Given an input text such where all spaces are replaced by n _ :
Hello_world_?. Hello_other_sentenc3___. World___________.
I want to keep the _ between words, but I want to stick each punctuation back to the last word of a sentence without any space between last word and punctuation. I want to use the the punctuation as pivot of my regex.
I wrote the following JS-Regex:
str = str.replace(/(_| )*([:punct:])*( |_)/g, "$2$3");
This fails, since it returns :
Hello_world_?. Hello_other_sentenc3_. World_._
Why it doesn't works ? How to delete all "_" between the last word and the punctuation ?
http://jsfiddle.net/9c4z5/
Try the following regex, which makes use of a positive lookahead:
str = str.replace(/_+(?=\.)/g, "");
It replaces all underscores which are immediately followed by a punctuation character with the empty string, thus removing them.
If you want to match other punctuation characters than just the period, replace the \. part with an appropriate character class.
JavaScript doesn't have :punct: in its regex implementation. I believe you'd have to list out the punctuation characters you care about, perhaps something like this:
str = str.replace(/(_| )+([.,?])/g, "$2");
That is, replace any group of _ or space that is immediately followed by punctation with just the punctuation.
Demo: http://jsfiddle.net/9c4z5/2/
Related
I’m trying to better understand RegEx and apostrophes/single quotes.
If I use this code:
const regex = /\b[\'\w]+\b/g
console.log(phrase.match(regex))
Then
let phrase = "'pickle'" // becomes pickle (single quotes disappear)
let phrase = "can't" // becomes can't (apostrophe remains)
I thought I knew what all regex do:
/text/g means everything between the slashes and g means global, to
keep searching after the first hit.
\b is word boundary, spaces on each side
w+ means alphanumerics, and the '+' indicates it can be for more
than 1 character
[\w\']+ means A-Za-z0-9 and apostrophe of any length.
But I'd like to get this:
let phrase = "'pickle'" // becomes 'pickle' (with single quotes)
What am I missing? I experimented with
const regex2 = /\b[\w+\']\b/g;
console.log(phrase.match(regex2))
let phrase = "can't"
But that becomes ["'", "t"] ... why? I understand now that the + is after the \w, the \' stands alone, but why "t" and where did the "can" go?
I tried
const regex3 = /\b\'[\w+]\'\b/g;
console.log(phrase.match(regex3))
But I get "null". Why?
The question is basically "How do I get word boundaries including apostrophes". Right?
If so, then the regex you have /\b[\'\w]+\b/g explicitly looks for \b for boundary which will match a non word character (like space or apostrohpe) followed by a letter or viceversa. Like this: https://regex101.com/r/7Pxsru/1, (I added a few more words so that the boundary is clearly seen)
If you would like to get "'pickle'" and "can't" then simply don't look for \b, like this /[\w+\']+/g, see the demo: https://regex101.com/r/FNjlEq/1
The two regexes you propose mean the following:
/\b[\w+\']\b/g: Look for a boundary letter then any word letter any number of times (note that this has no effect since it is inside a []) OR an apostrophe then a boundary.
/\b\'[\w+]\'\b/g: Look for a boundary letter by an apostrophe and any word any number of times (note that there is no need to be inside a []) then followed by an apostrophe and a word boundary.
const regex2 = /\b[\w+\']\b/g;
In this one, since the + in inside of [], it is matching a literal + character, so you're searching for a word boundary, followed by either a single alphanumeric character, a +, or a ', following by a word boundary.
You probably want:
\b(\w+|\')\b
which looks for a word boundary, followed by either at least one alphanumeric character or a single quote.
It would probably help to look at regex101 so you can see what the regex is actually doing: https://regex101.com/r/aJPWAB/1
I am trying to split a string in JS on spaces except when the space is in a quote. However, an incomplete quote should be maintained. I'm not skilled in regex wizardry, and have been using the below regex:
var list = text.match(/[^\s"]+|"([^"]*)"/g)
However, if I provide input like sdfj "sdfjjk this will become ["sdfj","sdfjjk"] rather than ["sdfj",""sdfjjk"].
You can use
var re = /"([^"]*)"|\S+/g;
By using \S (=[^\s]) we just drop the " from the negated character class.
By placing the "([^"]*)" pattern before \S+, we make sure substrings in quotes are not torn if they come before. This should work if the string contains well-paired quoted substrings and the last is unpaired.
