I'm trying to parse the following kind of string:
[key:"val" key2:"val2"]
where there are arbitrary key:"val" pairs inside. I want to grab the key name and the value.
For those curious I'm trying to parse the database format of task warrior.
Here is my test string:
[description:"aoeu" uuid:"123sth"]
which is meant to highlight that anything can be in a key or value aside from space, no spaces around the colons, and values are always in double quotes.
In node, this is my output:
[deuteronomy][gatlin][~]$ node
> var re = /^\[(?:(.+?):"(.+?)"\s*)+\]$/g
> re.exec('[description:"aoeu" uuid:"123sth"]');
[ '[description:"aoeu" uuid:"123sth"]',
'uuid',
'123sth',
index: 0,
input: '[description:"aoeu" uuid:"123sth"]' ]
But description:"aoeu" also matches this pattern. How can I get all matches back?
Continue calling re.exec(s) in a loop to obtain all the matches:
var re = /\s*([^[:]+):\"([^"]+)"/g;
var s = '[description:"aoeu" uuid:"123sth"]';
var m;
do {
m = re.exec(s);
if (m) {
console.log(m[1], m[2]);
}
} while (m);
Try it with this JSFiddle: https://jsfiddle.net/7yS2V/
str.match(pattern), if pattern has the global flag g, will return all the matches as an array.
For example:
const str = 'All of us except #Emran, #Raju and #Noman were there';
console.log(
str.match(/#\w*/g)
);
// Will log ["#Emran", "#Raju", "#Noman"]
To loop through all matches, you can use the replace function:
var re = /\s*([^[:]+):\"([^"]+)"/g;
var s = '[description:"aoeu" uuid:"123sth"]';
s.replace(re, function(match, g1, g2) { console.log(g1, g2); });
This is a solution
var s = '[description:"aoeu" uuid:"123sth"]';
var re = /\s*([^[:]+):\"([^"]+)"/g;
var m;
while (m = re.exec(s)) {
console.log(m[1], m[2]);
}
This is based on lawnsea's answer, but shorter.
Notice that the `g' flag must be set to move the internal pointer forward across invocations.
str.match(/regex/g)
returns all matches as an array.
If, for some mysterious reason, you need the additional information comes with exec, as an alternative to previous answers, you could do it with a recursive function instead of a loop as follows (which also looks cooler :).
function findMatches(regex, str, matches = []) {
const res = regex.exec(str)
res && matches.push(res) && findMatches(regex, str, matches)
return matches
}
// Usage
const matches = findMatches(/regex/g, str)
as stated in the comments before, it's important to have g at the end of regex definition to move the pointer forward in each execution.
We are finally beginning to see a built-in matchAll function, see here for the description and compatibility table. It looks like as of May 2020, Chrome, Edge, Firefox, and Node.js (12+) are supported but not IE, Safari, and Opera. Seems like it was drafted in December 2018 so give it some time to reach all browsers, but I trust it will get there.
The built-in matchAll function is nice because it returns an iterable. It also returns capturing groups for every match! So you can do things like
// get the letters before and after "o"
let matches = "stackoverflow".matchAll(/(\w)o(\w)/g);
for (match of matches) {
console.log("letter before:" + match[1]);
console.log("letter after:" + match[2]);
}
arrayOfAllMatches = [...matches]; // you can also turn the iterable into an array
It also seem like every match object uses the same format as match(). So each object is an array of the match and capturing groups, along with three additional properties index, input, and groups. So it looks like:
[<match>, <group1>, <group2>, ..., index: <match offset>, input: <original string>, groups: <named capture groups>]
For more information about matchAll there is also a Google developers page. There are also polyfills/shims available.
If you have ES9
(Meaning if your system: Chrome, Node.js, Firefox, etc supports Ecmascript 2019 or later)
Use the new yourString.matchAll( /your-regex/g ).
