I'm having trouble with removing all characters up to and including the 3 third slash in JavaScript. This is my string:
http://blablab/test
The result should be:
test
Does anybody know the correct solution?
To get the last item in a path, you can split the string on / and then pop():
var url = "http://blablab/test";
alert(url.split("/").pop());
//-> "test"
To specify an individual part of a path, split on / and use bracket notation to access the item:
var url = "http://blablab/test/page.php";
alert(url.split("/")[3]);
//-> "test"
Or, if you want everything after the third slash, split(), slice() and join():
var url = "http://blablab/test/page.php";
alert(url.split("/").slice(3).join("/"));
//-> "test/page.php"
var string = 'http://blablab/test'
string = string.replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'')
alert(string)
This is a regular expression. I will explain below
The regex is /[\s\S]*\//
/ is the start of the regex
Where [\s\S] means whitespace or non whitespace (anything), not to be confused with . which does not match line breaks (. is the same as [^\r\n]).
* means that we match anywhere from zero to unlimited number of [\s\S]
\/ Means match a slash character
The last / is the end of the regex
var str = "http://blablab/test";
var index = 0;
for(var i = 0; i < 3; i++){
index = str.indexOf("/",index)+1;
}
str = str.substr(index);
To make it a one liner you could make the following:
str = str.substr(str.indexOf("/",str.indexOf("/",str.indexOf("/")+1)+1)+1);
You can use split to split the string in parts and use slice to return all parts after the third slice.
var str = "http://blablab/test",
arr = str.split("/");
arr = arr.slice(3);
console.log(arr.join("/")); // "test"
// A longer string:
var str = "http://blablab/test/test"; // "test/test";
You could use a regular expression like this one:
'http://blablab/test'.match(/^(?:[^/]*\/){3}(.*)$/);
// -> ['http://blablab/test', 'test]
A string’s match method gives you either an array (of the whole match, in this case the whole input, and of any capture groups (and we want the first capture group)), or null. So, for general use you need to pull out the 1th element of the array, or null if a match wasn’t found:
var input = 'http://blablab/test',
re = /^(?:[^/]*\/){3}(.*)$/,
match = input.match(re),
result = match && match[1]; // With this input, result contains "test"
let str = "http://blablab/test";
let data = new URL(str).pathname.split("/").pop();
console.log(data);
Related
i need to replace phone number in string on \n new line.
My string: Jhony Jhons,jhon#gmail.com,380967574366
I tried this:
var str = 'Jhony Jhons,jhon#gmail.com,380967574366'
var regex = /[0-9]/g;
var rec = str.trim().replace(regex, '\n').split(','); //Jhony Jhons,jhon#gmail.com,
Number replace on \n but after using e-mail extra comma is in the string need to remove it.
Finally my string should look like this:
Jhony Jhons,jhon#gmail.com\n
You can try this:
var str = 'Jhony Jhons,jhon#gmail.com,380967574366';
var regex = /,[0-9]+/g;
str.replace(regex, '\n');
The snippet above may output what you want, i.e. Jhony Jhons,jhon#gmail.com\n
There's a lot of ways to that, and this is so easy, so try this simple answer:-
var str = 'Jhony Jhons,jhon#gmail.com,380967574366';
var splitted = str.split(","); //split them by comma
splitted.pop(); //removes the last element
var rec = splitted.join() + '\n'; //join them
You need a regex to select the complete phone number and also the preceding comma. Your current regex selects each digit and replaces each one with an "\n", resulting in a lot of "\n" in the result. Also the regex does not match the comma.
Use the following regex:
var str = 'Jhony Jhons,jhon#gmail.com,380967574366'
var regex = /,[0-9]+$/;
// it replaces all consecutive digits with the condition at least one digit exists (the "[0-9]+" part)
// placed at the end of the string (the "$" part)
// and also the digits must be preceded by a comma (the "," part in the beginning);
// also no need for global flag (/g) because of the $ symbol (the end of the string) which can be matched only once
var rec = str.trim().replace(regex, '\n'); //the result will be this string: Jhony Jhons,jhon#gmail.com\n
var str = "Jhony Jhons,jhon#gmail.com,380967574366";
var result = str.replace(/,\d+/g,'\\n');
console.log(result)
I'm trying to obtain all possible matches from a string using regex with javascript. It appears that my method of doing this is not matching parts of the string that have already been matched.
