path_of_code=blablaba&link=kjslfdjs...";
var index = contentJS.indexOf("&path_of_code"),
substr = contentJS.substr(index+1),
res = substr.substr(0, substr.indexOf("&"));
console.log(res)
but the second task I didn't.
You can use filter() and startsWith()
splittedResult = splittedResult.filter(i => i.startsWith('path_of_code='));