I'm trying to find if a given string of digits contains a sequence of three identical digits.
using a for loop, each digit in the string gets its own representation of a three digit sequence which is then checked against the string using Regex:
var str = "6854777322"
for(var i=0; i<str.length; i++)
{
seqToCompare = str[i] + str[i] + str[i];
var re = new RegExp(seqToCompare, "g");
if(str.match(re).length == 1)
{
match = str[i];
}
}
console.log(match)
The result should be seven (if I put 777 in seqToCompare, it would work), but it looks like the concatenation causes it to fail. Console shows "cannot read property length for null".
You can test it here - https://jsfiddle.net/kwnL7vLs/
I tried .toString, setting seqToCompare in Regex format and even parsing it as int (out of desperation for not knowing what to do anymore...)
Rather than looping over each character, you can use a simple regex to get a digit that is repeated 3 times:
/(\d)(?=\1{2})/
(\d) - Here we match a digit and group it in captured group #1
(?=\1{2}) is lookahead that asserts same captured group #1 is repeated twice ahead of current position
RegEx Demo
anubhava's answer is the way to go, as it's more efficient and simpler. However, if you're wondering why your code specifically is giving an error, it's because you try to find the length property of the return value of str.match(), even when no match is found.
Try this instead:
var str = "6854777322"
for(var i=0; i<str.length; i++)
{
seqToCompare = str[i] + str[i] + str[i];
var re = new RegExp(seqToCompare, "g");
if(str.match(re))
{
match = str[i];
}
}
console.log(match)
Related
I would like to get all possible matches of a string with forward slashes '/' using regex.
I would like to regex that matches all the possibilities of a string between slashes but excludes a part which has no ending '/'
For example a string /greatgrandparent/grandparent/parent/child
it should return something like this:
/greatgrandparent/
/greatgrandparent/grandparent/
/greatgrandparent/grandparent/parent/
The following regex that will get each word that begins with a / and a positive lookahead for the / character is this /\/\w+(?=\/)/g
You can use the match() function that will place each word it finds in an array. You can then loop through the array to combine the different results. Check out the snippet below.
var str = `/greatgrandparent/grandparent/parent/child`;
var strArr = str.match(/\/\w+(?=\/)/g);
console.log(strArr);
var strLoop = ``;
for (i = 0; i < strArr.length; i++) {
strLoop += strArr[i];
document.write(`${strLoop}<br>`);
}
I've been struggling getting my regex function to work as intended. My goal is to iterate endlessly over a string (until no match is found) and remove all duplicate, adjacent characters. Aside from checking if 2 characters (adjacent of each other) are equal, the regex should only remove the match when one of the pair is uppercase.
e.g. the regex should only remove 'Xx' or 'xX'.
My current regex only removes matches where a lowercase character is followed by any uppercase character.
(.)(([a-z]{0})+[A-Z])
How can I implement looking for the same adjacent character and the pattern of looking for an uppercase character followed by an equal lowercase character?
You'd either have to list out all possible combinations, eg
aA|Aa|bB|Bb...
Or implement it more programatically, without regex:
let str = 'fooaBbAfoo';
outer:
while (true) {
for (let i = 0; i < str.length - 1; i++) {
const thisChar = str[i];
const nextChar = str[i + 1];
if (/[a-z]/i.test(thisChar) && thisChar.toUpperCase() === nextChar.toUpperCase() && thisChar !== nextChar) {
str = str.slice(0, i) + str.slice(i + 2);
continue outer;
}
}
break;
}
console.log(str);
Looking for the same adjacent character: /(.)\1/
Looking for an uppercase character followed by an equal lowercase character isn't possible in JavaScript since it doesn't support inline modifiers. If they were regex should be: /(.)(?!\1)(?i:\1)/, so it matches both 'xX' or 'Xx'
code for detecting repeating letter in a string.
var str="paraven4sr";
var hasDuplicates = (/([a-zA-Z])\1+$/).test(str)
alert("repeating string "+hasDuplicates);
I am getting "false" as output for the above string "paraven4sr". But this code works correctly for the strings like "paraaven4sr". i mean if the character repeated consecutively, code gives output as "TRUE". how to rewrite this code so that i ll get output as "TRUE" when the character repeats in a string
JSFIDDLE
var str="paraven4sr";
var hasDuplicates = (/([a-zA-Z]).*?\1/).test(str)
alert("repeating string "+hasDuplicates);
The regular expression /([a-zA-Z])\1+$/ is looking for:
([a-zA-Z]]) - A letter which it captures in the first group; then
\1+ - immediately following it one or more copies of that letter; then
$ - the end of the string.
Changing it to /([a-zA-Z]).*?\1/ instead searches for:
([a-zA-Z]) - A letter which it captures in the first group; then
.*? - zero or more characters (the ? denotes as few as possible); until
\1 - it finds a repeat of the first matched character.
If you have a requirement that the second match must be at the end-of-the-string then you can add $ to the end of the regular expression but from your text description of what you wanted then this did not seem to be necessary.
Try this:
var str = "paraven4sr";
function checkDuplicate(str){
for(var i = 0; i < str.length; i++){
var re = new RegExp("[^"+ str[i] +"]","g");
if(str.replace(re, "").length >= 2){
return true;
}
}
return false;
}
alert(checkDuplicate(str));
Here is jsfiddle
To just test duplicate alphanumeric character (including underscore _):
console.log(/(\w)\1+/.test('aab'));
Something like this?
String.prototype.count=function(s1) {
return (this.length - this.replace(new RegExp(s1,"g"), '').length) / s1.length;
}
"aab".count("a") > 1
EDIT: Sorry, just read that you are not searching for a function to find whether a letter is found more than once but to find whether a letter is a duplicate. Anyway, I leave this function here, maybe it can help. Sorry ;)
I am trying to create a regex which will ultimately be used with Google Forms to validate a texarea input.
