Develop Reference

Finding characters with spaces

I was trying last week to find parts of a text containing specific words delimited by punctuation characters. That works well.
[^.?!:]*\b(why|how)\b[^.?!]*[.?!]
On the following sentence "How did you do it? bla bla bla! why did you do it?", it's giving me the following output :
"How did you do it?"
"why did you do it?"
Now I am trying to add the hyphen character : I want to detect if there is an hyphen with spaces around (a new sentence delimiter):
"The man went walking upstairs - why was he there?
That would return me : "why was he there?"
It would follow the following rules:
hello - bye -> this would be the only one to be matched
hello-bye -> not matched
hello -bye -> not matched
hello- bye -> not matched
Using the negation, I tried to add that part :
[^.?!:\\s\\-\\s] => ignore everything that ends with a "." or a "?" or a "!" or a ":" or a " - "
I doesn't work, but as I am pretty bad using regex, I am probably missing something obvious.
var regex = /[^.?!:\\s\\-\\s]*\b(why|how)\b[^.?!]*[.?!]/igm
var text = "Here I am - why did you want to see me?"
var match;
while ((match = regex.exec(text)) != null) {
console.log(match);
}
Output :
Here I am - why did you want to see me?
Expected output :
why did you want to see me?

There are two issues that I see:
backslashes (use single inside a regex literal, double in constructor) and
Sequence is used inside a character class (replace [^.?!:\s\-\s] with (?:(?!\s-\s)[^.?!:])*).
You may use
var regex = /(?:(?!\s-\s)[^.?!:])*\b((?:why|how)\b[^.?!]*)[.?!]/ig
where (?:(?!\s-\s)[^.?!:])* is a tempered greedy token matching any character other than ^.?!: that is not starting a whitespace+-+whitespace pattern.
var regex = /(?:(?!\s-\s)[^.?!:])*\b((?:why|where|pourquoi|how)\b[^.?!]*)[.?!]/ig;
var text = "L'Inde a déjà acheté nos rafales, pourquoi la France ne le -dirait-elle pas ?";
var match;
while ((match = regex.exec(text)) != null) {
console.log(match[1]);
}

[ ] is always a character class, which means that at one position, you can match one character. The "negation" in your example is in fact probably not even doing what you thing it does.
What you probably want to match is either the beginning of a string, the end of a sentence, or a dash with two spaces around, so just replace it with (^|[.?!]| - )\b((why|how)...etc). You will need some post processing of the result, as JavaScript does not support look-behind assertions as far as I know.

Given your 4 examples, this works.
/\s-\s(\w*)/g
Test it here - https://regex101.com/r/YQhRBI/1
I'm matching ANY character within the question portion. If you want to match specific key words, you'd swap the (\w*) with ([why|how|who|what|where|when])
I think if you had a paragraph, you'd have to be sure to find a way to terminate the answer portion with a specific delimiter. If this was more along the lines of a question/answer per new line, then you'd need only to end the regex with an end-of-line anchor.

Extract specific chars from a string using a regex

I need to split an email address and take out the first character and the first character after the '#'
I can do this as follows:
'bar#foo'.split('#').map(function(a){ return a.charAt(0); }).join('')
--> bf
Now I was wondering if it can be done using a regex match, something like this
'bar#foo'.match(/^(\w).*?#(\w)/).join('')
--> bar#fbf
Not really what I want, but I'm sure I miss something here! Any suggestions ?

Why use a regex for this? just use indexOf to get the char at any given position:
var addr = 'foo#bar';
console.log(addr[0], addr[addr.indexOf('#')+1])
To ensure your code works on all browsers, you might want to use charAt instead of []:
console.log(addr.charAt(0), addr.charAt(addr.indexOf('#')+1));
Either way, It'll work just fine, and This is undeniably the fastest approach
If you are going to persist, and choose a regex, then you should realize that the match method returns an array containing 3 strings, in your case:
/^(\w).*?#(\w)/
["the whole match",//start of string + first char + .*?# + first string after #
"groupw 1 \w",//first char
"group 2 \w"//first char after #
]
So addr.match(/^(\w).*?#(\w)/).slice(1).join('') is probably what you want.

