I am trying to get string using RegEx; here is the string:
window.runParams = {};
window.runParams = {blablabla};
How to get the second string {blablabla}? I am using REGEX:
(?<=window.runParams = ").*(?=;)
But that gets the first string {}.
If you want to get string with braces eg: {blablabla}
window.runParams = ({\w+})
If you want to get only the string inside braces eg: blablabla
window.runParams = {(\w+)}
Value of group 1 is your result
The following pattern captures only curly brackets with word character content:
(?<=window.runParams = ){\w+}(?=;)
and will only capture:
{blablabla}
when run against the text:
window.runParams = {};
window.runParams = {blablabla};
See results here:
https://regex101.com/r/mTwA64/1
try modifying your regex so it only accepts matches with non-empty curly brackets \{.+\} such as
(?<=window\.runParams = )(\{.+\})(?=;)
...there's probably ways to simplify the regex further, depending on you problem...my guess is you don't need the lookahead/lookbehind, e.g. in the example given \{.+\} will do just fine (returns {blablabla}) ....but it really depends on the format and content of your file...also remember braces, dots etc have a special meaning in regexes so you probably would want to escape them
Related
I have an input string and I would like to replace the values inside the first curly braces (including the curly braces) after a specific keyword.
For example in this example I would like to replace the values inside the curly braces which comes after Honda.
[{brand:”Toyota”, model:”Corolla”},
{brand:”BMW”, model:”X3”},
{brand:”Honda”, model:”{year:2022, name:Accord}”}]
So the end result will be something like:
[{brand:”Toyota”, model:”Corolla”},
{brand:”BMW”, model:”X3”},
{brand:”Honda”, model:”Civic”}]
It looks like when using Java I can use replaceAll method to utilize a regex expression but I couldn't figure what should be the regex expression for this use case.
I tried to create a regex like this one below but result was wrong
String output = input.replaceAll("Honda.*(?<=\\{).*?(?=\\})","Civic");
[{brand:”Toyota”, model:”Corolla”},
{brand:”BMW”, model:”X3”},
{brand:”Civic}”}]
Look-behinds are problematic because (in Java) they must be fixed-length.
An alternative is to capture the prefix (the brand) and then repeat that as $1 in the replacement string.
This regex will capture the brand. Notice the surrounding parentheses.
(\{brand:"Honda", model:")
And this will match the model (excluding the trailing quote):
[^"]*
Concatenate the two parts. Put them in a string literal, escaping every \ and ":
"(\\{brand:\"Honda\", model:\")[^\"]*"
Optionally, add \\s* to make it robust against arbitrary whitespace:
"(\\{\\s*brand\\s*:\\s*\"Honda\"\\s*,\\s*model\\s*:\\s*\")[^\"]*"
Make sure the replacement string starts with $1:
"$1Civic"
Working example:
class Main
{
public static void main(String[] args)
{
String input = "[{brand:\"Honda\", model:\"{year:2022, name:Accord}\"}]";
System.out.println(input);
String output = input.replaceAll(
"(\\{\\s*brand\\s*:\\s*\"Honda\"\\s*,\\s*model\\s*:\\s*\")[^\"]*",
"$1Civic");
System.out.println(output);
}
}
Output:
[{brand:"Honda", model:"{year:2022, name:Accord}"}]
[{brand:"Honda", model:"Civic"}]
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?
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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
Given string:
some_function(inputId = "select_something"),
(...)
some_other_function(inputId = "some_other_label")
I would like to arrive at:
some_function(inputId = ns("select_something")),
(...)
some_other_function(inputId = ns("some_other_label"))
The key change here is the element ns( ... ) that surrounds the string available in the "" after the inputId
Regex
So far, I have came up with this regex:
:%substitute/\(inputId\s=\s\)\(\"[a-zA-Z]"\)/\1ns(/2/cgI
However, when deployed, it produces an error:
E488: Trailing characters
A simpler version of that regex works, the syntax:
:%substitute/\(inputId\s=\s\)/\1ns(/cgI
would correctly inser ns( after finding inputId = and create string
some_other_function(inputId = ns("some_other_label")
Challenge
I'm struggling to match the remaining part of the string, ex. "select_something") and return it as:
"select_something")).
You have many problems with your regex.
[a-zA-Z] will only match one letter. Presumably you want to match everything up to the next ", so you'll need a \+ and you'll also need to match underscores too. I would recommend \w\+. Unless more than [a-zA-Z_] might be in the string, in which case I would do .\{-}.
You have a /2 instead of \2. This is why you're getting E488.
