Regex - Capture anything within parentheses and brackets - javascript

I'm really bad at regex and I'm not sure what to do with this. I want to capture anything within () or [] (including the brackets) and nothing after. For example if I type [this is text] I want it to return exactly that. Also I have a json full of terms the user types. If the term is not on the json then it shouldn't print. This is the snippet of code which relates to the regex.
let sf_re = /(?:(,)\s+|\s+(xx)\s+)|\+|(,)/
if (document.querySelector(".images")){
document.querySelector(".images").innerHTML = ""
for(var i=0;i<item.length;i++){
if(/(\[(.*?)\]|\((.*?)\))/.test(item[i])){
let text = item[i]
let note = document.createElement("span")
note.innerHTML = String(text)
document.querySelector(".images").appendChild(note)
}
Here is an example of what happens
The only thing that should show is [cat cat cat]. "dog" should not appear at all because it's not on my list. In regexr it seems to work fine. I'm not sure what to add.
Edit: I think that my original post had insufficient information. The user types into an input bar which is split into an array using .split(). My main goal is to allow the user to add small lines of text. For example if my json has the terms "cat", "dog", and "pig" and the user types those terms, then it will return something. This is what I get using the suggestions below. Notice how "f" returns an arrow the first time, but not the second time. I'm not sure why this happens. I may be using "match" wrong. I tried this and I get an error "cannot read property of undefined":
let regex = /(\[(.*?)\]|\((.*?)\))/
if (document.querySelector(".images")){
document.querySelector(".images").innerHTML = ""
for(var i=0;i<item.length;i++){
if(item[i].match(regex)[0]){
let text = item[i]
let note = document.createElement("span")
note.innerHTML = String(text)
document.querySelector(".images").appendChild(note)
}

Also I have a json full of terms the user types. If the term is not on the json then it shouldn't print.
You can use
obj.hasOwnProperty(prop);
This will let you know whether the object contains the specified prop. Here is an Example -
var x = {
y: 10
};
alert(x.hasOwnProperty("y")); //true
alert(x.hasOwnProperty("z")); //false

The regex is [\[|\(](.*)[\]\)]
Explanation: read '[' or '(', then read anything until there is a ']' or a ')'
var regexp = /[\[|\(](.*)[\]\)]/;
var match = ("[cat cat cat]dog").match(regexp);
console.log(match[1]);

Here are two simple regexs you can use. One matches on brackets and the other matches on parenthesis.
The third is a combination that checks for either. Keep in mind, however, that you'll only ever be able to one of each for [ ] and ( ). If you're accepting multiple of either kind, then the regex will break since it'll return all characters between the outermost ones.
Using match it gives back an array, so if you know you're only getting a single result, you can grab the first item, like in result and result2. If not, you can just deal with the array of results like in result3.
const regex = /\[.*\]/
const regex2 = /\(.*\)/
const regex3 = /(\[.*\])|(\(.*\))/g
const result = ("[cat cat cat]dog").match(regex3)[0];
const result2 = ("banana(cat cat)horse").match(regex3)[0];
const result3 = ("alfred[cat cat cat]banana(dog dog)").match(regex3);
console.log(result); //[cat cat cat]
console.log(result2); //(cat cat)
console.log(result3); // [ '[cat cat cat]', '(dog dog)' ]
The . character matches anything, and the * will repeat 0 or more times. The \ character can be used to escape [, ], ( and ) as necessary (and any other special regex character). The g at the end will check across the entire string.
Hopefully that should be enough to get you un-stuck.

Related

How to split a string by one delimiter but having a particular format as described below

