Develop Reference

javascript regex insert new element into expression

I am passing a URL to a block of code in which I need to insert a new element into the regex. Pretty sure the regex is valid and the code seems right but no matter what I can't seem to execute the match for regex!
//** Incoming url's
//** url e.g. api/223344
//** api/11aa/page/2017
//** Need to match to the following
//** dir/api/12ab/page/1999
//** Hence the need to add dir at the front
var url = req.url;
//** pass in: /^\/api\/([a-zA-Z0-9-_~ %]+)(?:\/page\/([a-zA-Z0-9-_~ %]+))?$/
var re = myregex.toString();
//** Insert dir into regex: /^dir\/api\/([a-zA-Z0-9-_~ %]+)(?:\/page\/([a-zA-Z0-9-_~ %]+))?$/
var regVar = re.substr(0, 2) + 'dir' + re.substr(2);
var matchedData = url.match(regVar);
matchedData === null ? console.log('NO') : console.log('Yay');
I hope I am just missing the obvious but can anyone see why I can't match and always returns NO?
Thanks

Let's break down your regex
^\/api\/ this matches the beginning of a string, and it looks to match exactly the string "/api"
([a-zA-Z0-9-_~ %]+) this is a capturing group: this one specifically will capture anything inside those brackets, with the + indicating to capture 1 or more, so for example, this section will match abAB25-_ %
(?:\/page\/([a-zA-Z0-9-_~ %]+)) this groups multiple tokens together as well, but does not create a capturing group like above (the ?: makes it non-captuing). You are first matching a string exactly like "/page/" followed by a group exactly like mentioned in the paragraph above (that matches a-z, A-Z, 0-9, etc.
?$ is at the end, and the ? means capture 0 or more of the precending group, and the $ matches the end of the string
This regex will match this string, for example: /api/abAB25-_ %/page/abAB25-_ %
You may be able to take advantage of capturing groups, however, and use something like this instead to get similar results: ^\/api\/([a-zA-Z0-9-_~ %]+)\/page\/\1?$. Here, we are using \1 to reference that first capturing group and match exactly the same tokens it is matching. EDIT: actually, this probably won't work, since the text after /api/ and the text after /page/ will most likely be different, carrying on...
Afterwards, you are are adding "dir" to the beginning of your search, so you can now match someting like this: dir/api/abAB25-_ %/page/abAB25-_ %
You have also now converted the regex to a string, so like Crayon Violent pointed out in their comment, this will break your expected funtionality. You can fix this by using .source on your regex: var matchedData = url.match(regVar.source); https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/source
Now you can properly match a string like this: dir/api/11aa/page/2017 see this example: https://repl.it/Mj8h

As mentioned by Crayon Violent in the comments, it seems you're passing a String rather than a regular expression in the .match() function. maybe try the following:
url.match(new RegExp(regVar, "i"));
to convert the string to a regular expression. The "i" is for ignore case; don't know that's what you want. Learn more here:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp

Regex - Extract digits from a url

I have this url:
http://example.com/things/stuff/532453?morethings&stuff=things&ver=1
I need just that number in the middle there. Closest I got was
(\d*)?\?
but this includes the question mark. Basiclly all numbers that come before the ? all the way to the slash so the ouput is 532453.

Try the following regex (?!\/)\d+(?=\?):
url = "http://example.com/things/stuff/532453?morethings&stuff=things"
url.match(/(?!\/)\d+(?=\?)/) # outputs 532453
This regex will attempt to match any series of digits only after a / and before ? by using negative/positive lookahead without returning the / or ? as part of the match.
A quick test within developer tools:
# create a list of example urls to test against (only one should match regex)
urls = ["http://example.com/things/stuff/532453?morethings&stuff=things",
"http://example.com/things/stuff?morethings&stuff=things",
"http://example.com/things/stuff/123a?morethings&stuff=things"]
urls.forEach(function(value) {
console.log(value.match(/(?!\/)\d+(?=\?)/));
})
# returns the following:
["532453"]
null
null

Just use this:
([\d]+)
You can check this link out: https://regex101.com/r/hR2eY7/1
if you use javascript:
/([\d]+)/g

Try this :
url = "http://example.com/things/stuff/532453?morethings&stuff=things"
number = url.match(/(\d+)\?/g)[0].slice(0,-1)
Though the approach is slightly naive, it works. It grabs numbers with ? at the end then removes the ? from the end using slice.

