This regular expression looks for words with 3 or less characters so that a non-breaking space can be placed in before them.
smallwords = /(\s|^)(([a-zA-Z-_(]{1,2}('|’)*[a-zA-Z-_,;]{0,1}?\s)+)/gi, // words with 3 or less characters
Is there a way, to make the expression only apply itself to 2 words in a row?
Example
Currently, the string:
Singapore, the USA and Vietnam.
will be turned into:
Singapore, the USA and Vietnam.
if the expression only applied to 2 words in a row it would show
Singapore, the USA and Vietnam.
here's the full script:
ragadjust = function (s, method) {
if (document.querySelectorAll) {
var eles = document.querySelectorAll(s),
elescount = eles.length,
smallwords = /(\s|^)(([a-zA-Z-_(]{1,2}('|’)*[a-zA-Z-_,;]{0,1}?\s)+)/gi, // words with 3 or less characters
while (elescount-- > 0) {
var ele = eles[elescount],
elehtml = ele.innerHTML;
if (method == 'small-words' || method == 'all')
// replace small words
elehtml = elehtml.replace(smallwords, function(contents, p1, p2) {
return p1 + p2.replace(/\s/g, ' ');
});
ele.innerHTML = elehtml;
}
}
};
This is from RagAdjust
I know that this is not what you are asking for, but I figured a code review wouldn't hurt:
I think the word boundary \b is better, in this case, than \s|^.
You have the A-Z and a-z characters in your match, yet you are use the i case insensitive operator.
{0,1}? is redundant - either use the ? to make it optional, or use {0,1} to make it match zero or one times.
If your are going to have a dash in your character set put it at the end so that you don't have an ambiguous regex, for example this [a-z_-] is much better than [a-z-_].
If you don't need to capture a value, use the non-capturing parenthesis (?:).
So, here's your cleaned up regex:
/\b((?:[a-z_(-]{1,2}(?:'|’)*[a-z_,;-]?\s)+)/gi
I'm pretty sure the '|’ bit is some sort of typo when you pasted this in from your editor. Not sure what it is supposed to be.
This doesn't quite solve the issue the way you suggested but it does reduce the number of non breaking spaces that end up in the string. But it might give you some insight. Because you have the trailing g on both regex replacements, you're doing global replace. If you instead loop it with some max number of fixes, things work out a little differently.
Try changing the max number of replacements. I think the other thing that happens here (in my modified code) is that after you make one replacement, the spaces and small words are gone because you jammed in a nbsp which may or may not solve the issue you're trying to get around.
Here's my replacement function (simplified from your original). The basic mod is to remove the g from the regex's and add the loop. You should check out the codepen to see the full deal
var new_ragadjust = function (contents) {
MAX_NUMBER_OF_REPLACEMENTS = 5;
smallwords = /(\s|^)(([a-zA-Z-_(]{1,2}('|’)*[a-zA-Z-_,;]{0,1}?\s)+)/i; // words with 3 or less characters
var ii = 0;
var c = contents;
for (;ii < MAX_NUMBER_OF_REPLACEMENTS; ++ii) {
c = c.replace(smallwords, function(contents, p1, p2) {
return p1 + p2.replace(/\s/, ' ');
});
}
return c;
};
Codepen
http://cdpn.io/DKLtc
Also, to see the difference, you need to inspect elements to actually see where the nbsps end up (as you probably already knew).
Related
I'm using RPG Maker MV which is a game creator that uses JavaScript to create plugins. I have a plugin in JavaScript already, however I'm trying to edit a part of the plugin so that it basically checks if a certain string exists in a character in the game and if it does, then sets specific variables to numbers within that string.
for (var i = 0; i < page.list.length; i++) {
if (page.list[i].code == 108 && page.list[i].parameters[0].contains("<post:" + (n) + "," + (n) + ">")) {
var post = page.list[i].parameters[0];
var array = post.split(',');
this._origMovement.x = Number(array[1]);
this._origMovement.y = Number(array[1]);
break;
};
};
So I know the first 2 lines work and contains works when I only put a specific string. However I can't figure out how to check for 2 numbers that are separated by a comma and wrapped in '<>' tags, without knowing what the numbers would be.