I have a string as:
const str = 'My [Link format](https://google.com) demo'
I want the word array to be like:
['My', '[Link format](https://google.com)', 'demo']
What to do in javascript?
I was trying using split() and str.match(). Nothing worked yet.
This is a simple split on a space as a delimiter, but we us a negative lookahead to check for the combination of open and closed square brackets [] and round brackets ()
const str = 'My [Link format](https://google.com) demo'
console.log(str.split(/\s+(?![^\[]*\])(?![^\(]*\))/));
We also allow for spaces in the URL portion, even though it has a low chance of having spaces, it could still happen
Try it here: https://jsfiddle.net/m4q6e9x7/
["My", "[Link format](https://google.com)", "demo"]
In the fiddle I've tried to show to two separate negative lookaheads for the combination of the types of brackets: (I've put a space in the round brackets to prove the concept)
const str = 'My [Link format](http s://google.com) demo'
ignore space between []
console.log(str.split(/\s+(?![^\[]*\])/));
["My", "[Link format](http", "s://google.com)", "demo"]
ignore space between ()
console.log(str.split(/\s+(?![^\(]*\))/));
["My", "[Link", "format](http s://google.com)", "demo"]
So we can easily combine the two criteria because we need both of them to not match.
Because [] and () need to be escaped, it might be easier to see the regex if we modify and test for spaces between braces {}
const str = 'My {Link format}(https://google.com) demo'
console.log(str.split(/\s+(?![^{]*})/));
["My", "{Link format}(https://google.com)", "demo"]
Both solutions assume, that the string has correct form (meaning basically no space between ']' and '(', no ']' characters inside [...] and similar intuitions. You didn't really provide information about what the input string can be other than your concrete example – so solutions work well in this and very similar cases. Second is very easily modified as needed, first is easily extended to check if the string is in fact not correct.
Solution using Regular Expressions
Below code finds everything before first '[', everything in '[...](...)' pattern (note: first ... must not contain ']', and second – ')', but I assume this would make for an incorrect input in the first place), and everything after that.
So
let regex = /(.*)(\[.*\]\(.*\))(.*)/
let res = str.match(regex).splice(1,3)
gives res as
['My ', '[Link format](https://google.com)', ' demo']
From there, you can trim every entry in this array ('My ' => 'My') for example using a trim function like so:
res.map((val) => val.trim());
Look here for explanation of what the array obtained from .match() method represents, but generally except index 0 it contains capture groups, meaning the parts of string corresponding to parts of regex surrounded by parentheses.
If you are not familiar with Regular Expressions (regexes) in JS, or at all, you will find many online resources about the topic easily. After grasping the basics, regex101 is a nice tool to experiment with regexes and explore their capabilities. When using it, you should probably choose EcmaSCRIPT/JS flavor from the menu on the left.
Equivalent solution without regex
Equivalent solution is to find where is the first '[' manually, as well as where the '[...](...)' pattern ends. Than splice the parts (before '[', pattern, and after pattern) from the string, and probably trim them. So just loop over characters of the string in search of '[' and than ']', '(', ')'. Note that in this case you can easily and granularily decide what to do if the string has unexpected/incorrect form.
TODO: I will probably sketch some code when I have time for it
Regex is your friend!
const regexMdLinks = /!?\[([^\]]*)\]\(([^\)]+)\)/gm
// Example md file contents
const str = `My [Link format](https://google.com) demo My [Link format2](https://google.com/2) demo2`
let regex_splitted = str.split(regexMdLinks);
let arr = [];
//1. Item will be the text (or empty text)
//2. Item is the link text
//3. Item is the url
for(let i = 0; i < regex_splitted.length; i++){
if(i % 3 == 0){ //Split normal text
arr.push(...regex_splitted[i].split(" ").filter(i => i));
} else if(i % 3 == 1){//Add brackets around link text
arr.push("["+regex_splitted[i]+"]");
} else {
arr.push("("+regex_splitted[i]+")");
}
}
console.log(arr)

Obtain arguments from a string seperated by a space and convert an argument in an array format to an array

I have arguments that will be passed by the user for a command. Each argument for a command will be seperated with a space, which will represent a new argument. Example: "arg1 arg2 arg3" converts to ["arg1", "arg2", "arg3"] where the output is a JS array. This can be done with a simple .split(" ").
However, my problem begin when trying to format an array as a command argument. My goal is to allow the user to enter an agument in the format of an array (e.g. Starts with [ may contain multiple elements seperated by a , and ends with a ]) so for example: "arg1 [elem1, elem2] arg3" converts to ["arg1", ["elem1", "elem2"], "arg3"] where the inner and outer array is a JS array.
I have tried using JSON.Parse() however, each element would require the user to have " at the start of each element which is too complex for the user and non essential to be inputting. Also, the elements may not always intend to be a string and may be Boolean, Number or a custom type.
As of currently, this has been my best solution but misses some requirements and also is non functional when an array has a space inside.
s.split(/[\[\]]|\s+/).filter(arg => arg.length > 1);
I have come up with some other solutions but all are missing one thing or another in the required specification set above. A solution that can handle nested arrays would be nice however it is non-essential and could make the solution alot more complex than it needs to be.
Let's assume no funny characters as the input. Also nesting not allowed.
var str = "arg1 [ elem1 , elem2,elem3 ] arg3";
console.log(str)
// removing white spaces from the [ array ]
str = str.replace(/\s*,\s*/g, ',');
str = str.replace(/\[\s*/g, '[');
str = str.replace(/\s*\]/g, ']');
// now split on words
var arr = str.split(/\s+/);
arr = arr.map(function(elem) {
// if begins with [ it is assumed to be an array to be splitted
return elem.charAt(0) == '[' ? elem.slice(1, -1).split(",") : elem;
})
console.log(arr)

