I'm trying to figure out a way to get all my Group 1 matches into an array without using a loop with matchAll().
Here's what I have thus far, but it only yields the first match:
let str = "123ABC, 123ABC"
let results = str.matchAll(/123(ABC)/gi);
let [group1] = results;
alert(group1[1]);
How can I get the results of the matchAll into one single array? Aka:
// ABC, ABC
const str = "123ABC, 123ABC"
const results = Array.from(
str.matchAll(/123(ABC)/gi),
([_, g1]) => g1
)
console.log(results)
If you only need the abc part of the string then you don't need to use matchAll method. You can easily get the results you want simply using the positive lookbehind regex expresion with the match method.
let str = "123ABC, 123ABC"
let results = str.match(/(?<=123)ABC/gi);
console.log(results)
// ["ABC","ABC"]
Here is some more information on these types of regex expressions Lookahead and lookbehind
You can use Array.from to convert results to array and perform map in one go:
const matches = Array.from(results, match => match[1])
Related
I have string with slash separated contains function names.
e.g.
my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName()
Within above string I want only function name i.e. getCustomer(), getControllerName() & getCsvFileName()
I searched some regex like:
let res = myString.match(/(?<=(function\s))(\w+)/g);
but its returning result as null.
Update:
Now I want to get function names without parentheses () i.e. getCustomer, getControllerName & getCsvFileName
Please help me in this
const str = "my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName()"
let tokens = [];
for (element of str.split("/"))
if (element.endsWith("()"))
tokens.push(element.split("=")[1] ?? element.split("=")[0])
console.log(tokens);
General idea: split the string along slashes, and for each of these tokens, if the token ends with () (as per Nick's suggestion), split the token along =. Append the second index of the token split along = if it exists, otherwise append the first.
A "smaller" version (using purely array methods) could be:
const str = "my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName()"
let tokens = str.split("/")
.filter(element => element.endsWith("()"))
.map(element => element.split("=")[1] ?? element.split("=")[0]);
console.log(tokens);
You can split the string that has parentheses () first like /.*?\([^)]*\)/g.
This will give array of results, and after that you can iterate the array data and for each item, you can split the = and / before function name with the help of item.split(/=|\//).
Then push the filtered function name into empty array functionNames.
Working Example:
const string = `my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName()`;
const functionNames = [];
string.match(/.*?\([^)]*\)/g).forEach(item => {
const splitString = item.split(/=|\//);
const functionName = splitString[splitString.length - 1];
functionNames.push(functionName);
});
console.log(functionNames);
As per, MDN docs the match() method returns null if it does not find a match for the provided regex in the provided search string.
The regular expression which you have provided,/(?<=(function\s))(\w+)/g matches any word that has 'function ' before it. (NOTE: a space after the word function)
Your search string my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName() does not include 'function ' before any characters. That is why you got null as result of match() method.
let yourString = 'my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName()';
let myReferenceString = 'SAMPLETEXTfunction sayHi()/function sayHello()';
let res = yourString.match(/(?<=(function\s))(\w+)/g);
let res2 = myReferenceString.match(/(?<=(function\s))(\w+)/g);
console.log("Result of your string", res);
console.log("Result of my string", res2);
My solution here,
let myreferenceString = 'my_doc/desktop/customer=getCustomer()/getCsvFileName()/controller=getControllerName()'
let res = myreferenceString.match(/((?<==)(\w+\(\)))|((?<=\/)(\w+\(\)))/g);
console.log("Result", res);
NOTE: I have used the 'Positive Look Behind regex operator', This is not supported in browsers like Safari and IE. Please do reasearch about this before considering this approach.
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}`)
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'm trying to match a certain word on a string and only if it doesn't exist i want to match the another one using the OR | operator ....but the match is ignoring that... how can i insure that the behavior works :
const str = 'Soraka is an ambulance 911'
const regex = RegExp('('+'911'+'|'+'soraka'+')','i')
console.log(str.match(regex)[0]) // should get 911 instead
911 occurs late in the string, whereas Soraka occurs earlier, and the regex engine iterates character-by-character, so Soraka gets matched first, even though it's on the right-hand side of the alternation.
One option would be to match Soraka or 911 in captured lookaheads instead, and then with the regex match object, alternate between the two groups to get the one which is not undefined:
const check = (str) => {
const regex = /^(?=.*(911)|.*(Soraka))/;
const match = str.match(regex);
console.log(match[1] || match[2]);
};
check('Soraka is an ambulance 911');
check('foo 911');
check('foo Soraka');
You can use includes and find
You can pass the strings in the priority sequence, so as soon as find found any string in the original string it returns that strings back,
const str = 'Soraka is an ambulance 911'
const findStr = (...arg) => {
return [...arg].find(toCheck => str.includes(toCheck))
}
console.log(findStr("911", "Soraka"))
You can extend the findStr if you want your match to be case insensitive something like this
const str = 'Soraka is an ambulance 911'
const findStr = (...arg) => {
return [...arg].find(toCheck => str.toLowerCase().includes(toCheck.toLowerCase()))
}
console.log(findStr("Soraka", "911"))
If you want match to be whole word not the partial words than you can build dynamic regex and use it search value
const str = '911234 Soraka is an ambulance 911'
const findStr = (...arg) => {
return [...arg].find(toCheck =>{
let regex = new RegExp(`\\b${toCheck}\\b`,'i')
return regex.test(str)
})
}
console.log(findStr("911", "Soraka"))
Just use a greedy dot before a capturing group that matches 911 or Soraka:
/.*(911)|(Soraka)/
See the regex demo
The .* (or, if there are line breaks, use /.*(911)|(Soraka)/s in Chrome/Node, or /[^]*(911)|(Soraka)/ to support legacy EMCMScript versions) will ensure the regex index advances to the rightmost position when matching 911 or Soraka.
JS demo (borrowed from #CertainPerformance's answer):
const check = (str) => {
const regex = /.*(911)|(Soraka)/;
const match = str.match(regex) || ["","NO MATCH","NO MATCH"];
console.log(match[1] || match[2]);
};
check('Soraka is an ambulance 911');
check('Ambulance 911, Soraka');
check('foo 911');
check('foo Soraka');
check('foo oops!');
convert a regular expression to all possible cases
For Example:
1.9856[1-4] then it should return values like 98561,98562,98563,98564
2.98[4-5]65 then it should return values 98465,98565
You can use the regex /\[(\d)-(\d)\]/ and match to get the start and end of the range. Then use a for loop to replace the match with numbers from start to end
function getCombinations(str) {
const regex = /\[(\d)-(\d)\]/,
[match, start, end] = str.match(regex),
output = []
for (let i = +start; i <= +end; i++)
output.push(str.replace(match, i))
return output
}
console.log(getCombinations('1.9856[1-4]'))
console.log(getCombinations('2.98[4-5]65'))
This works for only single range in the input string. For multiple ranges, you can use exec to get multiple matches and replace based on the index of the match
You can use Regex to match exact numbers or you can use \d to match any digits.
For your first example where 9856[1-4] then it should return values like 98561,98562,98563,98564,
you can return all numbers in a string that begin with 9856 and ends with any number ranging from 1-4 like so:
let nums = "985612, 985622, 985633, 985644, 985655"
let regex = /9856[1-4]/g
let result = nums.match(regex)
console.log(result)
For the second example where 98[4-5]65 then it should return values 98465, 98565
let nums = "98465, 98565"
let regex = /98[4-5]{2}65/g
let result = nums.match(regex)
console.log(result)