Variables:
var string = 'A1B1Y:A1B2Y:A1B3Y:A1B4Z:A1B5Y:A1B6Y:A1B7Y:A1B8Z:A1B9Y:A1B10Y:A1B11Y';
var reg = /A[0-9]+B[0-9]+Y:A[0-9]+B[0-9]+Y/g;
Code:
var match = string.match(reg);
All matched results I get:
A1B1Y:A1B2Y
A1B5Y:A1B6Y
A1B9Y:A1B10Y
Matched results I want:
A1B1Y:A1B2Y
A1B2Y:A1B3Y
A1B5Y:A1B6Y
A1B6Y:A1B7Y
A1B9Y:A1B10Y
A1B10Y:A1B11Y
In my head, I want A1B1Y:A1B2Y to be a match along with A1B2Y:A1B3Y, even though A1B2Y in the string will need to be part of two matches.
Without modifying your regex, you can set it to start matching at the beginning of the second half of the match after each match using .exec and manipulating the regex object's lastIndex property.
var string = 'A1B1Y:A1B2Y:A1B3Y:A1B4Z:A1B5Y:A1B6Y:A1B7Y:A1B8Z:A1B9Y:A1B10Y:A1B11Y';
var reg = /A[0-9]+B[0-9]+Y:A[0-9]+B[0-9]+Y/g;
var matches = [], found;
while (found = reg.exec(string)) {
matches.push(found[0]);
reg.lastIndex -= found[0].split(':')[1].length;
}
console.log(matches);
//["A1B1Y:A1B2Y", "A1B2Y:A1B3Y", "A1B5Y:A1B6Y", "A1B6Y:A1B7Y", "A1B9Y:A1B10Y", "A1B10Y:A1B11Y"]
Demo
As per Bergi's comment, you can also get the index of the last match and increment it by 1 so it instead of starting to match from the second half of the match onwards, it will start attempting to match from the second character of each match onwards:
reg.lastIndex = found.index+1;
Demo
The final outcome is the same. Though, Bergi's update has a little less code and performs slightly faster. =]
You cannot get the direct result from match, but it is possible to produce the result via RegExp.exec and with some modification to the regex:
var regex = /A[0-9]+B[0-9]+Y(?=(:A[0-9]+B[0-9]+Y))/g;
var input = 'A1B1Y:A1B2Y:A1B3Y:A1B4Z:A1B5Y:A1B6Y:A1B7Y:A1B8Z:A1B9Y:A1B10Y:A1B11Y'
var arr;
var results = [];
while ((arr = regex.exec(input)) !== null) {
results.push(arr[0] + arr[1]);
}
I used zero-width positive look-ahead (?=pattern) in order not to consume the text, so that the overlapping portion can be rematched.
Actually, it is possible to abuse replace method to do achieve the same result:
var input = 'A1B1Y:A1B2Y:A1B3Y:A1B4Z:A1B5Y:A1B6Y:A1B7Y:A1B8Z:A1B9Y:A1B10Y:A1B11Y'
var results = [];
input.replace(/A[0-9]+B[0-9]+Y(?=(:A[0-9]+B[0-9]+Y))/g, function ($0, $1) {
results.push($0 + $1);
return '';
});
However, since it is replace, it does extra useless replacement work.
Unfortunately, it's not quite as simple as a single string.match.
The reason is that you want overlapping matches, which the /g flag doesn't give you.
You could use lookahead:
var re = /A\d+B\d+Y(?=:A\d+B\d+Y)/g;
But now you get:
string.match(re); // ["A1B1Y", "A1B2Y", "A1B5Y", "A1B6Y", "A1B9Y", "A1B10Y"]
The reason is that lookahead is zero-width, meaning that it just says whether the pattern comes after what you're trying to match or not; it doesn't include it in the match.
You could use exec to try and grab what you want. If a regex has the /g flag, you can run exec repeatedly to get all the matches:
// using re from above to get the overlapping matches
var m;
var matches = [];
var re2 = /A\d+B\d+Y:A\d+B\d+Y/g; // make another regex to get what we need
while ((m = re.exec(string)) !== null) {
// m is a match object, which has the index of the current match
matches.push(string.substring(m.index).match(re2)[0]);
}
matches == [
"A1B1Y:A1B2Y",
"A1B2Y:A1B3Y",
"A1B5Y:A1B6Y",
"A1B6Y:A1B7Y",
"A1B9Y:A1B10Y",
"A1B10Y:A1B11Y"
];
Here's a fiddle of this in action. Open up the console to see the results
Alternatively, you could split the original string on :, then loop through the resulting array, pulling out the the ones that match when array[i] and array[i+1] both match like you want.