How to split a part from a string when a special character is found

I have an array of string which looks like this:
ErrStr = [
\\abc\def\ghi; ,
\\klm\nop\qrs;
]
So, this array of strings will be dynamic, so what I want to extract is only the abc and klm from this array.
This piece of code is what joins the strings after filtering from the data:
let errorArr = filteredError.map(a => a.ErrStr.split('\\\\', '')).join('')
I tried the above code, it cuts the \\ and replaces with , but I am not able to figure out how do I remove the 2nd and 3rd part of the string. Thanks in advance.
Backslash in an Escape character in Javascript and has special meaning. To literally use it you need to, well, escape it beforehand with another backslash
const str1 = "\a";
console.log(str1); // "a" Where's my backslash?
const str2 = "\\a";
console.log(str2); // "\a" Aha, here it is
Therefore, first make sure your array has Strings (which currently what you provided are not strings therefore it's an invalid JSON)
And use:
const arr = [
"\\\\abc\\def\\ghi;",
"\\\\klm\\nop\\qrs;",
];
const codes = arr.map(s => /(?<=^\\\\)[^\\]+/.exec(s)).flat()
console.log(codes); // ["abc", "klm"]
Overview example on Regex101.com

How to split a long regular expression into multiple lines in JavaScript?

I have a very long regular expression, which I wish to split into multiple lines in my JavaScript code to keep each line length 80 characters according to JSLint rules. It's just better for reading, I think.
Here's pattern sample:
var pattern = /^(([^<>()[\]\\.,;:\s#\"]+(\.[^<>()[\]\\.,;:\s#\"]+)*)|(\".+\"))#((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$/;
Extending #KooiInc answer, you can avoid manually escaping every special character by using the source property of the RegExp object.
Example:
var urlRegex= new RegExp(''
+ /(?:(?:(https?|ftp):)?\/\/)/.source // protocol
+ /(?:([^:\n\r]+):([^#\n\r]+)#)?/.source // user:pass
+ /(?:(?:www\.)?([^\/\n\r]+))/.source // domain
+ /(\/[^?\n\r]+)?/.source // request
+ /(\?[^#\n\r]*)?/.source // query
+ /(#?[^\n\r]*)?/.source // anchor
);
or if you want to avoid repeating the .source property you can do it using the Array.map() function:
var urlRegex= new RegExp([
/(?:(?:(https?|ftp):)?\/\/)/ // protocol
,/(?:([^:\n\r]+):([^#\n\r]+)#)?/ // user:pass
,/(?:(?:www\.)?([^\/\n\r]+))/ // domain
,/(\/[^?\n\r]+)?/ // request
,/(\?[^#\n\r]*)?/ // query
,/(#?[^\n\r]*)?/ // anchor
].map(function(r) {return r.source}).join(''));
In ES6 the map function can be reduced to:
.map(r => r.source)
[Edit 2022/08] Created a small github repository to create regular expressions with spaces, comments and templating.
You could convert it to a string and create the expression by calling new RegExp():
var myRE = new RegExp (['^(([^<>()[\]\\.,;:\\s#\"]+(\\.[^<>(),[\]\\.,;:\\s#\"]+)*)',
'|(\\".+\\"))#((\\[[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.',
'[0-9]{1,3}\])|(([a-zA-Z\-0-9]+\\.)+',
'[a-zA-Z]{2,}))$'].join(''));
Notes:
when converting the expression literal to a string you need to escape all backslashes as backslashes are consumed when evaluating a string literal. (See Kayo's comment for more detail.)
RegExp accepts modifiers as a second parameter
/regex/g => new RegExp('regex', 'g')
[Addition ES20xx (tagged template)]
In ES20xx you can use tagged templates. See the snippet.
Note:
Disadvantage here is that you can't use plain whitespace in the regular expression string (always use \s, \s+, \s{1,x}, \t, \n etc).
(() => {
const createRegExp = (str, opts) =>
new RegExp(str.raw[0].replace(/\s/gm, ""), opts || "");
const yourRE = createRegExp`
^(([^<>()[\]\\.,;:\s#\"]+(\.[^<>()[\]\\.,;:\s#\"]+)*)|
(\".+\"))#((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|
(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$`;
console.log(yourRE);
const anotherLongRE = createRegExp`
(\byyyy\b)|(\bm\b)|(\bd\b)|(\bh\b)|(\bmi\b)|(\bs\b)|(\bms\b)|
(\bwd\b)|(\bmm\b)|(\bdd\b)|(\bhh\b)|(\bMI\b)|(\bS\b)|(\bMS\b)|
(\bM\b)|(\bMM\b)|(\bdow\b)|(\bDOW\b)
${"gi"}`;
console.log(anotherLongRE);
})();
Using strings in new RegExp is awkward because you must escape all the backslashes. You may write smaller regexes and concatenate them.
Let's split this regex
/^foo(.*)\bar$/
We will use a function to make things more beautiful later
function multilineRegExp(regs, options) {
return new RegExp(regs.map(
function(reg){ return reg.source; }
).join(''), options);