I would do this:
:%s/\(inputId = \)\(".\{-}\)"/\1ns(\2)/cgI
Or use the start match atom: (that is, \zs)
:%s/inputId = \zs\".\{-}"/ns(&)/cgI
You can use a negated character class "[^"]*" to match a quoted string:
%s/\(inputId\s*=\s*\)\("[^"]*"\)/\1ns(\2)/g
I'm trying to get my Dart web app to: (1) determine if a particular string matches a given regex, and (2) if it does, extract a group/segment out of the string.
Specifically, I want to make sure that a given string is of the following form:
http://myapp.example.com/#<string-of-1-or-more-chars>[?param1=1¶m2=2]
Where <string-of-1-or-more-chars> is just that: any string of 1+ chars, and where the query string ([?param1=1¶m2=2]) is optional.
So:
Decide if the string matches the regex; and if so
Extract the <string-of-1-or-more-chars> group/segment out of the string
Here's my best attempt:
String testURL = "http://myapp.example.com/#fizz?a=1";
String regex = "^http://myapp.example.com/#.+(\?)+\$";
RegExp regexp= new RegExp(regex);
Iterable<Match> matches = regexp.allMatches(regex);
String viewName = null;
if(matches.length == 0) {
// testURL didn't match regex; throw error.
} else {
// It matched, now extract "fizz" from testURL...
viewName = ??? // (ex: matches.group(2)), etc.
}
In the above code, I know I'm using the RegExp API incorrectly (I'm not even using testURL anywhere), and on top of that, I have no clue how to use the RegExp API to extract (in this case) the "fizz" segment/group out of the URL.
The RegExp class comes with a convenience method for a single match:
RegExp regExp = new RegExp(r"^http://myapp.example.com/#([^?]+)");
var match = regExp.firstMatch("http://myapp.example.com/#fizz?a=1");
print(match[1]);
Note: I used anubhava's regular expression (yours was not escaping the ? correctly).
Note2: even though it's not necessary here, it is usually a good idea to use raw-strings for regular expressions since you don't need to escape $ and \ in them. Sometimes using triple-quote raw-strings are convenient too: new RegExp(r"""some'weird"regexp\$""").
Try this regex:
String regex = "^http://myapp.example.com/#([^?]+)";
And then grab: matches.group(1)
String regex = "^http://myapp.example.com/#([^?]+)";
Then:
var match = matches.elementAt(0);
print("${match.group(1)}"); // output : fizz
I know I can exclude outside characters in a string using look-ahead and look-behind, but I'm not sure about characters in the center.
What I want is to get a match of ABCDEF from the string ABC 123 DEF.
Is this possible with a Regex string? If not, can it be accomplished another way?
EDIT
For more clarification, in the example above I can use the regex string /ABC.*?DEF/ to sort of get what I want, but this includes everything matched by .*?. What I want is to match with something like ABC(match whatever, but then throw it out)DEF resulting in one single match of ABCDEF.
As another example, I can do the following (in sudo-code and regex):
string myStr = "ABC 123 DEF";
string tempMatch = RegexMatch(myStr, "(?<=ABC).*?(?=DEF)"); //Returns " 123 "
string FinalString = myStr.Replace(tempMatch, ""); //Returns "ABCDEF". This is what I want
Again, is there a way to do this with a single regex string?
Since the regex replace feature in most languages does not change the string it operates on (but produces a new one), you can do it as a one-liner in most languages. Firstly, you match everything, capturing the desired parts:
^.*(ABC).*(DEF).*$
(Make sure to use the single-line/"dotall" option if your input contains line breaks!)
And then you replace this with:
$1$2
That will give you ABCDEF in one assignment.
Still, as outlined in the comments and in Mark's answer, the engine does match the stuff in between ABC and DEF. It's only the replacement convenience function that throws it out. But that is supported in pretty much every language, I would say.
Important: this approach will of course only work if your input string contains the desired pattern only once (assuming ABC and DEF are actually variable).
Example implementation in PHP:
$output = preg_replace('/^.*(ABC).*(DEF).*$/s', '$1$2', $input);
Or JavaScript (which does not have single-line mode):
var output = input.replace(/^[\s\S]*(ABC)[\s\S]*(DEF)[\s\S]*$/, '$1$2');
Or C#:
string output = Regex.Replace(input, #"^.*(ABC).*(DEF).*$", "$1$2", RegexOptions.Singleline);
A regular expression can contain multiple capturing groups. Each group must consist of consecutive characters so it's not possible to have a single group that captures what you want, but the groups themselves do not have to be contiguous so you can combine multiple groups to get your desired result.
Regular expression
(ABC).*(DEF)
Captures
ABC
DEF
See it online: rubular
Example C# code
string myStr = "ABC 123 DEF";
Match m = Regex.Match(myStr, "(ABC).*(DEF)");
if (m.Success)
{
string result = m.Groups[1].Value + m.Groups[2].Value; // Gives "ABCDEF"
// ...
}