If you don't have ES9
If you have an older system, here's a function for easy copy and pasting
function findAll(regexPattern, sourceString) {
let output = []
let match
// auto-add global flag while keeping others as-is
let regexPatternWithGlobal = RegExp(regexPattern,[...new Set("g"+regexPattern.flags)].join(""))
while (match = regexPatternWithGlobal.exec(sourceString)) {
// get rid of the string copy
delete match.input
// store the match data
output.push(match)
}
return output
}
example usage:
console.log( findAll(/blah/g,'blah1 blah2') )
outputs:
[ [ 'blah', index: 0 ], [ 'blah', index: 6 ] ]
Based on Agus's function, but I prefer return just the match values:
var bob = "> bob <";
function matchAll(str, regex) {
var res = [];
var m;
if (regex.global) {
while (m = regex.exec(str)) {
res.push(m[1]);
}
} else {
if (m = regex.exec(str)) {
res.push(m[1]);
}
}
return res;
}
var Amatch = matchAll(bob, /(&.*?;)/g);
console.log(Amatch); // yeilds: [>, <]
Iterables are nicer:
const matches = (text, pattern) => ({
[Symbol.iterator]: function * () {
const clone = new RegExp(pattern.source, pattern.flags);
let match = null;
do {
match = clone.exec(text);
if (match) {
yield match;
}
} while (match);
}
});
Usage in a loop:
for (const match of matches('abcdefabcdef', /ab/g)) {
console.log(match);
}
Or if you want an array:
[ ...matches('abcdefabcdef', /ab/g) ]
Here is my function to get the matches :
function getAllMatches(regex, text) {
if (regex.constructor !== RegExp) {
throw new Error('not RegExp');
}
var res = [];
var match = null;
if (regex.global) {
while (match = regex.exec(text)) {
res.push(match);
}
}
else {
if (match = regex.exec(text)) {
res.push(match);
}
}
return res;
}
// Example:
var regex = /abc|def|ghi/g;
var res = getAllMatches(regex, 'abcdefghi');
res.forEach(function (item) {
console.log(item[0]);
});
If you're able to use matchAll here's a trick:
Array.From has a 'selector' parameter so instead of ending up with an array of awkward 'match' results you can project it to what you really need:
Array.from(str.matchAll(regexp), m => m[0]);
If you have named groups eg. (/(?<firstname>[a-z][A-Z]+)/g) you could do this:
Array.from(str.matchAll(regexp), m => m.groups.firstName);
Since ES9, there's now a simpler, better way of getting all the matches, together with information about the capture groups, and their index:
const string = 'Mice like to dice rice';
const regex = /.ice/gu;
for(const match of string.matchAll(regex)) {
console.log(match);
}
// ["mice", index: 0, input: "mice like to dice rice", groups:
undefined]
// ["dice", index: 13, input: "mice like to dice rice",
groups: undefined]
// ["rice", index: 18, input: "mice like to dice
rice", groups: undefined]
It is currently supported in Chrome, Firefox, Opera. Depending on when you read this, check this link to see its current support.
Use this...
var all_matches = your_string.match(re);
console.log(all_matches)
It will return an array of all matches...That would work just fine....
But remember it won't take groups in account..It will just return the full matches...
I would definatly recommend using the String.match() function, and creating a relevant RegEx for it. My example is with a list of strings, which is often necessary when scanning user inputs for keywords and phrases.
// 1) Define keywords
var keywords = ['apple', 'orange', 'banana'];
// 2) Create regex, pass "i" for case-insensitive and "g" for global search
regex = new RegExp("(" + keywords.join('|') + ")", "ig");
=> /(apple|orange|banana)/gi
// 3) Match it against any string to get all matches
"Test string for ORANGE's or apples were mentioned".match(regex);
=> ["ORANGE", "apple"]
Hope this helps!
This isn't really going to help with your more complex issue but I'm posting this anyway because it is a simple solution for people that aren't doing a global search like you are.