If I understand correctly, you are quite close. Just don't join everything returned by match because the first element is the entire matched string.
'bar#foo'.match(/^(\w).*?#(\w)/).splice(1).join('')
--> bf

Using regex:
matched="",
'abc#xyz'.replace(/(?:^|#)(\w)/g, function($0, $1) { matched += $1; return $0; });
console.log(matched);
// ax

The regex match function returns an array of all matches, where the first one is the 'full text' of the match, followed by every sub-group. In your case, it returns this:
bar#f
b
f
To get rid of the first item (the full match), use slice:
'bar#foo'.match(/^(\w).*?#(\w)/).slice(1).join('\r')

Use String.prototype.replace with regular expression:
'bar#foo'.replace(/^(\w).*#(\w).*$/, '$1$2'); // "bf"

Or using RegEx
^([a-zA-Z0-9])[a-zA-Z0-9.!#$%&'*+\/=?^_`{|}~-]+#([a-zA-Z0-9-])[a-zA-Z0-9-]+(?:\.[a-zA-Z0-9-]+)*$
Fiddle

Need a regex that finds "string" but not "[string]"

I'm trying to build a regular expression that parses a string and skips things in brackets.
Something like
string = "A bc defg hi [hi] jkl mnop.";
The .match() should return "hi" but not [hi]. I've spent 5 hours running through RE's but I'm throwing in the towel.
Also this is for javascript or jquery if that matters.
Any help is appreciated. Also I'm working on getting my questions formatted correctly : )
EDIT:
Ok I just had a eureka moment and figured out that the original RegExp I was using actually did work. But when I was replaces the matches with the [matches] it simply replaced the first match in the string... over and over. I thought this was my regex refusing to skip the brackets but after much time of trying almost all of the solutions below, I realized that I was derping Hardcore.
When .replace was working its magic it was on the first match, so I quite simply added a space to the end of the result word as follows:
var result = string.match(regex);
var modifiedResult = '[' + result[0].toString() + ']';
string.replace(result[0].toString() + ' ', modifiedResult + ' ');
This got it to stop targeting the original word in the string and stop adding a new set of brackets to it with every match. Thank you all for your help. I am going to give answer credit to the post that prodded me in the right direction.

preprocess the target string by removing everything between brackets before trying to match your RE
string = "A bc defg hi [hi] jkl mnop."
tmpstring = string.replace(/\[.*\]/, "")
then apply your RE to tmpstring
correction: made the match for brackets eager per nhahtd comment below, and also, made the RE global
string = "A bc defg hi [hi] jkl mnop."
tmpstring = string.replace(/\[.*?\]/g, "")

You don't necessarily need regex for this. Simply use string manipulation:
var arr = string.split("[");
var final = arr[0] + arr[1].split("]")[1];
If there are multiple bracketed expressions, use a loop:
while (string.indexOf("[") != -1){
var arr = string.split("[");
string = arr[0] + arr.slice(1).join("[").split("]").slice(1).join("]");
}

Using only Regular Expressions, you can use:
hi(?!])
as an example.
Look here about negative lookahead: http://www.regular-expressions.info/lookaround.html
Unfortunately, javascript does not support negative lookbehind.
I used http://regexpal.com/ to test, abcd[hi]jkhilmnop as test data, hi(?!]) as the regex to find. It matched 'hi' without matching '[hi]'. Basically it matched the 'hi' so long as there was not a following ']' character.
This of course, can be expanded if needed. This has a benefit of not requiring any pre-processing for the string.

r"\[(.*)\]"
Just play arounds with this if you wanto to use regular expressions.

What do yo uwant to do with it? If you want to selectively replace parts like "hi" except when it's "[hi]", then I often use a system where I match what I want to avoid first and then what I want to watch; if it matches what I want to avoid then I return the match, otherwise I return the processed match.
Like this:
return string.replace(/(\[\w+\])|(\w+)/g, function(all, m1, m2) {return m1 || m2.toUpperCase()});
which, with the given string, returns:
"A BC DEFG HI [hi] JKL MNOP."
Thus: it replaces every word with uppercase (m1 is empty), except if the word is between square brackets (m1 is not empty).

This builds an array of all the strings contained in [ ]:
var regex = /\[([^\]]*)\]/;
var string = "A bc defg hi [hi] [jkl] mnop.";
var results=[], result;
while(result = regex.exec(string))
results.push(result[1]);
edit
To answer to the question, this regex returns the string less all is in [ ], and trim whitespaces:
"A bc defg [hi] mnop [jkl].".replace(/(\s{0,1})\[[^\]]*\](\s{0,1})/g,'$1')

Instead of skipping the match you can probably try something different - match everything but do not capture the string within square brackets (inclusive) with something like this:
var r = /(?:\[.*?[^\[\]]\])|(.)/g;
var result;
var str = [];
while((result = r.exec(s)) !== null){
if(result[1] !== undefined){ //true if [string] matched but not captured
str.push(result[1]);
}
}
console.log(str.join(''));
The last line will print parts of the string which do not match the [string] pattern. For example, when called with the input "A [bc] [defg] hi [hi] j[kl]u m[no]p." the code prints "A hi ju mp." with whitespaces intact.
You can try different things with this code e.g. replacing etc.