Demo:
var re = /"([^"]*)"|\S+/g;
var str = 'sdfj "sdfjjk';
document.body.innerHTML = JSON.stringify(str.match(re));
Note that to get the captured texts in-between quotes, you will need to use RegExp#exec in a loop (as String#match "drops" submatches).
UPDATE
No idea what downvoter thought when downvoting, but let me guess. The quotes are usually used around word characters. If there is a "wild" quote, it is still a quote right before/after a word.
So, we can utilize word boundaries like this:
"\b[^"]*\b"|\S+
See regex demo.
Here, "\b[^"]*\b" matches a " that is followed by a word character, then matches zero or more characters other than " and then is followed with a " that is preceded with a word character.
Moving further in this direction, we can make it as far as:
\B"\b[^"\n]*\b"\B|\S+
With \B" we require that " should be preceded with a non-word character, and "\B should be followed with a non-word character.
See another regex demo
A lot depends on what specific issue you have with your specific input!
Try the following:
text.match(/".*?"|[^\s]+/g).map(s => s.replace(/^"(.*)"$/, "$1"))
This repeatedly finds either properly quoted substrings (first), OR other sequences of non-whitespace. The map part is to remove the quotes around the quoted substrings.
> text = 'abc "def ghi" lmn "opq'
< ["abc", "def ghi", "lmn", ""opq"]
I need is the last match. In the case below the word test without the $ signs or any other special character:
Test String:
$this$ $is$ $a$ $test$
Regex:
\b(\w+)\b
The $ represents the end of the string, so...
\b(\w+)$
However, your test string seems to have dollar sign delimiters, so if those are always there, then you can use that instead of \b.
\$(\w+)\$$
var s = "$this$ $is$ $a$ $test$";
document.body.textContent = /\$(\w+)\$$/.exec(s)[1];
If there could be trailing spaces, then add \s* before the end.
\$(\w+)\$\s*$
And finally, if there could be other non-word stuff at the end, then use \W* instead.
\b(\w+)\W*$
In some cases a word may be proceeded by non-word characters, for example, take the following sentence:
Marvelous Marvin Hagler was a very talented boxer!
If we want to match the word boxer all previous answers will not suffice due the fact we have an exclamation mark character proceeding the word. In order for us to ensure a successful capture the following expression will suffice and in addition take into account extraneous whitespace, newlines and any non-word character.
[a-zA-Z]+?(?=\s*?[^\w]*?$)
https://regex101.com/r/D3bRHW/1
We are informing upon the following:
We are looking for letters only, either uppercase or lowercase.
We will expand only as necessary.
We leverage a positive lookahead.
We exclude any word boundary.
We expand that exclusion,
We assert end of line.
The benefit here are that we do not need to assert any flags or word boundaries, it will take into account non-word characters and we do not need to reach for negate.
var input = "$this$ $is$ $a$ $test$";
If you use var result = input.match("\b(\w+)\b") an array of all the matches will be returned next you can get it by using pop() on the result or by doing: result[result.length]
Your regex will find a word, and since regexes operate left to right it will find the first word.
A \w+ matches as many consecutive alphanumeric character as it can, but it must match at least 1.
A \b matches an alphanumeric character next to a non-alphanumeric character. In your case this matches the '$' characters.
What you need is to anchor your regex to the end of the input which is denoted in a regex by the $ character.
To support an input that may have more than just a '$' character at the end of the line, spaces or a period for instance, you can use \W+ which matches as many non-alphanumeric characters as it can:
\$(\w+)\W+$
Avoid regex - use .split and .pop the result. Use .replace to remove the special characters:
var match = str.split(' ').pop().replace(/[^\w\s]/gi, '');
DEMO
I'm able to match and highlight this Hebrew letter in JS:
var myText = $('#text').html();
var myHilite = myText.replace(/(\u05D0+)/g,"<span class='highlight'>$1</span>");
$('#text').html(myHilite);
fiddle
but can't highlight a word containing that letter at a word boundary:
/(\u05D0)\b/g
fiddle
I know that JS is bad at regex with Unicode (and server side is preferred), but I also know that I'm bad at regex. Is this a limit in JS or an error in my syntax?
I can't read Hebrew... does this regex do what you want?
