This question already has answers here:
How can I match overlapping strings with regex?
(6 answers)
Closed 3 years ago.
I want to match all occurrence in string.
Example:
pc+pc2/pc2+pc2*rr+pd
I want to check how many matched of pc2 value and regular expression is before and after special character exists.
var str = "pc+pc2/pc2+pc2*rr+pd";
var res = str.match(new RegExp("([\\W])pc2([\\W])",'g'));
But I got only +pc2/ and +pc2* and /pc2+ not get in this.
Problem is in first match / is removed. So after that, it is starting to check from pc2+pc2*rr+pd. That's why /pc2+ value does not get in the match.
How do I solve this problem?
You need some sort of recursive regex to achieve what you're trying to get, you can use exec to manipulate lastIndex in case of value in string is p
let regex1 = /\Wpc2\W/g;
let str1 = 'pc+pc2/pc2+pc2*rr+pd';
let array1;
let op = []
while ((array1 = regex1.exec(str1)) !== null) {
op.push(array1[0])
if(str1[regex1.lastIndex] === 'p'){
regex1.lastIndex--;
}
}
console.log(op)
Related
This question already has answers here:
Get the first integers in a string with JavaScript
(5 answers)
Closed 6 months ago.
How to find the first number from string in javascript?
var string = "120-250";
var string = "120,250";
var string = "120 | 250";
Here is an example that may help you understand.
Use the search() method to get the index of the first number in the string.
The search method takes a regular expression and returns the index of the first match in the string.
const str = 'one 2 three 4'
const index = str.search(/[0-9]/);
console.log(index); // 4
const firstNum = Number(str[index]);
console.log(firstNum); // 2
Basic regular expression start of string followed by numbers /^\d+/
const getStart = str => str.match(/^\d+/)?.[0];
console.log(getStart("123,456"));
console.log(getStart("123-456"));
console.log(getStart("123|456"));
console.log(getStart("xxx,xxx"));
Or parseInt can be used, but it will drop leading zeros.
const getStart = str => parseInt(str, 10);
console.log(getStart("123,456"));
console.log(getStart("123-456"));
console.log(getStart("123|456"));
console.log(getStart("xxx,xxx"));
This question already has answers here:
How to get the nth occurrence in a string?
(14 answers)
Closed 2 years ago.
For example, I have a string like following
var string = "test1;test2;test3;test4;test5";
I want the following substring from the above string, I don't know the startIndex, the only thing I can tell substring should start after the second semicolon to till the end.
var substring = "test3;test4;test5";
Now I want to have substring like following
var substring2 = "test4;test5"
How to achieve this in JavaScript
You mean this?
const string = "test1;test2;test3;test4;test5";
const arr = string.split(";")
console.log(arr.slice(2).join(";")); // from item #2
console.log(arr.slice(-2).join(";")) // last 2 items
If the string is very long, you may want to use one of these versions
How to get the nth occurrence in a string?
As a function
const string = "test1;test2;test3;test4;test5";
const restOfString = (string,pos) => {
const arr = string.split(";")
return arr.slice(pos).join(";"); // from item #pos
};
console.log(restOfString(string,2))
console.log(restOfString(string,3))
Try to use a combination of string split and join to achieve this.
var s = "test1;test2;test3;test4;test5";
var a = s.split(";")
console.log(a.slice(3).join(";"))
This question already has answers here:
How to check if character is a letter in Javascript?
(17 answers)
Closed 3 years ago.
Hello I am having trouble trying to use Regex to check if each character in string is an alphabet.
First let me introduce the problem itself.
There is a string mixed with special chars and alphabets and suppose to return the number of alphabets only.
My code/pseudo code for problem is :
//Create var to hold count;
var count = 0;
//Loop thru str
for(let char of str){
//Check if char is a alphabet
***if(char === /[A-Za-z]/gi){***
//if so add to count
count ++;
}
//return count;
return count;
}
How can I use Regex in a conditional statement to check if each char is an alphabet????
Please help!
const pattern = /[a-z]/i
const result = [...'Abc1'].reduce((count,c) => pattern.test(c) ? count+1 : count, 0)
console.log(result) // 3
This question already has answers here:
Is there a javascript method to find substring in an array of strings?
(5 answers)
How to check if a string contains text from an array of substrings in JavaScript?
(24 answers)
Closed 4 years ago.
I have a variable like,
var url = "/login/user";
And I have an array like,
var x = ["login", "resetpassword", "authenticate"];
Now I need to check, whether that url string is present in an array of string. As we can see that login is present in an array but when i do x.indexOf(url), it always receive false because the field url has rest other letters also. So now how can I ingnore those letters while checking a string in an array and return true?
Use .some over the array instead:
var url = "/login/user";
var x = ["login", "resetpassword", "authenticate"];
if (x.some(str => url.includes(str))) {
console.log('something in X is included in URL');
}
Or, if the substring you're looking for is always between the first two slashes in the url variable, then extract that substring first, and use .includes:
var url = "/login/user";
var x = ["login", "resetpassword", "authenticate"];
var foundStr = url.split('/')[1];
if (x.includes(foundStr)) {
console.log('something in X is included in URL');
}
One way is to split url with / character and than use some
var url = "/login/user";
var x = ["login", "resetpassword", "authenticate"];
let urlSplitted = url.split('/')
let op = urlSplitted.some(e=> x.includes(e))
console.log(op)
You could join the given words with a pipe (as or operator in regex), generate a regular expression and test against the string.
This works as long as you do not have some characters with special meanings.
var url = "/login/user",
x = ["login", "resetpassword", "authenticate"];
console.log(new RegExp(x.join('|')).test(url));
This question already has answers here:
Find and replace nth occurrence of [bracketed] expression in string
(4 answers)
Closed 5 years ago.
This question been asked before, but I did not succeed in solving the problem.
I have a string that contains numbers, e.g.
var stringWithNumbers = "bla_3_bla_14_bla_5";
I want to replace the nth occurence of a number (e.g. the 2nd) with javascript. I did not get farer than
var regex = new RegExp("([0-9]+)");
var replacement = "xy";
var changedString = stringWithNumbers.replace(regex, replacement);
This only changes the first number.
It was suggested to use back references like $1, but this did not help me.
The result should, for example, be
"bla_3_bla_xy_bla_5" //changed 2nd occurence
You may define a regex that matches all occurrences and pass a callback method as the second argument to the replace method and add some custom logic there:
var mystr = 'bla_3_bla_14_bla_5';
function replaceOccurrence(string, regex, n, replace) {
var i = 0;
return string.replace(regex, function(match) {
i+=1;
if(i===n) return replace;
return match;
});
}
console.log(
replaceOccurrence(mystr, /\d+/g, 2, 'NUM')
)
Here, replaceOccurrence(mystr, /\d+/g, 2, 'NUM') takes mystr, searches for all digit sequences with /\d+/g and when it comes to the second occurrence, it replaces with a NUM substring.
var stringWithNumbers = "bla_3_bla_14_bla_5";
var n = 1;
var changedString = stringWithNumbers.replace(/[0-9]+/g,v => n++ == 2 ? "xy" : v);
console.log(changedString);