In the following input string:
{$foo}foo bar \\{$blah1}oh{$blah2} even more{$blah3} but not{$blarg}{$why_not_me}
I am trying to match all instances of {$SOMETHING_HERE} that are not preceded by an unescaped backslash.
Example:
I want it to match {$SOMETHING} but not \{$SOMETHING}.
But I do want it to match \\{$SOMETHING}
Attempts:
All of my attempts so far will match what I want except for tags right next to each other like {$SOMETHING}{$SOMETHING_ELSE}
Here is what I currently have:
var input = '{$foo}foo bar \\{$blah1}oh{$blah2} even more{$blah3} but not{$blarg}{$why_not_me}';
var results = input.match(/(?:[^\\]|^)\{\$[a-zA-Z_][a-zA-Z0-9_]*\}/g);
console.log(results);
Which outputs:
["{$foo}", "h{$blah2}", "e{$blah3}", "t{$blarg}"]
Goal
I want it to be :
["{$foo}", "{$blah2}", "{$blah3}", "{$blarg}", "{$why_not_me}"]
Question
Can anybody point me in the right direction?
The problem here is that you need a lookbehind, which JavaScript Regexs don't support
basically you need "${whatever} if it is preceded by a double slash but not a single slash" which is what the lookbehind does.
You can mimic simple cases of lookbehinds, but not sure if it will help in this example. Give it a go: http://blog.stevenlevithan.com/archives/mimic-lookbehind-javascript
edit
Btw, I don't think you can do this a 'stupid way' either because if you have [^\\]\{ you'll match any character that is not a backslash before the brace. You really need the lookbehind to do this cleanly.
Otherwise you can do
(\\*{\$[a-zA-Z_][a-zA-Z0-9_]*\})
Then just count the number of backslashes in the resulting tokens.
When all else fails, split, join/replace the crap out of it.
Note: the first split/join is actually the cleanup portion. That kills \{<*>}
Also, I didn't account for the stuff inside the brackets since there's code for that already.
var input = '{$foo}foo bar \\{$blah1}oh{$blah2} even more\\\\{$blah3} but not{$blarg}{$why_not_me}';
input.split(/(?:[^\\])\\\{[^\}]*\}/).join('').replace(/\}[^\{]*\{/g,'},{').split(/,/));
This seems to do what I want:
var input = '{$foo}foo bar \\{$blah1}oh{$blah2} even more\\\\{$blah3} but not{$blarg}{$why_not_me}';
var results = [];
input.replace(/(\\*)\{\$[a-z_][a-z0-9_]*\}/g, function($0,$1){
$0 = $0.replace(/^\\\\/g,'');
var result = ($0.indexOf('\\') === 0 ? false : $0);
if(result) {
results.push(result);
}
})
console.log(results);
Which gives:
["{$foo}", "{$blah2}", "{$blah3}", "{$blarg}", "{$why_not_me}"]
Related
Let's say I have a string: "We.need..to...split.asap". What I would like to do is to split the string by the delimiter ., but I only wish to split by the first . and include any recurring .s in the succeeding token.
Expected output:
["We", "need", ".to", "..split", "asap"]
In other languages, I know that this is possible with a look-behind /(?<!\.)\./ but Javascript unfortunately does not support such a feature.
I am curious to see your answers to this question. Perhaps there is a clever use of look-aheads that presently evades me?
I was considering reversing the string, then re-reversing the tokens, but that seems like too much work for what I am after... plus controversy: How do you reverse a string in place in JavaScript?
Thanks for the help!
Here's a variation of the answer by guest271314 that handles more than two consecutive delimiters:
var text = "We.need.to...split.asap";
var re = /(\.*[^.]+)\./;
var items = text.split(re).filter(function(val) { return val.length > 0; });
It uses the detail that if the split expression includes a capture group, the captured items are included in the returned array. These capture groups are actually the only thing we are interested in; the tokens are all empty strings, which we filter out.
EDIT: Unfortunately there's perhaps one slight bug with this. If the text to be split starts with a delimiter, that will be included in the first token. If that's an issue, it can be remedied with:
var re = /(?:^|(\.*[^.]+))\./;
var items = text.split(re).filter(function(val) { return !!val; });
(I think this regex is ugly and would welcome an improvement.)
You can do this without any lookaheads:
var subject = "We.need.to....split.asap";
var regex = /\.?(\.*[^.]+)/g;
var matches, output = [];
while(matches = regex.exec(subject)) {
output.push(matches[1]);
}
document.write(JSON.stringify(output));
It seemed like it'd work in one line, as it did on https://regex101.com/r/cO1dP3/1, but had to be expanded in the code above because the /g option by default prevents capturing groups from returning with .match (i.e. the correct data was in the capturing groups, but we couldn't immediately access them without doing the above).
See: JavaScript Regex Global Match Groups
An alternative solution with the original one liner (plus one line) is:
document.write(JSON.stringify(
"We.need.to....split.asap".match(/\.?(\.*[^.]+)/g)
.map(function(s) { return s.replace(/^\./, ''); })
));
Take your pick!
Note: This answer can't handle more than 2 consecutive delimiters, since it was written according to the example in the revision 1 of the question, which was not very clear about such cases.
var text = "We.need.to..split.asap";
// split "." if followed by "."
var res = text.split(/\.(?=\.)/).map(function(val, key) {
// if `val[0]` does not begin with "." split "."
// else split "." if not followed by "."
return val[0] !== "." ? val.split(/\./) : val.split(/\.(?!.*\.)/)
});
// concat arrays `res[0]` , `res[1]`
res = res[0].concat(res[1]);
document.write(JSON.stringify(res));
I am developing a small character encoder generator where the user input their text and on the click of a button, it outputs the encoded version.