I've simplified the regex in the answer to be clearer (this is not a solution to your exact problem).
var re = /^(.+?):"(.+)"$/
var regExResult = re.exec('description:"aoeu"');
var purifiedResult = purify_regex(regExResult);
// We only want the group matches in the array
function purify_regex(reResult){
// Removes the Regex specific values and clones the array to prevent mutation
let purifiedArray = [...reResult];
// Removes the full match value at position 0
purifiedArray.shift();
// Returns a pure array without mutating the original regex result
return purifiedArray;
}
// purifiedResult= ["description", "aoeu"]
That looks more verbose than it is because of the comments, this is what it looks like without comments
var re = /^(.+?):"(.+)"$/
var regExResult = re.exec('description:"aoeu"');
var purifiedResult = purify_regex(regExResult);
function purify_regex(reResult){
let purifiedArray = [...reResult];
purifiedArray.shift();
return purifiedArray;
}
Note that any groups that do not match will be listed in the array as undefined values.
This solution uses the ES6 spread operator to purify the array of regex specific values. You will need to run your code through Babel if you want IE11 support.
Here's a one line solution without a while loop.
The order is preserved in the resulting list.
The potential downsides are
It clones the regex for every match.
The result is in a different form than expected solutions. You'll need to process them one more time.
let re = /\s*([^[:]+):\"([^"]+)"/g
let str = '[description:"aoeu" uuid:"123sth"]'
(str.match(re) || []).map(e => RegExp(re.source, re.flags).exec(e))
[ [ 'description:"aoeu"',
'description',
'aoeu',
index: 0,
input: 'description:"aoeu"',
groups: undefined ],
[ ' uuid:"123sth"',
'uuid',
'123sth',
index: 0,
input: ' uuid:"123sth"',
groups: undefined ] ]
My guess is that if there would be edge cases such as extra or missing spaces, this expression with less boundaries might also be an option:
^\s*\[\s*([^\s\r\n:]+)\s*:\s*"([^"]*)"\s*([^\s\r\n:]+)\s*:\s*"([^"]*)"\s*\]\s*$
If you wish to explore/simplify/modify the expression, it's been
explained on the top right panel of
regex101.com. If you'd like, you
can also watch in this
link, how it would match
against some sample inputs.
Test
const regex = /^\s*\[\s*([^\s\r\n:]+)\s*:\s*"([^"]*)"\s*([^\s\r\n:]+)\s*:\s*"([^"]*)"\s*\]\s*$/gm;
const str = `[description:"aoeu" uuid:"123sth"]
[description : "aoeu" uuid: "123sth"]
[ description : "aoeu" uuid: "123sth" ]
[ description : "aoeu" uuid : "123sth" ]
[ description : "aoeu"uuid : "123sth" ] `;
let m;
while ((m = regex.exec(str)) !== null) {
// This is necessary to avoid infinite loops with zero-width matches
if (m.index === regex.lastIndex) {
regex.lastIndex++;
}
// The result can be accessed through the `m`-variable.
m.forEach((match, groupIndex) => {
console.log(`Found match, group ${groupIndex}: ${match}`);
});
}
RegEx Circuit
jex.im visualizes regular expressions:
const re = /^\[(?:(.+?):"(.+?)"\s*)+\]$/g
const matches = [...re.exec('[description:"aoeu" uuid:"123sth"]').entries()]
console.log(matches)
Basically, this is ES6 way to convert Iterator returned by exec to a regular Array
Here is my answer:
var str = '[me nombre es] : My name is. [Yo puedo] is the right word';
var reg = /\[(.*?)\]/g;
var a = str.match(reg);
a = a.toString().replace(/[\[\]]/g, "").split(','));
Related
I have a list of data that I need to put a ' symbol at the start of the line and at the end of the line. So the original data look like
11223334444xxx55555
11xxx223334444555xxx55
11xxxx22333444xxx455555
11xxxxx22333444455555
11xxxxxx223334444555xxx55
and I want all the line to look like
11223334444yyyy55555
11yyyy223334444555yyyy55
11yyyyx22333444yyyy455555
11yyyyxx22333444455555
11yyyyyyyy223334444555yyyy55
that is 'yyyy' replace 'xxx', how I write my code? Both typescript and javascript are perfect.