Then it needs to extract those numbers and assign one to this._origMovement.x and the other to this._origMovement.y.
Any help would be greatly appreciated.
This is one of those rare cases where I'd use a regular expression. If you haven't come across regular expressions before I suggest reading an introduction to them, such as this one: https://regexone.com/
In your case, you probable want something like this:
var myRegex = /<post:(\d+),(\d+)>/;
var matches = myParameter.match(myRegex);
this._origMovement.x = matches[1]; //the first number
this._origMovement.y = matches[2]; //the second number
The myRegex variable is a regular expression that looks for the pattern you describe, and has 2 capture groups which look for a string of one or more digits (\d+ means "one or more digits"). The result of the .match() call gives you an array containing the entire match and the results of the capture groups.
If you want to allow for decimal numbers, you'll need to use a different capture group that allows for a decimal point, such as ([\d\.]+), which means "a sequence of one or more digits and decimal points", or more sophisticated, (\d+\.?\d*), which is "a sequence of one or more digits, following by an optional decimal point, followed by zero or more digits).
There are lots of good tutorials around to help you write good regular expressions, and sites that will help you live-test your expressions to make sure they work correctly. They're a powerful tool, but be careful not to over-use them!
Got it to work. For anyone who may ever be interested, the code is below.
for (var i = 0; i < page.list.length; i++) {
if (page.list[i].code == 108 && page.list[i].parameters[0].contains("<post:")) {
var myRegex = /<post:(\d+),(\d+)>/;
var matches = page.list[i].parameters[0].match(myRegex);
this._origMovement.x = matches[1]; //the first number
this._origMovement.y = matches[2]; //the second number
break;
}
};
My string is: (as(dh(kshd)kj)ad)... ()()
How is it possible to count the parentheses with a regular expression? I would like to select the string which begins at the first opening bracket and ends before the ...
Applying that to the above example, that means I would like to get this string: (as(dh(kshd)kj)ad)
I tried to write it, but this doesn't work:
var str = "(as(dh(kshd)kj)ad)... ()()";
document.write(str.match(/(.*)/m));
As I said in the comments, contrary to popular belief (don't believe everything people say) matching nested brackets is possible with regex.
The downside of using it is that you can only do it up to a fixed level of nesting. And for every additional level you wish to support, your regex will be bigger and bigger.
But don't take my word for it. Let me show you. The regex \([^()]*\) matches one level. For up to two levels see the regex here. To match your case, you'd need:
\(([^()]*|\(([^()]*|\([^()]*\))*\))*\)
It would match the bold part: (as(dh(kshd)kj)ad)... ()()
Check the DEMO HERE and see what I mean by fixed level of nesting.
And so on. To keep adding levels, all you have to do is change the last [^()]* part to ([^()]*|\([^()]*\))* (check three levels here). As I said, it will get bigger and bigger.
See Tim's answer for why this won't work, but here's a function that'll do what you're after instead.
function getFirstBracket(str){
var pos = str.indexOf("("),
bracket = 0;
if(pos===-1) return false;
for(var x=pos; x<str.length; x++){
var char = str.substr(x, 1);
bracket = bracket + (char=="(" ? 1 : (char==")" ? -1 : 0));
if(bracket==0) return str.substr(pos, (x+1)-pos);
}
return false;
}
getFirstBracket("(as(dh(kshd)kj)ad)... ()(");
There is a possibility and your approach was quite good:
Match will give you an array if you had some hits, if so you can look up the array length.
var str = "(as(dh(kshd)kj)ad)... ()()",
match = str.match(new RegExp('.*?(?:\\(|\\)).*?', 'g')),
count = match ? match.length : 0;
This regular expression will get all parts of your text that include round brackets. See http://gskinner.com/RegExr/ for a nice online regex tester.
Now you can use count for all brackets.
match will deliver a array that looks like:
["(", "as(", "dh(", "kshd)", "kj)", "ad)", "... (", ")", "(", ")"]
Now you can start sorting your results:
var newStr = '', open = 0, close = 0;
for (var n = 0, m = match.length; n < m; n++) {
if (match[n].indexOf('(') !== -1) {
open++;
newStr += match[n];
} else {
if (open > close) newStr += match[n];
close++;
}
if (open === close) break;
}
... and newStr will be (as(dh(kshd)kj)ad)
This is probably not the nicest code but it will make it easier to understand what you're doing.
