I have some issues, I need to "limit" character for specific word with special character (10 characters)
example in a textarea :
The #dog is here, I need a #rest and this is not #availableeeeeeeee for now
the word "availableeeeeeeee" needs to be cut when I reach 10 characters
Desired results
The #dog is here, I need a #rest and this is not #availablee for now
My question is how to limit characters for each word that containing a hashtag?
Thanks
1. Regex Solution:
You can use .replace() method with the following regex /(#\w{10})\[\w\d\]+/g, it will remove the extra characters:
str = str.replace(/(#\w{10})[\w\d]+/g, '$1');
Demo:
var str = "The #dog is here, I need a #rest and this is not #availableeeeeeeee for now";
str = str.replace(/(#\w{10})[\w\d]+/g, '$1');
console.log(str);
Note:
This regex matches the words starting with # using a matching group to get only the first 10 characters.
Full match #availableeeeeeeee
Group 1. n/a #availablee
And the .replace() call will keep only the matched group from the regex and skip the extra characters.
Note that you need to attach this code in the onchange event handler of your textarea.
2. split() Solution:
If you want to go with a solution that doesn't use Regex, you can use .split() method with Array.prototype.map() like this:
str = str.split(" ").map(function(item){
return item.startsWith("#") && item.length > 11 ? item.substr(0,11) : item;
}).join(" ");
Demo:
var str = "The #dog is here, I need a #rest and this is not #availableeeeeeeee for now";
str = str.split(" ").map(function(item){
return item.startsWith("#") && item.length > 11 ? item.substr(0,11) : item;
}).join(" ");
console.log(str);
a simple solution with javascript could be, to split text area all words into array. iterate it and validate word length.
var value = $('#text').val();
var maxSize = 10;
var words = value.trim().replace(regex, ' ').split(' ');
for(var wlength= 0 ; wlength < words.length; wlength++)
{
if(words[wlength] > maxSize)
{
alert('size exceeds max allowed');
}
}
you can try not allowing typing itself after 10 characters for any word by regular expression inline validation in HTML directly.
Well, I think you can try the following.
Using split() method will cut the string in words, then forEach word if it startsWith a '#', we substr it up to 10 + 1 characters. Finally, join everybody to obtain the final result :).
string="The #dog is here, I need a #rest and this is not #availableeeeeeeee for now"
var result = []
string.split(" ").forEach(function(item){
if (item.startsWith("#")){
result.push(item.substr(0,11));
} else result.push(item);
});
console.log(result.join(" "));
Related
I have this string:
var s = '/channels/mtb/videos?page=2&per_page=100&fields=uri%2Cname%2Cdescription%2Cduration%2Cwidth%2Cheight%2Cprivacy%2Cpictures.sizes&sort=date&direction=asc&filter=embeddable&filter_embeddable=true'
I want to repace per_page number (in this case 100, but it can be any number from 1-100, maybe more?)
I can select first part of the string with:
var s1 = s.substr(0, s.lastIndexOf('per_page=')+9)
which give me:
/channels/mtb/videos?page=2&per_page=
but how would I select next '&' after that so I can replace number occurrence?
dont assume same order of parameters!
You can use following regex to replace the content you want.
regex:- /per_page=[\d]*/g(this is only for your requirement)
var new_no=12; //change 100 to 12
var x='/channels/mtb/videos?page=2&per_page=100&fields=uri%2Cname%2Cdescription%2Cduration%2Cwidth%2Cheight%2Cprivacy%2Cpictures.sizes&sort=date&direction=asc&filter=embeddable&filter_embeddable=true';
var y=x.replace(/per_page=[\d]*/g,'per_page='+new_no);
console.log(y);
Explanation:-
/per_page=[\d]*/g
/ ----> is for regex pattern(it inform that from next character onward whatever it encounter will be regex pattern)
per_page= ----> try to find 'per_page=' in string
[\d]* ----> match 0 or more digit (it match until non digit encounter)
/g ---->/ to indicate end of regex pattern and 'g' is for global means find in all string(not only first occurrence)
Use replace with a regular expression to find the numbers after the text per_page=. Like this:
s.replace(/per_page=\d+/,"per_page=" + 33)
Replace the 33 with the number you want.
Result:
"/channels/mtb/videos?page=2&per_page=33&fields=uri%2Cname%2Cdescription%2Cduration%2Cwidth%2Cheight%2Cprivacy%2Cpictures.sizes&sort=date&direction=asc&filter=embeddable&filter_embeddable=true"
Start with the index from the lastIndexOf-per_page instead of 0.
Get the index of the first & and create a substr s2 to the end.
Then concat s1 + nr + s2.
I would not use regex, because it is much slower for this simple stuff.
With Array.filter you can do this, where one split the text into key/value pairs, and filter out the one that starts with per_page=.