Sorry, my bad. I want develop an extension to do it, and above just an example. Many answers below just miss select full text part.
const textEditor = vscode.window.activeTextEditor;
if (!textEditor) {
return; // No open text editor
}
var firstLine = textEditor.document.lineAt(0);
var lastLine = textEditor.document.lineAt(textEditor.document.lineCount - 1);
var textRange = new vscode.Range(0,
firstLine.range.start.character,
textEditor.document.lineCount - 1,
lastLine.range.end.character);
textEditor.edit(function (editBuilder) {
editBuilder.replace(textRange, '$1');
});
});
replace function above just has one replace argument, not two, how can I replace it?
Try this:
let checkD =
11223334444xxx55555
11xxx223334444555xxx55
11xxxx22333444xxx455555
11xxxxx22333444455555
11xxxxxx223334444555xxx55
;
checkD.replace(/xxx/g, 'yyyy');
You can try it with regex. Read more at the MDN.
Here is an array example:
let data = [
"11223334444xxx55555",
"11xxx223334444555xxx55",
"11xxxx22333444xxx455555",
"11xxxxx22333444455555",
"11xxxxxx223334444555xxx55"
];
let max = data.length;
for (let i = 0; i < max; i++) {
let regex = new RegExp("xxx", "g")
data[i] = data[i].replace(regex, "yyyy")
}
console.log(data);
Here is a single string example:
let data = `11223334444xxx55555
11xxx223334444555xxx55
11xxxx22333444xxx455555
11xxxxx22333444455555
11xxxxxx223334444555xxx55`;
let regex = new RegExp("xxx", "g")
data = data.replace(regex, "yyyy")
console.log(data);
I know it's late, but I was struggling with this sort of an issue and got myself to resolve it, using JS.
If I understood your question right.
You want to replace three 'x' letters with four 'y' letters.
I.e., turn this
xxxxxx1xxx2xx into this yyyyyyyy1yyyy2xx
const textEditor = vscode.window.activeTextEditor;
if (!textEditor) {
vscode.window.showErrorMessage("Editor Does Not Exist");
return;
}
var m;
let fullText = editor.document.getText();
const regex = /xxx/gm; // 'g' flag is for global search & 'm' flag is for multiline.
//searching for previously declared xxx in regex and replacing it with 'yyyy'.
let textReplace = fullText.replace(regex, `yyyy`);
//Creating a new range with startLine, startCharacter & endLine, endCharacter.
let invalidRange = new vscode.Range(0, 0, editor.document.lineCount, 0);
// To ensure that above range is completely contained in this document.
let validFullRange = editor.document.validateRange(invalidRange);
while ((m = regex.exec(fullText)) !== null) {
// This is necessary to avoid infinite loops with zero-width matches
if (m.index === regex.lastIndex) {
regex.lastIndex++;
}
editor.edit(editBuilder => {
editBuilder.replace(validFullRange, textReplace);
}).catch(err => console.log(err));
}
Regexp g flag is for global search to match all occurrences, without it, it'll only check for the first.
Regexp m flag is for multiline, it makes ^ and $ match at line beginnings and line endings(instead of at string), respectively.
Reference: global - JavaScript | MDN ,
multiline - JavaScript | MDN
Also, consider looking at the VSCode API Doc for Range & Validate Range
that is 'yyyy' replace 'xxx', how I write my code
split and join is what I use e.g.
str.split('yyyy').join('xxx');
I want to match a number in a string:
'abc#2003, or something else #2017'
I want to get result [2003, 2007] with match function.
let strReg = 'abc#2003, or something else #2017';
let reg = new RegExp(/(?=(#\d+))\1/);
strReg.match(reg) //[ '#2003 ', '#2017 ' ]
let reg1 = new RegExp(/(?=#(\d+))\1/)
strReg.match(reg1) //null, but I expect ［2003, 2007］
the result mains '\1' match after '?=', ?=()\1 works, ?=#()\1 not.
javascript only supports backwards, how should I do to match '#' but ignore it?
