Develop Reference

RegExp to filter characters after the last dot

For example, I have a string "esolri.gbn43sh.earbnf", and I want to remove every character after the last dot(i.e. "esolri.gbn43sh"). How can I do so with regular expression?
I could of course use non-RegExp way to do it, for example:
"esolri.gbn43sh.earbnf".slice("esolri.gbn43sh.earbnf".lastIndexOf(".")+1);
But I want a regular expression.
I tried /\..*?/, but that remove the first dot instead.
I am using Javascript. Any help is much appreciated.

I would use standard js rather than regex for this one, as it will be easier for others to understand your code
var str = 'esolri.gbn43sh.earbnf'
console.log(
str.slice(str.lastIndexOf('.') + 1)
)

Pattern Matching
Match a dot followed by non-dots until the end of string
let re = /\.[^.]*$/;
Use this with String.prototype.replace to achieve the desired output
'foo.bar.baz'.replace(re, ''); // 'foo.bar'
Other choices
You may find it is more efficient to do a simple substring search for the last . and then use a string slicing method on this index.
let str = 'foo.bar.baz',
i = str.lastIndexOf('.');
if (i !== -1) // i = -1 means no match
str = str.slice(0, i); // "foo.bar"

replace multiple consecutive hyphens with one

Is this possible in Javascript?
I have this from Java, but can not be used in Javascript.
s/-{2,}/-/g
Is there another way that works?

Yes it is. You can use the same regex with the Javascript replace() method.
s.replace(find, replacement)
// where 's' is your string object
Example:
var r = 'foo--bar---baz'.replace(/-{2,}/g, '-');
console.log(r); // "foo-bar-baz"

You can just do this:
var newStr = "hi--this is----good".replace(/-+/g,'-'); // hi-this is-good
-+ matches more than 1 - and replaces them with a single -
Your regex is also valid. Except you cannot use s modifier.

query.trim().replace(/\s\s+/g, '-').replace(/[\s.;,?%]/, '-')

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.

myString.replace( VARIABLE, "") ...... but globally

How can I use a variable to remove all instances of a substring from a string?
(to remove, I'm thinking the best way is to replace, with nothing, globally... right?)
if I have these 2 strings,
myString = "This sentence is an example sentence."
oldWord = " sentence"
then something like this
myString.replace(oldWord, "");
only replaces the first instance of the variable in the string.
but if I add the global g like this myString.replace(/oldWord/g, ""); it doesn't work, because it thinks oldWord, in this case, is the substring, not a variable. How can I do this with the variable?

Well, you can use this:
var reg = new RegExp(oldWord, "g");
myString.replace(reg, "");
or simply:
myString.replace(new RegExp(oldWord, "g"), "");

You have to use the constructor rather than the literal syntax when passing variables. Stick with the literal syntax for literal strings to avoid confusing escape syntax.
var oldWordRegEx = new RegExp(oldWord,'g');
myString.replace(oldWordRegEx,"");

No need to use a regular expression here: split the string around matches of the substring you want to remove, then join the remaining parts together:
myString.split(oldWord).join('')
In the OP's example:
var myString = "This sentence is an example sentence.";
var oldWord = " sentence";
console.log(myString.split(oldWord).join(''));

According to the docs at MDN, you can do this:
var re = /apples/gi;
var str = 'Apples are round, and apples are juicy.';
var newstr = str.replace(re, 'oranges');
console.log(newstr); // oranges are round, and oranges are juicy.
where /gi tells it to do a global replace, ignoring case.

How to replace underscores with spaces using a regex in Javascript

How can I replace underscores with spaces using a regex in Javascript?
var ZZZ = "This_is_my_name";

If it is a JavaScript code, write this, to have transformed string in ZZZ2:
var ZZZ = "This_is_my_name";
var ZZZ2 = ZZZ.replace(/_/g, " ");
also, you can do it in less efficient, but more funky, way, without using regex:
var ZZZ = "This_is_my_name";
var ZZZ2 = ZZZ.split("_").join(" ");

Regular expressions are not a tool to replace texts inside strings but just something that can search for patterns inside strings. You need to provide a context of a programming language to have your solution.
I can tell you that the regex _ will match the underscore but nothing more.
For example in Groovy you would do something like:
"This_is_my_name".replaceAll(/_/," ")
===> This is my name
but this is just language specific (replaceAll method)..

var str1="my__st_ri_ng";
var str2=str1.replace(/_/g, ' ');

Replace "_" with " "
The actual implementation depends on your language.
In Perl it would be:
s/_/ /g
But the truth is, if you are replacing a fixed string with something else, you don't need a regular expression, you can use your language/library's basic string replacement algorithms.
Another possible Perl solution would be:
tr/_/ /

To replace the underscores with spaces in a string, call the replaceAll() method, passing it an underscore and space as parameters, e.g. str.replaceAll('_', ' '). The replaceAll method will return a new string where each underscore is replaced by a space.
const str = 'apple_pear_melon';
// ✅ without regular expression
const result1 = str.replaceAll('_', ' ');
console.log(result1); // 👉️ "apple pear melon"
// ✅ with regular expression
const result2 = str.replace(/_+/g, ' ');
console.log(result2); // 👉️ "apple pear melon"