问题
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?
回答1:
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/
回答2:
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 was there';
console.log(
str.match(/@\w*/g)
);
// Will log ["@Emran", "@Raju", "@Noman"]回答3:
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); });
回答4:
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.
回答5:
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.
回答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: [>, <]
回答7:
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) ]
回答8:
We are finally beginning to see a built-in matchAll function, see here for the description and compatibility table. It looks like as of April 2019, Chrome and Firefox are supported but not IE, Edge, Opera, or Node.js. 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.
回答9:
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;
}
var regex = /abc|def|ghi/g;
var res = getAllMatches(regex, 'abcdefghi');
res.forEach(function (item) {
console.log(item[0]);
});
回答10:
If your system (Chrome/Node.js/Firefox) supports ES9, use new the a_string.matchAll(regex). If you have an older system, here's a function for easy copy and pasting
function findAll(regexPattern, sourceString) {
let output = []
let match
// make sure the pattern has the global flag
let regexPatternWithGlobal = RegExp(regexPattern,"g")
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 ] ]
回答11:
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.
回答12:
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...
回答13:
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!
回答14:
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.
回答15:
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 ] ]
回答16:
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:
回答17:
If you want a more functional approach and to avoid looping you can just call a function until the result is null and on each capture you slice the string to the position of the captures group.
// The MatchAll Function
function matchAll (regexp, input, matches = []) {
const regex = regexp.exec(input)
if (regex === null) return matches
// Filter out any undefined results
const matched = regex.filter(i => i)
// Destruct some common used values
const { index } = regex
const [ full, g1, g2, g3] = matched
// Slice the input string to last match
const string = input.slice(index + full.length)
// Do something with the captured groups
// Push this into an array
matches.push({
prop: 'H' + g1 + g3 + g3 + 'ary ' + g2
})
// Return
return matchAll(regexp, string)
}
// Record of matches
const matches = []
// The RegExp, we are looking for some random letters / words in string
const regExp = new RegExp(/(i{1}).*(did).*(l{1})/)
// An example string to parse
const testString = `Jeffrey Epstein didn't kill himself!`
// Run
matchAll(regExp, testString, matches)
// Returned Result
console.log(matches)回答18:
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(','));
来源：https://stackoverflow.com/questions/6323417/regex-to-extract-all-matches-from-string-using-regexp-exec