Match only the line which end with specific char

How to match the line which does not contain the final dot (full stop/period), in order to add it afterwards.
Someword someword someword.
Someword someword someword
Someword someword someword.
These are my unsuccessful attempts:
.+(?=\.)
.+[^.]
--- update
This works for me:
.+\w+(?:\n)
https://regex101.com/r/sR0aD7/1

The following should match a string that ends with anything but dot: [^.]$ - "anything but dot" and end-of-text marker.

How to match the line which does not contain the final dot (full stop/period),
You can use negative lookahead like this:
/(?!\.$)/
OR else you can also inverse test:
if (!/\.$/.test(input)) { console.log("line is not ending with dot"); }

Regular expression is one way i think you can use this method also --->
function lastCharacter(sentence){
var length = sentence.length;
return sentence.charAt(length-1);
}
Example :-
Input ---> Hey JavaScript is damm good.
Use ---> lastCharacter('Hey JavaScript is damm good.');
Output ---> '.'
In other cases you can check with if condition for dot('.').

Just use something like this: [^\.]$
$ - Indicates end of line.
[^...] - Indicates selecting lines not containing "..."
\. - This is the escaped "." Character. It needs to be escaped because . is anything.
Pulling this together, you get a regular expression .+[^\.]$ which will match your line. You will need the m flag (Multiline) for this to work (I believe)

This works for me:
.+\w+(?:\n)
https://regex101.com/r/sR0aD7/1

How to find in javascript with regular expression string from url?

Good evening, How can I find in javascript with regular expression string from url address for example i have url: http://www.odsavacky.cz/blog/wpcproduct/mikronebulizer/ and I need only string between last slashes (/ /) http://something.cz/something/string/ in this example word that i need is mikronebulizer. Thank you very much for you help.

You could use a regex match with a group.
Use this:
/([\w\-]+)\/$/.exec("http://www.odsavacky.cz/blog/wpcproduct/mikronebulizer/")[1];
Here's a jsfiddle showing it in action
This part: ([\w\-]+)
Means at least 1 or more of the set of alphanumeric, underscore and hyphen and use it as the first match group.
Followed by a /
And then finally the: $
Which means the line should end with this
The .exec() returns an array where the first value is the full match (IE: "mikronebulizer/") and then each match group after that.
So .exec()[1] returns your value: mikronebulizer

Simply:
url.match(/([^\/]*)\/$/);
Should do it.
If you want to match (optionally) without a trailing slash, use:
url.match(/([^\/]*)\/?$/);
See it in action here: http://regex101.com/r/cL3qG3

If you have the url provided, then you can do it this way:
var url = 'http://www.odsavacky.cz/blog/wpcproduct/mikronebulizer/';
var urlsplit = url.split('/');
var urlEnd = urlsplit[urlsplit.length- (urlsplit[urlsplit.length-1] == '' ? 2 : 1)];
This will match either everything after the last slash, if there's any content there, and otherwise, it will match the part between the second-last and the last slash.

Something else to consider - yes a pure RegEx approach might be easier (heck, and faster), but I wanted to include this simply to point out window.location.pathName.
function getLast(){
// Strip trailing slash if present
var path = window.location.pathname.replace(/\/$?/, '');
return path.split('/').pop();
}

Alternatively you could get using split:
var pieces = "http://www.odsavacky.cz/blog/wpcproduct/mikronebulizer/".split("/");
var lastSegment = pieces[pieces.length - 2];
// lastSegment == mikronebulizer

var url = 'http://www.odsavacky.cz/blog/wpcproduct/mikronebulizer/';
if (url.slice(-1)=="/") {
url = url.substr(0,url.length-1);
}
var lastSegment = url.split('/').pop();
document.write(lastSegment+"<br>");

Regex to get string between curly braces

Unfortunately, despite having tried to learn regex at least one time a year for as many years as I can remember, I always forget as I use them so infrequently. This year my new year's resolution is to not try and learn regex again - So this year to save me from tears I'll give it to Stack Overflow. (Last Christmas remix).
I want to pass in a string in this format {getThis}, and be returned the string getThis. Could anyone be of assistance in helping to stick to my new year's resolution?
Related questions on Stack Overflow:
How can one turn regular quotes (i.e. ', ") into LaTeX/TeX quotes (i.e. `', ``'')
Regex: To pull out a sub-string between two tags in a string
Regex to replace all \n in a String, but no those inside [code] [/code] tag

Try
/{(.*?)}/
That means, match any character between { and }, but don't be greedy - match the shortest string which ends with } (the ? stops * being greedy). The parentheses let you extract the matched portion.
Another way would be
/{([^}]*)}/
This matches any character except a } char (another way of not being greedy)