Stack snippet
var s = '/channels/mtb/videos?page=2&per_page=100&fields=uri%2Cname%2Cdescription%2Cduration%2Cwidth%2Cheight%2Cprivacy%2Cpictures.sizes&sort=date&direction=asc&filter=embeddable&filter_embeddable=true'
var kv_pairs = s.split('&');
var s2 = s.replace((kv_pairs.filter(w => w.startsWith('per_page=')))[0],'per_page=' + 123);
//console.log(s2);
var matches = /(.*\bper_page=)(\d+)(.*)/;
if (matches) {
s = matches[0] + newValue + matches[2];
}
Here, I want to give each character a space except for the last alphabetic one:
var str = "test"
var result = "";
for (var i = 0; i < str.length; i++) {
result += (/*condition*/) ? str[i]
: str[i] + " ";
}
console.log(result);
So it prints "t e s t".
I tried this (i === str.length - 1) but it didn't work when a string had period(.) as it's last character ("test.") while I wanna target only alphabetics.
You can use a regular expression to remove all non-alphabetical characters first, and then do a split/join combination to insert the spaces (or use another regex):
var str = "don't test.";
var result = str.replace(/[^a-z]/gi, '').split('').join(' ');
console.log(result);
"testasdf. asdf asdf asd.d?".replace(/./g,"$& ").replace(/([A-Za-z]) ([^A-Za-z]*)$/, '$1$2')
the first replace add a space to all char
the second replace removes the space after the last letter
console.log("testasdf?".replace(/./g,"$& ").replace(/([A-Za-z]) ([^A-Za-z]*)$/, '$1$2'));
console.log("Super test ! Yes".replace(/./g,"$& ").replace(/([A-Za-z]) ([^A-Za-z]*)$/, '$1$2'));
There is such a feature like lookahead assertions.
So it could be
str.replace(/(\w)(?=.*\w)/, "$1")
Using a regex replace function, you can space out all of your characters like this:
"test string, with several words.".replace(/\w{2,}/g, match => match.split('').join(' '))
Explanation
\w{2,} match 2 or more alphabetic characters
match.split('').join(' ') split each match into characters, and rejoin with spaces between
I have a string and want to add a colon after every 2nd character (but not after the last set), eg:
12345678
becomes
12:34:56:78
I've been using .replace(), eg:
mystring = mystring.replace(/(.{2})/g, NOT SURE WHAT GOES HERE)
but none of the regex for : I've used work and I havent been able to find anything useful on Google.
Can anyone point me in the right direction?
Without the need to remove any trailing colons:
mystring = mystring.replace(/..\B/g, '$&:')
\B matches a zero-width non-word boundary; in other words, when it hits the end of the string, it won't match (as that is considered to be a word boundary) and therefore won't perform the replacement (hence no trailing colon, either).
$& contains the matched substring (so you don't need to use a capture group).
mystring = mystring.replace(/(..)/g, '$1:').slice(0,-1)
This is what comes to mind immediately. I just strip off the final character to get rid of the colon at the end.
If you want to use this for odd length strings as well, you just need to make the second character optional. Like so:
mystring = mystring.replace(/(..?)/g, '$1:').slice(0,-1)
If you're looking for approach other than RegEx, try this:
var str = '12345678';
var output = '';
for(var i = 0; i < str.length; i++) {
output += str.charAt(i);
if(i % 2 == 1 && i > 0) {
output += ':';
}
}
alert(output.substring(0, output.length - 1));
Working JSFiddle
A somewhat different approach without regex could be using Array.prototype.reduce:
Array.prototype.reduce.call('12345678', function(acc, item, index){
return acc += index && index % 2 === 0 ? ':' + item : item;
}, ''); //12:34:56:78
mystring = mytring.replace(/(.{2})/g, '\:$1').slice(1)
try this
Easy, just match every group of up-to 2 characters and join the array with ':'
mystring.match(/.{1,2}/g).join(':')
var mystring = '12345678';
document.write(mystring.match(/.{1,2}/g).join(':'))
no string slicing / trimming required.
It's easier if you tweak what you're searching for to avoid an end-of-line colon(using negative lookahead regex)
mystring = mystring.replace(/(.{2})(?!$)/g, '\$1:');
mystring = mystring.replace(/(.{2})/g, '$1\:')
Give that a try
I like my approach the best :)
function colonizer(strIn){
var rebuiltString = '';
strIn.split('').forEach(function(ltr, i){
(i % 2) ? rebuiltString += ltr + ':' : rebuiltString += ltr;
});
return rebuiltString;
}
alert(colonizer('Nicholas Abrams'));
Here is a demo
http://codepen.io/anon/pen/BjjNJj
I am trying to remove some spaces from a few dynamically generated strings. Which space I remove depends on the length of the string. The strings change all the time so in order to know how many spaces there are, I iterate over the string and increment a variable every time the iteration encounters a space. I can already remove all of a specific type of character with str.replace(' ',''); where 'str' is the name of my string, but I only need to remove a specific occurrence of a space, not all the spaces. So let's say my string is
var str = "Hello, this is a test.";
How can I remove ONLY the space after the word "is"? (Assuming that the next string will be different so I can't just write str.replace('is ','is'); because the word "is" might not be in the next string).