I take it that you want an array of the results, so...
var s = "abc#2003, or something else #2017 not the 2001 though";
var re = /#(\d+)/g;
var result = [];
var match = re.exec(s);
while (match !== null) {
result.push(parseInt(match[1]));
match = re.exec(s);
}
console.log(result);
Outputs:
Array [ 2003, 2017 ]
match(0) is the entire match, match(1) is the captured group.
Also, see How do you access the matched groups in a JavaScript regular expression?
Inspired by javascript regex - look behind alternative?, if you want to do it as almost a one-liner:
var re = /(\d+)(?=#)/g; /* write the regex backwards */
var result = [];
s.split('').reverse().join('').match(re).forEach(function (el) { result.push(parseInt(el.split('').reverse().join(''))); });
console.log(result.reverse());
Caveat: Who wrote this programing saying? “Always code as if the guy who ends up maintaining your code will be a violent psychopath who knows where you live.”
Small change to your code does the job as follows:
/#(\d+)/g
number followed by # will be remembered as you required.
Probably something simple but i am trying to return the capture groups from this regex...
const expression = /^\/api(?:\/)?([^\/]+)?\/users\/([^\/]+)$/g
The code i am using to do this is the following...
const matchExpression = (expression, pattern) => {
let match;
let matches = [];
while((match = expression.exec(pattern)) != null) {
matches.push(match[1]);
};
return matches;
};
I am expecting the following result when matched against /api/v1/users/1...
['v1', '1']
But instead only seem to get one result which is always the first group.
The expression itself is fine and has been tested across multiple services but can't seem to figure out why this is not working as expected.
Any help would be hugely appreciated.
You must make sure you add the second capturing group contents to the resulting array:
while((match = expression.exec(pattern)) != null) {
matches.push(match[1]);
matches.push(match[2]); // <- here
};
Since you are matching an entire string, you can use a /^\/api(?:\/)?([^\/]+)?\/users\/([^\/]+)$/ regex (since you are matching a whole string you need no g global modifier) and reduce the code to:
const matchExpression = (expression, pattern) => {
let matches = pattern.match(expression);
if (matches) {
matches = matches.slice(1);
}
return matches;
};
The point is that you can use String#match with a regex without global modifier to access capturing group contents.
Demo:
var expr = /^\/api(?:\/)?([^\/]+)?\/users\/([^\/]+)$/;
var matches = "/api/v1/users/1".match(expr);
if (matches) {
console.log(matches.slice(1));
}
disclaimer - absolutely new to regexes....
I have a string like this:
subject=something||x-access-token=something
For this I need to extract two values. Subject and x-access-token.
As a starting point, I wanted to collect two strings: subject= and x-access-token=. For this here is what I did:
/[a-z,-]+=/g.exec(mystring)
It returns only one element subject=. I expected both of them. Where i am doing wrong?
The g modifier does not affect exec, because exec only returns the first match by specification. What you want is the match method:
mystring.match(/[a-z,-]+=/g)
No regex necessary. Write a tiny parser, it's easy.
function parseValues(str) {
var result = {};
str.split("||").forEach(function (item) {
var parts = item.split("=");
result[ parts[0] /* key */ ] = parts[1]; /* value */
});
return result;
}
usage
var obj = parseValues("subject=something||x-access-token=something-else");
// -> {subject: "something", x-access-token: "something-else"}
var subj = obj.subject;
// -> "something"
var token = obj["x-access-token"];
// -> "something-else"
Additional complications my arise when there is an escaping schema involved that allows you to have || inside a value, or when a value can contain an =.
You will hit these complications with regex approach as well, but with a parser-based approach they will be much easier to solve.
You have to execute exec twice to get 2 extracted strings.
According to MDN: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/exec
If your regular expression uses the "g" flag, you can use the exec() method multiple times to find successive matches in the same string.