The rule is,
Input area can have one or more URLs (http or https)
Each URL must be separated either by one or more new lines
Each line which has text, must be a single valid URL
Last URL may have or may not have new line character/s after it
Till now, I have written this regex ^(https?://.+[\r\n]+)*(https?://.+[\r\n]+?)$ but the problem is that if a line has more than 1 url, it validates that too.
Here is my testing playground: http://goo.gl/YPdvBH.
Here is what you are looking for
Demo , Demo with your URLS
function validate(ele) {
str = ele.value;
str = str.replace(/\r/g, "");
while (/\s\n/.test(str)) {
str = str.replace(/\s\n/g, "\n");
}
while (/\n\n/.test(str)) {
str = str.replace(/\n\n/g, "\n");
}
ele.value = str;
str = str.replace(/\n/g, "_!_&_!_").split("_!_&_!_")
var result = [], counter = 0;
for (var i = 0; i < str.length; i++) {
str[i] = str[i].replace(/(?:(?:^|\n)\s+|\s+(?:$|\n))/g, '').replace(/\s+/g, ' ');
if(str[i].length !== 0){
if (isValidAddress(str[i])) {
result.push(str[i]);
}
counter += 1;
}
}
function isValidAddress(s) {
return /^(https?|ftp):\/\/(((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:)*#)?(((\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5]))|((([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.)+(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.?)(:\d*)?)(\/((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)+(\/(([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)*)*)?)?(\?((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|[\uE000-\uF8FF]|\/|\?)*)?(\#((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|\/|\?)*)?$/i.test(s)
}
return (result.length === str.length);
}
var ele = document.getElementById('urls');
validate(ele);
This is closer to the regex you are looking for:
^(https?://[\S]+[\r\n]+)*(https?://[\S]+[\r\n]+?)$
The difference between your regex and this one is that you use .+ which will match all characters except newline whereas I use [\S]+ (note it is a capital S) which will match all non-whitespace characters. So, this doesn't match more than one token on one line. Hence, on each line you can match at max one token and that must be of the form that you have defined.
For a regex to match a single URL, look at this question on StackOverflow:
What is the best regular expression to check if a string is a valid URL?
I don't know whether google-forms have a length limit. But if they have, it is sure to almost bounce into it.
If i understand right - in your regexp missing m flag for multiline, so you need something like this
/^(https?://.+this your reg exp for one url)$/m
sample with regexp from Javascript URL validation regex
/^(ht|f)tps?:\/\/[a-z0-9-\.]+\.[a-z]{2,4}\/?([^\s<>\#%"\,\{\}\\|\\\^\[\]`]+)?$/m
base on the following string
...here..
..there...
.their.here.
How can i remove the . on the beginning and end of string like the trim that removes all spaces, using javascript
the output should be
here
there
their.here
These are the reasons why the RegEx for this task is /(^\.+|\.+$)/mg:
Inside /()/ is where you write the pattern of the substring you want to find in the string:
/(ol)/ This will find the substring ol in the string.
var x = "colt".replace(/(ol)/, 'a'); will give you x == "cat";
The ^\.+|\.+$ in /()/ is separated into 2 parts by the symbol | [means or]
^\.+ and \.+$
^\.+ means to find as many . as possible at the start.
^ means at the start; \ is to escape the character; adding + behind a character means to match any string containing one or more that character
\.+$ means to find as many . as possible at the end.
$ means at the end.
The m behind /()/ is used to specify that if the string has newline or carriage return characters, the ^ and $ operators will now match against a newline boundary, instead of a string boundary.
The g behind /()/ is used to perform a global match: so it find all matches rather than stopping after the first match.
To learn more about RegEx you can check out this guide.
Try to use the following regex
var text = '...here..\n..there...\n.their.here.';
var replaced = text.replace(/(^\.+|\.+$)/mg, '');
Here is working Demo
Use Regex /(^\.+|\.+$)/mg
^ represent at start
\.+ one or many full stops
$ represents at end
so:
var text = '...here..\n..there...\n.their.here.';
alert(text.replace(/(^\.+|\.+$)/mg, ''));
Here is an non regular expression answer which utilizes String.prototype
String.prototype.strim = function(needle){
var first_pos = 0;
var last_pos = this.length-1;
//find first non needle char position
for(var i = 0; i<this.length;i++){
if(this.charAt(i) !== needle){
first_pos = (i == 0? 0:i);
break;
}
}
//find last non needle char position
for(var i = this.length-1; i>0;i--){
if(this.charAt(i) !== needle){
last_pos = (i == this.length? this.length:i+1);
break;
}
}
return this.substring(first_pos,last_pos);
}
alert("...here..".strim('.'));
alert("..there...".strim('.'))
alert(".their.here.".strim('.'))
alert("hereagain..".strim('.'))
and see it working here : http://jsfiddle.net/cettox/VQPbp/
Slightly more code-golfy, if not readable, non-regexp prototype extension:
String.prototype.strim = function(needle) {
var out = this;
while (0 === out.indexOf(needle))
out = out.substr(needle.length);
while (out.length === out.lastIndexOf(needle) + needle.length)
out = out.slice(0,out.length-needle.length);
return out;
}
var spam = "this is a string that ends with thisthis";
alert("#" + spam.strim("this") + "#");
Fiddle-ige
Use RegEx with javaScript Replace
var res = s.replace(/(^\.+|\.+$)/mg, '');
We can use replace() method to remove the unwanted string in a string
Example:
var str = '<pre>I'm big fan of Stackoverflow</pre>'
str.replace(/<pre>/g, '').replace(/<\/pre>/g, '')
console.log(str)
output:
Check rules on RULES blotter