/(\S*[\u05D0]+\S*)/g
Your first regex, /(\u05D0+)/g matches on only the character you are interested in.
Your second regex, /(\u05D0)\b/g, matches only when the character you are interested in is the last-only (or last-repeated) character before a word boundary...so that doesn't won't match that character in the beginning or middle of a word.
EDIT:
Look at this anwer
utf-8 word boundary regex in javascript
Using the info from that answer, I come up with this regex, is this correct?
/([\u05D0])(?=\s|$)/g
What about using the following regexp which uses all cases of a word in a sentence:
/^u05D0\s|\u05D0$|\u05D0\s|^\u05D0$/
it actually uses 4 regexps with the OR operator ('|').
Either the string starts with your exact word followed by a space
OR your string has space + your word + space
OR your string ends with space + your word
OR your string is the exact word only.
I am trying to match all words with single or multiple spaces. my expression
(\w+\s*)* is not working
edit 1:
Let say i have a sentence in this form
[[do "hi i am bob"]]
[[do "hi i am Bob"]]
now I have to replace this with
cool("hi i am bob") or
cool("hi i am Bob")
I do not care about replacing multiple spaces with single .
I can achieve this for a single word like
\[\[do\"(\w+)\"\]\] and replacing regex cool\(\"$1\") but this does not look like an effective solution and does not match multiple words ....
I apologies for incomplete question
any help will be aprecciated
Find this Regular Expression:
/\[\[do\s+("[\w\s]+")\s*\]\]/
And do the following replacement:
'cool($1)'
The only special thing that's being done here is using character classes to our advantage with
[\w\s]+
Matches one or more word or space characters (a-z, A-Z, 0-9, _, and whitespace). That';; eat up your internal stuff no problem.
'[[do "hi i am Bob"]]'.replace(/\[\[do\s+("[\w\s]+")\s*\]\]/, 'cool($1)')
Spits out
cool("hi i am Bob")
Though - if you want to add punctuation (which you probably will), you should do it like this:
/\[\[do\s+("[^"]+")\s*\]\]/
Which will match any character that's not a double quote, preserving your substring. There are more complicated ones to allow you to deal with escaped quotation marks, but I think that's outside the scope of this question.
To match "all words with single or multiple spaces", you cannot use \s*, as it will match even no spaces.
On the other hand, it looks like you want to match even "hi", which is one word with no spaces.
You probably want to match one or more words separated by spaces. If so, use regex pattern
(\w+(?:$|\s+))+
or
\w+(\s+\w+)*
I'm not sure, but maybe this is what you're trying to get:
"Hi I am bob".match(/\b\w+\b/g); // ["Hi", "I", "am", "bob"]
Use regex pattern \w+(\s+\w+)* as follows:
m = s.match(/\w+(\s+\w+)*/g);
Simple. Match all groups of characters that are not white spaces
var str = "Hi I am Bob";
var matches = str.match(/[^ ]+/g); // => ["Hi", "I", "am", "Bob"]
What your regex is doing is:
/([a-zA-Z0-9_]{1,}[ \r\v\n\t\f]{0,}){0,}/
That is, find the first match of one or more of A through Z bother lower and upper along with digits and underscore, then followed by zero or more space characters which are:
A space character
A carriage return character
A vertical tab character
A new line character
A tab character
A form feed character
Then followed by zero or more of A through Z bother lower and upper along with digits and underscore.
\s matches more than just simple spaces, you can put in a literal space, and it will work.
I believe you want:
/(\w+ +\w+)/g
Which all matches of one or more of A through Z bother lower and upper along with digits and underscore, followed by one or more spaces, then followed by one or more of A through Z bother lower and upper along with digits and underscore.
This will match all word-characters separated by spaces.
If you just want to find all clusters of word characters, without punctuation or spaces, then, you would use:
/(\w+)/g
Which will find all word-characters that are grouped together.
var regex=/\w+\s+/g;
Live demo: http://jsfiddle.net/GngWn/
[Update] I was just answering the question, but based on the comments this is more likely what you're looking for:
var regex=/\b\w+\b/g;
\b are word boundaries.
Demo: http://jsfiddle.net/GngWn/2/
[Update2] Your edit makes it a completely different question:
string.replace(/\[\[do "([\s\S]+)"\]\]/,'cool("$1")');
Demo: http://jsfiddle.net/GngWn/3/