Usually, people extract all strings matching the pattern one by one with a while loop. Please execute following code in browser console to see how it works.
var regex = /[a-z,-]+=/g;
var string = "subject=something||x-access-token=something";
while(matched = regex.exec(string)) console.log(matched);
You can convert the string into a valid JSON string, then parse it to retrieve an object containing the expected data.
var str = 'subject=something||x-access-token=something';
var obj = JSON.parse('{"' + str.replace(/=/g, '":"').replace(/\|\|/g, '","') + '"}');
console.log(obj);
I don't think you need regexp here, just use the javascript builtin function "split".
var s = "subject=something1||x-access-token=something2";
var r = s.split('||'); // r now is an array: ["subject=something1", "x-access-token=something2"]
var i;
for(i=0; i<r.length; i++){
// for each array's item, split again
r[i] = r[i].split('=');
}
At the end you have a matrix like the following:
y x 0 1
0 subject something1
1 x-access-token something2
And you can access the elements using x and y:
"subject" == r[0][0]
"x-access-token" == r[1][0]
"something2" == r[1][1]
If you really want to do it with a pure regexp:
var input = 'subject=something1||x-access-token=something2'
var m = /subject=(.*)\|\|x-access-token=(.*)/.exec(input)
var subject = m[1]
var xAccessToken = m[2]
console.log(subject);
console.log(xAccessToken);
However, it would probably be cleaner to split it instead:
console.log('subject=something||x-access-token=something'
.split(/\|\|/)
.map(function(a) {
a = a.split(/=/);
return { key: a[0], val: a[1] }
}));
I'm trying to use node to do some regex on a css file.
Here's my javascript:
var fs = require ('fs');
fs.readFile('test.css','utf8',function(error,css){
if(error){
console.log("I'm sorry, something went terribly wrong :o Here's the message: "+error);
}
var matches = css.match(/([a-zA-Z-]+):\s*([0-9]+)(vh|VH|vw|VW)/g);
console.log(matches[2][1]);
});
Expected output when I run it:
Actual output:
As you can see it does not put every match in its own array as expected, it just puts everything in one giant array without any sub-arrays.
Anything I can do?
match doesn't give you the detailed match results in this case:
If the regular expression includes the g flag, the method returns an Array containing all matched substrings rather than match objects.
You can use exec instead:
var regex = /([a-zA-Z-]+):\s*([0-9]+)(vh|VH|vw|VW)/g;
var css = "body{\nfont-size: 10vw;\n height: 500vh\n}";
var match;
while (match = r.exec(css)){
console.log(match)
}
Which gives this output:
["font-size: 10vw", "font-size", "10", "vw", index: 6, input: "body{↵font-size: 10vw;↵ height: 500vh↵}"]
["height: 500vh", "height", "500", "vh", index: 24, input: "body{↵font-size: 10vw;↵ height: 500vh↵}"]
var reg = /([a-zA-Z-]+):\s*([0-9]+)(vh|VH|vw|VW)/g;
var matches = [];
var m;
while ((m = reg.exec(css)) !== null) {
if (m.index === reg.lastIndex) {
reg.lastIndex++;
}
matches.push(m);
}
console.log(matches);
It's actually expected behaviour. If you use string.match method with global flag, parenthesis do not create groups inside matches:
var str = "javascript is cool";
var result = str.match( /JAVA(SCRIPT)/g );
console.log( result[0] ); // javascript
console.log( result.length ); //1
Your case is using regexp.exec(str). It can find all matches and parenthesis groups in it.
var str = 'javascript is cool. Use javascript';
var regexp = /java(script)/g;
while (result = regexp.exec(str)) {
console.log(result.length); //2
console.log(result[0]); //javascript
console.log(result[1]); //script
}
Based on above answers I managed to come up with the following code:
while(match = css.match(/([a-zA-Z-]+):\s*([0-9]+)(vh|VH|vw|VW)/)){
matches.push(match);
css = css.replace(match[0],'');
}
Thanks a ton for the help everyone!