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>`);
}
Related
I have this string:
a\,bcde,fgh,ijk\,lmno,pqrst\,uv
I need a JavaScript function that will split the string by every , but only those that don't have a \ before them
How can this be done?
Here's the shortest thing I could come up with:
'a\\,bcde,fgh,ijk\\,lmno,pqrst\\,uv'.replace(/([^\\]),/g, '$1\u000B').split('\u000B')
The idea behind is to find every place where comma isn't prefixed with a backslash, replace those with string that is uncommon to come up in your strings and then split by that uncommon string.
Note that backslashes before commas have to be escaped using another backslash. Otherwise, javascript treats form \, as escaped comma and produce simply a comma out of it! In other words if you won't escape the backslash, javascript sees this: a\,bcde,fgh,ijk\,lmno,pqrst\,uv as this a,bcde,fgh,ijk,lmno,pqrst,uv.
Since regular expressions in JavaScript does not support lookbehinds, I'm not going to cook up a giant hack to mimic this behavior. Instead, you can just split() on all commas (,) and then glue back the pieces that shouldn't have been split in the first place.
Quick 'n' dirty demo:
var str = 'a\\,bcde,fgh,ijk\\,lmno,pqrst\\,uv'.split(','), // Split on all commas
out = []; // Output
for (var i = 0, j = str.length - 1; i < j; i++) { // Iterate all but last (last can never be glued to non-existing next)
var curr = str[i]; // This piece
if (curr.charAt(curr.length - 1) == '\\') { // If ends with \ ...
curr += ',' + str[++i]; // ... glue with next and skip next (increment i)
}
out.push(curr); // Add to output
}
Another ugly hack around the lack of look-behinds:
function rev(s) {
return s.split('').reverse().join('');
}
var s = 'a\\,bcde,fgh,ijk\\,lmno,pqrst\\,uv';
// Enter bizarro world...
var r = rev(s);
// Split with a look-ahead
var rparts = r.split(/,(?!\\)/);
// And put it back together with double reversing.
var sparts = [ ];
while(rparts.length)
sparts.push(rev(rparts.pop()));
for(var i = 0; i < sparts.length; ++i)
$('#out').append('<pre>' + sparts[i] + '</pre>');
Demo: http://jsfiddle.net/ambiguous/QbBfw/1/
I don't think I'd do this in real life but it works even if it does make me feel dirty. Consider this a curiosity rather than something you should really use.
In case if need remove backslashes also:
var test='a\\.b.c';
var result = test.replace(/\\?\./g, function (t) { return t == '.' ? '\u000B' : '.'; }).split('\u000B');
//result: ["a.b", "c"]
In 2022 most of browsers support lookbehinds:
https://caniuse.com/js-regexp-lookbehind
Safari should be your only concern.
With a lookbehind you can split your string this way:
"a\\,bcde,fgh,ijk\\,lmno,pqrst\\,uv".split(/(?<!\\),/)
// => ['a\\,bcde', 'fgh', 'ijk\\,lmno', 'pqrst\\,uv']
You can use regex to do the split.
Here is the link to regex in javascript http://www.w3schools.com/jsref/jsref_obj_regexp.asp
Here is the link to other post where the author have used regex for split Javascript won't split using regex
From the first link if you note you can create a regular expression using
?!n Matches any string that is not followed by a specific string n
[,]!\\
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)
in google script I am trying to replace a %string basing on the character following it.
I've tried using:
var indexOfPercent = newString.indexOf("%");
and then check the character of indexOfPercent+1, but indexOf returns only the first occurrence of '%'.
How can I get all occurrences? Maybe there is easier way to do that (regular expressions)?
EDIT:
Finally I want to replace all my % occurrences to %%, but not if percent sign was part of %# or %#.
To sum up: my string: Test%# Test2%s Test3%. should look like: Test%# Test2%s Test3%%.
I've tried using something like this:
//?!n Matches any string that is not followed by a specific string n
//(x|y) Find any of the alternatives specified
var newString = newString.replace(\%?![s]|\%?![%], "%%")
but it didn't find any strings. I am not familiar with regex's, so maybe it is a simple mistake.
Thanks
Try this code:
// replace all '%'
var StrPercent = '%100%ffff%';
var StrNoPersent = StrPercent.replace(/\%/g,'');
Logger.log(StrNoPersent); // 100ffff
Look for more info here
Edit
In your case you need RegEx with the character not followed by group of characters. Similiar question was asked here:
Regular expressions - how to match the character '<' not followed by ('a' or 'em' or 'strong')?
Thy this code:
function RegexNotFollowedBy() {
var sample = ['Test%#',
'Test2%s',
'Test3%',
'%Test4%'];
var RegEx = /%(?!s|#)/g;
var Replace = "%%";
var str, newStr;
for (var i = 0; i < sample.length; i++) {
str = sample[i];
newStr = str.replace(RegEx, Replace);
Logger.log(newStr);
}
}
I'll explain expression /%(?!s|#)/g:
% -- look '%'
(text1|text2|text3...|textN) -- not followed by text1, 2 etc.
g -- look for any accurance of searched text
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<>\#%"\,\{\}\\|\\\^\[\]`]+)?$/m
I'm having trouble with removing all characters up to and including the 3 third slash in JavaScript. This is my string:
http://blablab/test
The result should be:
test
Does anybody know the correct solution?
To get the last item in a path, you can split the string on / and then pop():
var url = "http://blablab/test";
alert(url.split("/").pop());
//-> "test"
To specify an individual part of a path, split on / and use bracket notation to access the item:
var url = "http://blablab/test/page.php";
alert(url.split("/")[3]);
//-> "test"
Or, if you want everything after the third slash, split(), slice() and join():
var url = "http://blablab/test/page.php";
alert(url.split("/").slice(3).join("/"));
//-> "test/page.php"
var string = 'http://blablab/test'
string = string.replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'')
alert(string)
This is a regular expression. I will explain below
The regex is /[\s\S]*\//
/ is the start of the regex
Where [\s\S] means whitespace or non whitespace (anything), not to be confused with . which does not match line breaks (. is the same as [^\r\n]).
* means that we match anywhere from zero to unlimited number of [\s\S]
\/ Means match a slash character
The last / is the end of the regex
var str = "http://blablab/test";
var index = 0;
for(var i = 0; i < 3; i++){
index = str.indexOf("/",index)+1;
}
str = str.substr(index);
To make it a one liner you could make the following:
str = str.substr(str.indexOf("/",str.indexOf("/",str.indexOf("/")+1)+1)+1);
You can use split to split the string in parts and use slice to return all parts after the third slice.
var str = "http://blablab/test",
arr = str.split("/");
arr = arr.slice(3);
console.log(arr.join("/")); // "test"
// A longer string:
var str = "http://blablab/test/test"; // "test/test";
You could use a regular expression like this one:
'http://blablab/test'.match(/^(?:[^/]*\/){3}(.*)$/);
// -> ['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);