I checked documentation on .replace, but there are no other parameters that it accepts so I can't tell it just to replace the nth instance of a space.
If you want to go by indexes of the spaces:
var str = 'Hello, this is a test.';
function replace(str, indexes){
return str.split(' ').reduce(function(prev, curr, i){
var separator = ~indexes.indexOf(i) ? '' : ' ';
return prev + separator + curr;
});
}
console.log(replace(str, [2,3]));
http://jsfiddle.net/96Lvpcew/1/
As it is easy for you to get the index of the space (as you are iterating over the string) , you can create a new string without the space by doing:
str = str.substr(0, index)+ str.substr(index);
where index is the index of the space you want to remove.
I came up with this for unknown indices
function removeNthSpace(str, n) {
var spacelessArray = str.split(' ');
return spacelessArray
.slice(0, n - 1) // left prefix part may be '', saves spaces
.concat([spacelessArray.slice(n - 1, n + 1).join('')]) // middle part: the one without the space
.concat(spacelessArray.slice(n + 1)).join(' '); // right part, saves spaces
}
Do you know which space you want to remove because of word count or chars count?
If char count, you can Rafaels Cardoso's answer,
If word count you can split them with space and join however you want:
var wordArray = str.split(" ");
var newStr = "";
wordIndex = 3; // or whatever you want
for (i; i<wordArray.length; i++) {
newStr+=wordArray[i];
if (i!=wordIndex) {
newStr+=' ';
}
}
I think your best bet is to split the string into an array based on placement of spaces in the string, splice off the space you don't want, and rejoin the array into a string.
Check this out:
var x = "Hello, this is a test.";
var n = 3; // we want to remove the third space
var arr = x.split(/([ ])/); // copy to an array based on space placement
// arr: ["Hello,"," ","this"," ","is"," ","a"," ","test."]
arr.splice(n*2-1,1); // Remove the third space
x = arr.join("");
alert(x); // "Hello, this isa test."
Further Notes
The first thing to note is that str.replace(' ',''); will actually only replace the first instance of a space character. String.replace() also accepts a regular expression as the first parameter, which you'll want to use for more complex replacements.
To actually replace all spaces in the string, you could do str.replace(/ /g,""); and to replace all whitespace (including spaces, tabs, and newlines), you could do str.replace(/\s/g,"");
To fiddle around with different regular expressions and see what they mean, I recommend using http://www.regexr.com
A lot of the functions on the JavaScript String object that seem to take strings as parameters can also take regular expressions, including .split() and .search().
I have a string with this format:
#someID#tn#company#somethingNew#classing#somethingElse#With
There might be unlimited #-separated words, but definitely the whole string begins with #
I have written the following regexp, though it matches it, but I cannot get each #-separated word, and what I get is the last recursion and the first (as well as the whole string). How can I get an array of every word in an element separately?
(?:^\#\w*)(?:(\#\w*)+) //I know I have ruled out second capturing group with ?: , though doesn't make much difference.
And here is my Javascript code:
var reg = /(?:^\#\w*)(?:(\#\w*)+)/g;
var x = null;
while(x = reg.exec("#someID#tn#company#somethingNew#classing#somethingElse#With"))
{
console.log(x);
}
And here is the result (Firebug, console):
["#someID#tn#company#somet...sing#somethingElse#With", "#With"]
0
"#someID#tn#company#somet...sing#somethingElse#With"
1
"#With"
index
0
input
"#someID#tn#company#somet...sing#somethingElse#With"
EDIT :
I want an output like this with regular expression if possible:
["#someID", "#tn", #company", "#somethingNew", "#classing", "#somethingElse", "#With"]
NOTE that I want a RegExp solution. I know about String.split() and String operations.
You can use:
var s = '#someID#tn#company#somethingNew#classing#somethingElse#With'
if (s.substr(0, 1) == "#")
tok = s.substr(1).split('#');
//=> ["someID", "tn", "company", "somethingNew", "classing", "somethingElse", "With"]
You could try this regex also,
((?:#|#)\w+)
DEMO
Explanation:
() Capturing groups. Anything inside this capturing group would be captured.
(?:) It just matches the strings but won't capture anything.
#|# Literal # or # symbol.
\w+ Followed by one or more word characters.
OR
> "#someID#tn#company#somethingNew#classing#somethingElse#With".split(/\b(?=#|#)/g);
[ '#someID',
'#tn',
'#company',
'#somethingNew',
'#classing',
'#somethingElse',
'#With' ]
It will be easier without regExp:
var str = "#someID#tn#company#somethingNew#classing#somethingElse#With";
var strSplit = str.split("#");
for(var i = 1; i < strSplit.length; i++) {
strSplit[i] = "#" + strSplit[i];
}
console.log(strSplit);
// ["#someID", "#tn", "#company", "#somethingNew", "#classing", "#somethingElse", "#With"]