}
And now let's rock
var r = multilineRegExp([
/^foo/, // we can add comments too
/(.*)/,
/\bar$/
]);
Since it has a cost, try to build the real regex just once and then use that.
Thanks to the wonderous world of template literals you can now write big, multi-line, well-commented, and even semantically nested regexes in ES6.
//build regexes without worrying about
// - double-backslashing
// - adding whitespace for readability
// - adding in comments
let clean = (piece) => (piece
.replace(/((^|\n)(?:[^\/\\]|\/[^*\/]|\\.)*?)\s*\/\*(?:[^*]|\*[^\/])*(\*\/|)/g, '$1')
.replace(/((^|\n)(?:[^\/\\]|\/[^\/]|\\.)*?)\s*\/\/[^\n]*/g, '$1')
.replace(/\n\s*/g, '')
);
window.regex = ({raw}, ...interpolations) => (
new RegExp(interpolations.reduce(
(regex, insert, index) => (regex + insert + clean(raw[index + 1])),
clean(raw[0])
))
);
Using this you can now write regexes like this:
let re = regex`I'm a special regex{3} //with a comment!`;
Outputs
/I'm a special regex{3}/
Or what about multiline?
'123hello'
.match(regex`
//so this is a regex
//here I am matching some numbers
(\d+)
//Oh! See how I didn't need to double backslash that \d?
([a-z]{1,3}) /*note to self, this is group #2*/
`)
[2]
Outputs hel, neat!
"What if I need to actually search a newline?", well then use \n silly!
Working on my Firefox and Chrome.
Okay, "how about something a little more complex?"
Sure, here's a piece of an object destructuring JS parser I was working on:
regex`^\s*
(
//closing the object
(\})|
//starting from open or comma you can...
(?:[,{]\s*)(?:
//have a rest operator
(\.\.\.)
|
//have a property key
(
//a non-negative integer
\b\d+\b
|
//any unencapsulated string of the following
\b[A-Za-z$_][\w$]*\b
|
//a quoted string
//this is #5!
("|')(?:
//that contains any non-escape, non-quote character
(?!\5|\\).
|
//or any escape sequence
(?:\\.)
//finished by the quote
)*\5
)
//after a property key, we can go inside
\s*(:|)
|
\s*(?={)
)
)
((?:
//after closing we expect either
// - the parent's comma/close,
// - or the end of the string
\s*(?:[,}\]=]|$)
|
//after the rest operator we expect the close
\s*\}
|
//after diving into a key we expect that object to open
\s*[{[:]
|
//otherwise we saw only a key, we now expect a comma or close
\s*[,}{]
).*)
$`
It outputs /^\s*((\})|(?:[,{]\s*)(?:(\.\.\.)|(\b\d+\b|\b[A-Za-z$_][\w$]*\b|("|')(?:(?!\5|\\).|(?:\\.))*\5)\s*(:|)|\s*(?={)))((?:\s*(?:[,}\]=]|$)|\s*\}|\s*[{[:]|\s*[,}{]).*)$/
And running it with a little demo?
let input = '{why, hello, there, "you huge \\"", 17, {big,smelly}}';
for (
let parsed;
parsed = input.match(r);
input = parsed[parsed.length - 1]
) console.log(parsed[1]);
Successfully outputs
{why
, hello
, there
, "you huge \""
, 17
,
{big
,smelly
}
}
Note the successful capturing of the quoted string.
I tested it on Chrome and Firefox, works a treat!
If curious you can checkout what I was doing, and its demonstration.
Though it only works on Chrome, because Firefox doesn't support backreferences or named groups. So note the example given in this answer is actually a neutered version and might get easily tricked into accepting invalid strings.
There are good answers here, but for completeness someone should mention Javascript's core feature of inheritance with the prototype chain. Something like this illustrates the idea:
RegExp.prototype.append = function(re) {
return new RegExp(this.source + re.source, this.flags);
};
let regex = /[a-z]/g
.append(/[A-Z]/)
.append(/[0-9]/);
console.log(regex); //=> /[a-z][A-Z][0-9]/g
The regex above is missing some black slashes which isn't working properly. So, I edited the regex. Please consider this regex which works 99.99% for email validation.
let EMAIL_REGEXP =
new RegExp (['^(([^<>()[\\]\\\.,;:\\s#\"]+(\\.[^<>()\\[\\]\\\.,;:\\s#\"]+)*)',
'|(".+"))#((\\[[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.',
'[0-9]{1,3}\])|(([a-zA-Z\\-0-9]+\\.)+',
'[a-zA-Z]{2,}))$'].join(''));
To avoid the Array join, you can also use the following syntax:
var pattern = new RegExp('^(([^<>()[\]\\.,;:\s#\"]+' +
'(\.[^<>()[\]\\.,;:\s#\"]+)*)|(\".+\"))#' +
'((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|' +
'(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$');
You can simply use string operation.
var pattenString = "^(([^<>()[\]\\.,;:\s#\"]+(\.[^<>()[\]\\.,;:\s#\"]+)*)|"+
"(\".+\"))#((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|"+
"(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$";