/\{([^}]+)\}/
/ - delimiter
\{ - opening literal brace escaped because it is a special character used for quantifiers eg {2,3}
( - start capturing
[^}] - character class consisting of
^ - not
} - a closing brace (no escaping necessary because special characters in a character class are different)
+ - one or more of the character class
) - end capturing
\} - the closing literal brace
/ - delimiter

If your string will always be of that format, a regex is overkill:
>>> var g='{getThis}';
>>> g.substring(1,g.length-1)
"getThis"
substring(1 means to start one character in (just past the first {) and ,g.length-1) means to take characters until (but not including) the character at the string length minus one. This works because the position is zero-based, i.e. g.length-1 is the last position.
For readers other than the original poster: If it has to be a regex, use /{([^}]*)}/ if you want to allow empty strings, or /{([^}]+)}/ if you want to only match when there is at least one character between the curly braces. Breakdown:
/: start the regex pattern
{: a literal curly brace
(: start capturing
[: start defining a class of characters to capture
^}: "anything other than }"
]: OK, that's our whole class definition
*: any number of characters matching that class we just defined
): done capturing
}: a literal curly brace must immediately follow what we captured
/: end the regex pattern

Try this:
/[^{\}]+(?=})/g
For example
Welcome to RegExr v2.1 by #{gskinner.com}, #{ssd.sd} hosted by Media Temple!
will return gskinner.com, ssd.sd.

Try this
let path = "/{id}/{name}/{age}";
const paramsPattern = /[^{}]+(?=})/g;
let extractParams = path.match(paramsPattern);
console.log("extractParams", extractParams) // prints all the names between {} = ["id", "name", "age"]

Here's a simple solution using javascript replace
var st = '{getThis}';
st = st.replace(/\{|\}/gi,''); // "getThis"
As the accepted answer above points out the original problem is easily solved with substring, but using replace can solve the more complicated use cases
If you have a string like "randomstring999[fieldname]"
You use a slightly different pattern to get fieldname
var nameAttr = "randomstring999[fieldname]";
var justName = nameAttr.replace(/.*\[|\]/gi,''); // "fieldname"

This one works in Textmate and it matches everything in a CSS file between the curly brackets.
\{(\s*?.*?)*?\}
selector {.
.
matches here
including white space.
.
.}
If you want to further be able to return the content, then wrap it all in one more set of parentheses like so:
\{((\s*?.*?)*?)\}
and you can access the contents via $1.
This also works for functions, but I haven't tested it with nested curly brackets.

You want to use regex lookahead and lookbehind. This will give you only what is inside the curly braces:
(?<=\{)(.*?)(?=\})

i have looked into the other answers, and a vital logic seems to be missing from them . ie, select everything between two CONSECUTIVE brackets,but NOT the brackets
so, here is my answer
\{([^{}]+)\}

Regex for getting arrays of string with curly braces enclosed occurs in string, rather than just finding first occurrence.
/\{([^}]+)\}/gm

var re = /{(.*)}/;
var m = "{helloworld}".match(re);
if (m != null)
console.log(m[0].replace(re, '$1'));
The simpler .replace(/.*{(.*)}.*/, '$1') unfortunately returns the entire string if the regex does not match. The above code snippet can more easily detect a match.

Try this one, according to http://www.regextester.com it works for js normaly.
([^{]*?)(?=\})

This one matches everything even if it finds multiple closing curly braces in the middle:
\{([\s\S]*)\}
Example:
{
"foo": {
"bar": 1,
"baz": 1,
}
}

You can use this regex recursion to match everythin between, even another {} (like a JSON text) :
\{([^()]|())*\}

Even this helps me while trying to solve someone's problem,
Split the contents inside curly braces ({}) having a pattern like,
{'day': 1, 'count': 100}.
For example:
#include <iostream>
#include <regex>
#include<string>
using namespace std;
int main()
{
//string to be searched
string s = "{'day': 1, 'count': 100}, {'day': 2, 'count': 100}";
// regex expression for pattern to be searched
regex e ("\\{[a-z':, 0-9]+\\}");
regex_token_iterator<string::iterator> rend;
regex_token_iterator<string::iterator> a ( s.begin(), s.end(), e );
while (a!=rend) cout << " [" << *a++ << "]";
cout << endl;
return 0;
}
Output:
[{'day': 1, 'count': 100}] [{'day': 2, 'count': 100}]

Your can use String.slice() method.
let str = "{something}";
str = str.slice(1,-1) // something