getting contents of string between digits

have a regex problem :(
what i would like to do is to find out the contents between two or more numbers.
var string = "90+*-+80-+/*70"
im trying to edit the symbols in between so it only shows up the last symbol and not the ones before it. so trying to get the above variable to be turned into 90+80*70. although this is just an example i have no idea how to do this. the length of the numbers, how many "sets" of numbers and the length of the symbols in between could be anything.
many thanks,
Steve,

The trick is in matching '90+-+' and '80-+/' seperately, and selecting only the number and the last constant.
The expression for finding the a number followed by 1 or more non-numbers would be
\d+[^\d]+
To select the number and the last non-number, add parens:
(\d+)[^\d]*([^\d])
Finally add a /g to repeat the procedure for each match, and replace it with the 2 matched groups for each match:
js> '90+*-+80-+/*70'.replace(/(\d+)[^\d]*([^\d])/g, '$1$2');
90+80*70
js>

Or you can use lookahead assertion and simply remove all non-numerical characters which are not last: "90+*-+80-+/*70".replace(/[^0-9]+(?=[^0-9])/g,'');

You can use a regular expression to match the non-digits and a callback function to process the match and decide what to replace:
var test = "90+*-+80-+/*70";
var out = test.replace(/[^\d]+/g, function(str) {
return(str.substr(-1));
})
alert(out);
See it work here: http://jsfiddle.net/jfriend00/Tncya/
This works by using a regular expression to match sequences of non-digits and then replacing that sequence of non-digits with the last character in the matched sequence.

i would use this tutorial, first, then review this for javascript-specific regex questions.

This should do it -
var string = "90+*-+80-+/*70"
var result = '';
var arr = string.split(/(\d+)/)
for (i = 0; i < arr.length; i++) {
if (!isNaN(arr[i])) result = result + arr[i];
else result = result + arr[i].slice(arr[i].length - 1, arr[i].length);
}
alert(result);
Working demo - http://jsfiddle.net/ipr101/SA2pR/

Similar to #Arnout Engelen
var string = "90+*-+80-+/*70";
string = string.replace(/(\d+)[^\d]*([^\d])(?=\d+)/g, '$1$2');
This was my first thinking of how the RegEx should perform, it also looks ahead to make sure the non-digit pattern is followed by another digit, which is what the question asked for (between two numbers)
Similar to #jfriend00
var string = "90+*-+80-+/*70";
string = string.replace( /(\d+?)([^\d]+?)(?=\d+)/g
, function(){
return arguments[1] + arguments[2].substr(-1);
});
Instead of only matching on non-digits, it matches on non-digits between two numbers, which is what the question asked
Why would this be any better?
If your equation was embedded in a paragraph or string of text. Like:
This is a test where I want to clean up something like 90+*-+80-+/*70 and don't want to scrap the whole paragraph.
Result (Expected) :
This is a test where I want to clean up something like 90+80*70 and don't want to scrap the whole paragraph.
Why would this not be any better?
There is more pattern matching, which makes it theoretically slower (negligible)
It would fail if your paragraph had embedded numbers. Like:
This is a paragraph where Sally bought 4 eggs from the supermarket, but only 3 of them made it back in one piece.
Result (Unexpected):
This is a paragraph where Sally bought 4 3 of them made it back in one piece.

java script Regular Expressions patterns problem

My problem start with like-
var str='0|31|2|03|.....|4|2007'
str=str.replace(/[^|]\d*[^|]/,'5');
so the output becomes like:"0|5|2|03|....|4|2007" so it replaces 31->5
But this doesn't work for replacing other segments when i change code like this:
str=str.replace(/[^|]{2}\d*[^|]/,'6');
doesn't change 2->6.
What actually i am missing here.Any help?

I think a regular expression is a bad solution for that problem. I'd rather do something like this:
var str = '0|31|2|03|4|2007';
var segments = str.split("|");
segments[1] = "35";
segments[2] = "123";
Can't think of a good way to solve this with a regexp.

Here is a specific regex solution which replaces the number following the first | pipe symbol with the number 5:
var re = /^((?:\d+\|){1})\d+/;
return text.replace(re, '$15');
If you want to replace the digits following the third |, simply change the {1} portion of the regex to {3}
Here is a generalized function that will replace any given number slot (zero-based index), with a specified new number:
function replaceNthNumber(text, n, newnum) {
var re = new RegExp("^((?:\\d+\\|){"+ n +'})\\d+');
return text.replace(re, '$1'+ newnum);
}

Firstly, you don't have to escape | in the character set, because it doesn't have any special meaning in character sets.
Secondly, you don't put quantifiers in character sets.
And finally, to create a global matching expression, you have to use the g flag.

[^\|] means anything but a '|', so in your case it only matches a digit. So it will only match anything with 2 or more digits.
Second you should put the {2} outside of the []-brackets
I'm not sure what you want to achieve here.