var patten = new RegExp(pattenString);
I tried improving korun's answer by encapsulating everything and implementing support for splitting capturing groups and character sets - making this method much more versatile.
To use this snippet you need to call the variadic function combineRegex whose arguments are the regular expression objects you need to combine. Its implementation can be found at the bottom.
Capturing groups can't be split directly that way though as it would leave some parts with just one parenthesis. Your browser would fail with an exception.
Instead I'm simply passing the contents of the capture group inside an array. The parentheses are automatically added when combineRegex encounters an array.
Furthermore quantifiers need to follow something. If for some reason the regular expression needs to be split in front of a quantifier you need to add a pair of parentheses. These will be removed automatically. The point is that an empty capture group is pretty useless and this way quantifiers have something to refer to. The same method can be used for things like non-capturing groups (/(?:abc)/ becomes [/()?:abc/]).
This is best explained using a simple example:
var regex = /abcd(efghi)+jkl/;
would become:
var regex = combineRegex(
/ab/,
/cd/,
[
/ef/,
/ghi/
],
/()+jkl/ // Note the added '()' in front of '+'
);
If you must split character sets you can use objects ({"":[regex1, regex2, ...]}) instead of arrays ([regex1, regex2, ...]). The key's content can be anything as long as the object only contains one key. Note that instead of () you have to use ] as dummy beginning if the first character could be interpreted as quantifier. I.e. /[+?]/ becomes {"":[/]+?/]}
Here is the snippet and a more complete example:
function combineRegexStr(dummy, ...regex)
{
return regex.map(r => {
if(Array.isArray(r))
return "("+combineRegexStr(dummy, ...r).replace(dummy, "")+")";
else if(Object.getPrototypeOf(r) === Object.getPrototypeOf({}))
return "["+combineRegexStr(/^\]/, ...(Object.entries(r)[0][1]))+"]";
else
return r.source.replace(dummy, "");
}).join("");
}
function combineRegex(...regex)
{
return new RegExp(combineRegexStr(/^\(\)/, ...regex));
}
//Usage:
//Original:
console.log(/abcd(?:ef[+A-Z0-9]gh)+$/.source);
//Same as:
console.log(
combineRegex(
/ab/,
/cd/,
[
/()?:ef/,
{"": [/]+A-Z/, /0-9/]},
/gh/
],
/()+$/
).source
);
Personally, I'd go for a less complicated regex:
/\S+#\S+\.\S+/
Sure, it is less accurate than your current pattern, but what are you trying to accomplish? Are you trying to catch accidental errors your users might enter, or are you worried that your users might try to enter invalid addresses? If it's the first, I'd go for an easier pattern. If it's the latter, some verification by responding to an e-mail sent to that address might be a better option.
However, if you want to use your current pattern, it would be (IMO) easier to read (and maintain!) by building it from smaller sub-patterns, like this:
var box1 = "([^<>()[\]\\\\.,;:\s#\"]+(\\.[^<>()[\\]\\\\.,;:\s#\"]+)*)";
var box2 = "(\".+\")";
var host1 = "(\\[[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\])";
var host2 = "(([a-zA-Z\-0-9]+\\.)+[a-zA-Z]{2,})";
var regex = new RegExp("^(" + box1 + "|" + box2 + ")#(" + host1 + "|" + host2 + ")$");
#Hashbrown's great answer got me on the right track. Here's my version, also inspired by this blog.
function regexp(...args) {
function cleanup(string) {
// remove whitespace, single and multi-line comments
return string.replace(/\s+|\/\/.*|\/\*[\s\S]*?\*\//g, '');
}
function escape(string) {
// escape regular expression
return string.replace(/[-.*+?^${}()|[\]\\]/g, '\\$&');
}
function create(flags, strings, ...values) {
let pattern = '';
for (let i = 0; i < values.length; ++i) {
pattern += cleanup(strings.raw[i]); // strings are cleaned up
pattern += escape(values[i]); // values are escaped
}
pattern += cleanup(strings.raw[values.length]);
return RegExp(pattern, flags);
}
if (Array.isArray(args[0])) {
// used as a template tag (no flags)
return create('', ...args);
}
// used as a function (with flags)
return create.bind(void 0, args[0]);
}
Use it like this:
regexp('i')`
//so this is a regex
//here I am matching some numbers
(\d+)
//Oh! See how I didn't need to double backslash that \d?
([a-z]{1,3}) /*note to self, this is group #2*/
`
To create this RegExp object:
/(\d+)([a-z]{1,3})/i