With this approach there is no limit of nesting levels.
This is not possible with a JavaScript regex. Generally, regular expressions can't handle arbitrary nesting because that can no longer be described by a regular language.
Several modern regex flavors do have extensions that allow for recursive matching (like PHP, Perl or .NET), but JavaScript is not among them.
No. Regular expressions express regular languages. Finite automatons (FA) are the machines which recognise regular language. A FA is, as its name implies, finite in memory. With a finite memory, the FA can not remember an arbitrary number of parentheses - a feature which is needed in order to do what you want.
I suggest you use an algorithms involving an enumerator in order to solve your problem.
try this jsfiddle
var str = "(as(dh(kshd)kj)ad)... ()()";
document.write(str.match(/\((.*?)\.\.\./m)[1] );
I'm trying to write the code so it removes the "bad" words from the string (the text).
The word is "bad" if it has comma or any special sign thereafter. The word is not "bad" if it contains only a to z (small letters).
So, the result I'm trying to achieve is:
<script>
String.prototype.azwords = function() {
return this.replace(/[^a-z]+/g, "0");
}
var res = "good Remove remove1 remove, ### rem0ve? RemoVE gooood remove.".azwords();//should be "good gooood"
//Remove has a capital letter
//remove1 has 1
//remove, has comma
//### has three #
//rem0ve? has 0 and ?
//RemoVE has R and V and E
//remove. has .
alert(res);//should alert "good gooood"
</script>
Try this:
return this.replace(/(^|\s+)[a-z]*[^a-z\s]\S*(?!\S)/g, "");
It tries to match a word (that is surrounded by whitespaces / string ends) and contains any (non-whitespace) character but at least one that is not a-z. However, this is quite complicated and unmaintainable. Maybe you should try a more functional approach:
return this.split(/\s+/).filter(function(word) {
return word && !/[^a-z]/.test(word);
}).join(" ");
okay, first off you probably want to use the word boundary escape \b in your regex. Also, it's a bit tricky if you match the bad words, because a bad word might contain lower case chars, so your current regex will exclude anything which does have lowecase letters.
I'd be tempted to pick out the good words and put them in a new string. It's a much easier regex.
/\b[a-z]+\b/g
NB: I'm not totally sure that it'll work for the first and last words in the string so you might need to account for that as well. http://www.regextester.com/ is exceptionally useful.
EDIT: as you want punctiation after the word to be 'bad', this will actually do what I was suggesting
(^|\s)[a-z]+(\s|$)
Firstly I wouldn't recommend changing the prototype of String (or of any native object) if you can avoid because you leave yourself open to conflicts with other code that might define the same property in different ways. Much better to put custom methods like this on a namespaced object, though I'm sure some will disagree.
Second, is there any need to use RegEx completely? (Genuine question; not trying to be facetious.)
Here is an example of the function with plain old JS using a little bit of RegEx here and there. Easier to comment, debug, and reuse.
Here is the code:
var azwords = function(str) {
var arr = str.split(/\s+/),
len = arr.length,
i = 0,
res = "";
for (i; i < len; i += 1) {
if (!(arr[i].match(/[^a-z]/))) {
res += (!res) ? arr[i] : " " + arr[i];
}
}
return res;
}
var res = "good Remove remove1 remove, ### rem0ve? RemoVE gooood remove."; //should be "good gooood"
//Remove has a capital letter
//remove1 has 1
//remove, has comma
//### has three #
//rem0ve? has 0 and ?
//RemoVE has R and V and E
//remove. has .
alert(azwords(res));//should alert "good gooood";
Try this one:
var res = "good Remove remove1 remove, ### rem0ve? RemoVE gooood remove.";
var new_one = res.replace(/\s*\w*[#A-Z0-9,.?\\xA1-\\xFF]\w*/g,'');
//Output `good gooood`
Description:
\s* # zero-or-more spaces
\w* # zero-or-more alphanumeric characters
[#A-Z0-9,.?\\xA1-\\xFF] # matches any list of characters
\w* # zero-or-more alphanumeric characters
/g - global (run over all string)
This will find all the words you want /^[a-z]+\s|\s[a-z]+$|\s[a-z]+\s/g so you could use match.
this.match(/^[a-z]+\s|\s[a-z]+$|\s[a-z]+\s/g).join(" "); should return the list of valid words.