I've matched a string successfully, but I need to split it and add some new segments to URL. If it is possible by regex, How to match url and extract two strings like in the example below?
Current result:
["domain.com/collection/430000000000000"]
Desired result:
["domain.com/collection/", "430000000000000"]
Current code:
var reg = new RegExp('domain.com\/collection\/[0-9]+');
var str = 'http://localhost:3000/#/domain.com/collection/430000000000000?page=0&layout=grid';
console.log(str.match(reg));
You want Regex Capture Groups.
Put the parts you want to extract into braces like this, each part forming a matching group:
new RegExp('(domain.com\/collection\/)([0-9]+)')
Then after matching, you can extract each group content by index, with index 0 being the whole string match, 1 the first group, 2 the second etc. (thanks for the addendum, jcubic!).
This is done with exec() on the regex string like described here:
/\d(\d)\d/.exec("123");
// → ["123", "2"]
First comes the whole match, then the group matches in the sequence they appear in the pattern.
You can declare an array and then fill it with the required values that you can capture with parentheses (thus, making use of capturing groups):
var reg = /(domain.com\/collection)\/([0-9]+)/g;
// ^ ^ ^ ^
var str = 'http://localhost:3000/#/domain.com/collection/430000000000000?page=0&layout=grid';
var arr = [];
while ((m = reg.exec(str)) !== null) {
arr.push(m[1]);
arr.push(m[2]);
}
console.log(arr);
Output: ["domain.com/collection", "430000000000000"]
I have tried to display characters other than alphabets in the particular string but it is displaying only the first char.
var myArray = /[^a-zA-Z]+/g.exec("cdAbb#2547dbsbz78678");
The reason it is only displaying the first character is because with using exec and the g modifier (global), this method is meant to be used in a loop for getting all sub matches.
var str = "cdAbb#2547dbsbz78678";
var re = /[^a-zA-Z]+/g;
var myArray;
while (myArray = re.exec(str)) {
console.log(myArray[0]);
}
Output
#2547
78678
If you were wanting to combine the matches you could use the following.
var str = "cdAbb#2547dbsbz78678",
res = str.match(/[\W\d]+/g).join('');
# => "#254778678"
Or do a replacement
str = str.replace(/[a-z]+/gi, '');
You can do:
"cdAbb#2547dbsbz78678".match(/[^a-zA-Z]+/g).join('');
//=> #254778678
RegExp.exec with g (global) modifier needs to run in loop to give you all the matches.
My string has [1212,1212],[1212,11212],...
I'd like to extract each value into an array for example I'd want 1212,1212 as one pair and evaluate a series of steps.
Tried /[[0-9],[0-9]]/ but It wasn't doing the task as I wanted. Basically I'm a noob in Regex, could someone please help.
Thanks in advance
You need some modifications for your regular expression for it to work correctly:
/\[[0-9]+,[0-9]+\]/g
You need to escape square brackets [ because they have special meaning.
[0-9] matches only one digits, you need the + quantifier to match one or more digits and thus [0-9]+.
Use the global modifier g to extract all matches.
Then you can extract all the values into an array like this:
var input = "[1212,1212],[1212,11212]";
var pattern = /\[[0-9]+,[0-9]+\]/g;
var result = [];
var currentMatch;
while((currentMatch = pattern.exec(input)) != null) {
result.push(currentMatch.toString());
}
result;
Or if you don't need to find the matches successively one at a time, then you can use String.match() as #Matthew Mcveigh did:
var input = "[1212,1212],[1212,11212]";
var result = input.match(/\[[0-9]+,[0-9]+\]/g);
It seems like you just need to match one or more digits before and after a comma, so you could do the following:
"[1212,1212],[1212,11212]".match(/\d+,\d+/g)
Which will give you the array: ["1212,1212", "1212,11212"]
To extract the pairs:
var result = "[1212,1212],[1212,11212]".match(/\d+,\d+/g);
for (var i = 0; i < result.length; i++) {
var pair = result[i].match(/\d+/g),
left = pair[0], right = pair[1];
alert("left: " + left + ", right: " + right);
}
You need to escape the literal brackets that you want to match. You can also use \d to match "any digit", which makes it tidier. Also, you're only matching one digit. You need to match "one or more" (+ quantifier)
/\[\d+,\d+\]/g
That g modifier finds all matches in the string, otherwise only the first one is found.