Regular Expression with multiple words (in any order) without repeat

I'm trying to execute a search of sorts (using JavaScript) on a list of strings. Each string in the list has multiple words.
A search query may also include multiple words, but the ordering of the words should not matter.
For example, on the string "This is a random string", the query "trin and is" should match. However, these terms cannot overlap. For example, "random random" as a query on the same string should not match.
I'm going to be sorting the results based on relevance, but I should have no problem doing that myself, I just can't figure out how to build up the regular expression(s). Any ideas?
The query trin and is becomes the following regular expression:
/trin.*(?:and.*is|is.*and)|and.*(?:trin.*is|is.*trin)|is.*(?:trin.*and|and.*trin)/
In other words, don't use regular expressions for this.
It probably isn't a good idea to do this with just a regular expression. A (pure, computer science) regular expression "can't count". The only "memory" it has at any point is the state of the DFA. To match multiple words in any order without repeat you'd need on the order of 2^n states. So probably a really horrible regex.
(Aside: I mention "pure, computer science" regular expressions because most implementations are actually an extension, and let you do things that are non-regular. I'm not aware of any extensions, certainly none in JavaScript, that make doing what you want to do any less painless with a single pattern.)
A better approach would be to keep a dictionary (Object, in JavaScript) that maps from words to counts. Initialize it to your set of words with the appropriate counts for each. You can use a regular expression to match words, and then for each word you find, decrement the corresponding entry in the dictionary. If the dictionary contains any non-0 values at the end, or if somewhere a long the way you try to over-decrement a value (or decrement one that doesn't exist), then you have a failed match.
I'm totally not sure if I get you right there, so I'll just post my suggestion for it.
var query = "trin and is",
target = "This is a random string",
search = { },
matches = 0;
query.split( /\s+/ ).forEach(function( word ) {
search[ word ] = true;
});
Object.keys( search ).forEach(function( word ) {
matches += +new RegExp( word ).test( target );
});
// do something useful with "matches" for the query, should be "3"
alert( matches );
So, the variable matches will contain the number of unique matches for the query. The first split-loop just makes sure that no "doubles" are counted since we would overwrite our search object. The second loop checks for the individuals words within the target string and uses the nifty + to cast the result (either true or false) into a number, hence, +1 on a match or +0.
I was looking for a solution to this issue and none of the solutions presented here was good enough, so this is what I came up with:
function filterMatch(itemStr, keyword){
var words = keyword.split(' '), i = 0, w, reg;
for(; w = words[i++] ;){
reg = new RegExp(w, 'ig');
if (reg.test(itemStr) === false) return false; // word not found
itemStr = itemStr.replace(reg, ''); // remove matched word from original string
}
return true;
}
// test
filterMatch('This is a random string', 'trin and is'); // true
filterMatch('This is a random string', 'trin not is'); // false

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