Note that this took some time as a JSFiddle so it maybe more efficient to split and iterate your list.
The following function is meant to remove random articles (parts of speech) from the text. Eventually the percentages will be user-adjustable, and the regex more sophisticated to catch word boundaries better, etc. It is replacing (and about 50/50), but it's also squashing the spaces (which are matched, but not captured). I think I'm being really bone-headed here but I can't figure out the proper syntax... can anyone help?
function posArticles(t) {
var text = t;
var re = / (a|the|an) /g;
var rArray;
text = text.replace(re, function(_, m) {
if (Math.floor(Math.random()*101) < 50) return '';
else return m;
});
return text;
}
I realize that this has to do with the positional/optional arguments to the anon function, but I can't figure out which is the match and which is the capture and so forth.
There are numerous ways you could do this, but I think your best bet is to use \b – a zero-width match for a "word boundary." That guarantees that you're getting "the" and not "there" or whatever, but doesn't match the spaces around it.
Thus, use re = /\b([Aa]n?|[Tt]he)\b/;
I realize that this has to do with the positional/optional arguments
to the anon function, but I can't figure out which is the match and
which is the capture and so forth.
First argument passed to your callback function is whole match (ie: _ = ' the '). Next argument are your captured groups (m = 'the'). Callback function is replacing whole match, so if you are including spaces in your expresion, they will also be replaced.
if (Math.floor(Math.random()*101) < 50) return ' ';
return a space instead of empty string :)
Say we have a string
blue|blue|green|blue|blue|yellow|yellow|blue|yellow|yellow|
And we want to figure out whether the word "yellow" occurs in the last 5 words of the string, specifically by returning a capture group containing these occurences if any.
Is there a way to do that with a regex?
Update: I'm feeding a regex engine some rules. For various reasons I'm trying to work with the engine rather than go outside it, which would be my last resort.
/\b(yellow)\|(?=(?:\w+\|){0,4}$)/g
This will return one hit for each yellow| that's followed by fewer than five words (per your definition of "word"). This assumes the sequence always ends with a pipe; if that's not the case, you might want to change it to:
/\b(yellow)(?=(?:\|\w+){0,4}\|?$)/g
EDIT (in response to comment): The definition of a "word" in this solution is arbitrary, and doesn't really correspond to real-world usage. To allow for hyphenated words like "real-world" you could use this:
/\b(yellow)\|(?=(?:\w+(?:-\w+)*\|){0,4}$)/g
...or, for this particular job, you could define a word as one or more of any characters except pipes:
/\b(yellow)\|(?=(?:[^|]+\|){0,4}$)/g
No need to use a Regex for such a simple thing.
Simply split on the pipe, and check with indexOf:
var group = 'blue|blue|green|blue|blue|yellow|yellow|blue|yellow|yellow';
if ( group.split('|').slice(-5).indexOf('yellow') == -1 ) {
alert('Not there :(');
} else {
alert('Found!!!');
}
Note: indexOf is not natively supported in IE < 9, but support for it can be added very easily.
Can't think of a way to do this with a single regular expression, but you can form one for each of the last five positions and sum the matches.
var string = "blue|blue|green|blue|blue|yellow|yellow|blue|yellow|yellow|";
var regexes = [];
regexes.push(/(yellow)\|[^|]+\|[^|]+\|[^|]+\|[^|]+\|$/);
regexes.push(/(yellow)\|[^|]+\|[^|]+\|[^|]+\|$/);
regexes.push(/(yellow)\|[^|]+\|[^|]+\|$/);
regexes.push(/(yellow)\|[^|]+\|$/);
regexes.push(/(yellow)\|$/);
var count = 0;
var regex;
while (regex = regexes.shift()) {
if (string.match(regex)) {
count++;
}
}
console.log(count);
Should find four matches.