I've defined an object of the characters that need to be encoded like so:
map = {
'©' : '©',
'&' : '&'
},
And here is the loop that gets the values from the map and replaces them:
Object.keys(map).forEach(function (ico) {
var icoE = ico.replace(/([.?*+^$[\]\\(){}|-])/g, "\\$1");
raw = raw.replace( new RegExp(icoE, 'g'), map[ico] );
});
I am them simply outputting the result to a textarea. This all works fine, however the problem I'm facing is this.
© is replaced with © however the & symbol at the beginning of this is then converted to & so it ends up being ©.
I see why this is happening however I'm not sure how to go about ensuring that & is not replaced within character encoded strings.
Here is a JSFiddle for a live preview of what I mean:
http://jsfiddle.net/4m3nw/1/
Any help would be much appreciated
Prelude: Apart from regex, an idea worth considering is something like this JS function that already handles html entities. Now, on to the regex question.
HTML Special Characters, Negative Lookahead
In HTML, special characters can look not only like © but also like —, and they can have upper-case characters.
To replace ampersands that are not immediately followed by a hash or word characters and a semicolon, you can use something like this:
&(?!(?:#[0-9]+|[a-z]+);)
See the demo.
Make sure to use the i flag to activate case-insensitive mode
& matches the literal ampersand
The negative lookahead (?!(?:#[0-9]+|[a-z]+);) asserts that it is not followed by...
(?:#[0-9]+|[a-z]+) a hash and digits, | OR letters...
then a semicolon.
Reference
Lookahead and Lookbehind Zero-Length Assertions
Mastering Lookahead and Lookbehind
The problem is that since you process the same string you replace the &in ©. If you re-order your map then that seemingly solves the problem. However according to the ECMAScript specifications, this is not a given, so you would be relying on implementation details of the ECMAScript engine used.
What you can do to make sure it will always work is to swap the keys so that & is always processed first:
map = {
'©' : '©',
'&' : '&'
};
var keys = Object.keys(map);
keys[keys.indexOf('&')] = keys[0];
keys[0] = '&';
keys.forEach(function (ico) {
var icoE = ico.replace(/([.?*+^$[\]\\(){}|-])/g, "\\$1");
raw = raw.replace( new RegExp(icoE, 'g'), map[ico] );
});
Obviously you need to add checks for the &'s existence if it isn't always there.
jsFiddle Demo.
Probably the simplest code change is to reorder your map by putting the ampersand on top.
I hope I can explain myself clearly here and that this is not too much of a specific issue.
I am working on some javascript that needs to take a string, find instances of chars between square brackets, store any returned results and then remove them from the original string.
My code so far is as follows:
parseLine : function(raw)
{
var arr = [];
var regex = /\[(.*?)]/g;
var arr;
while((arr = regex.exec(raw)) !== null)
{
console.log(" ", arr);
arr.push(arr[1]);
raw = raw.replace(/\[(.*?)]/, "");
console.log(" ", raw);
}
return {results:arr, text:raw};
}
This seems to work in most cases. If I pass in the string [id1]It [someChar]found [a#]an [id2]excellent [aa]match then it returns all the chars from within the square brackets and the original string with the bracketed groups removed.
The problem arises when I use the string [id1]It [someChar]found [a#]a [aa]match.
It seems to fail when only a single letter (and space?) follows a bracketed group and starts missing groups as you can see in the log if you try it out. It also freaks out if i use groups back to back like [a][b] which I will need to do.
I'm guessing this is my RegEx - begged and borrowed from various posts here as I know nothing about it really - but I've had no luck fixing it and could use some help if anyone has any to offer. A fix would be great but more than that an explanation of what is actually going on behind the scenes would be awesome.
Thanks in advance all.
You could use the replace method with a function to simplify the code and run the regexp only once:
function parseLine(raw) {
var results = [];
var parsed = raw.replace(/\[(.*?)\]/g, function(match,capture) {
results.push(capture);
return '';
});
return { results : results, text : parsed };
}
The problem is due to the lastIndex property of the regex /\[(.*?)]/g; not resetting, since the regex is declared as global. When the regex has global flag g on, lastIndex property of RegExp is used to mark the position to start the next attempt to search for a match, and it is expected that the same string is fed to the RegExp.exec() function (explicitly, or implicitly via RegExp.test() for example) until no more match can be found. Either that, or you reset the lastIndex to 0 before feeding in a new input.
Since your code is reassigning the variable raw on every loop, you are using the wrong lastIndex to attempt the next match.
The problem will be solved when you remove g flag from your regex. Or you could use the solution proposed by Tibos where you supply a function to String.replace() function to do replacement and extract the capturing group at the same time.
You need to escape the last bracket: \[(.*?)\].
I've got some string that contain invisible characters, but they are in somewhat predictable places. Typically the surround the piece of text I want to extract, and then after the 2nd occurrence I want to keep the rest of the text.
I can't seem to figure out how to both key off of the invisible characters, and exclude them from my result. To match invisibles I've been using this regex: /\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F/ which does seem to work.
Here's an example: [invisibles]Keep as match 1[invisibles]Keep as match 2
Here's what I've been using so far without success:
/([\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+)(.+)([\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+)/(.+)
I've got the capture groups in there, but it's bee a while since I've had to use regex's in this way, so I know I'm missing something important. I was hoping to just make the invisible matches non-capturing groups, but it seems that JavaScript does not support this.
Something like this seems like what you want. The second regex you have pretty much works, but the / is in totally the wrong place. Perhaps you weren't properly reading out the group data.
var s = "\x0EKeep as match 1\x0EKeep as match 2";
var r = /[\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+(.+)[\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+(.+)/;
var match = s.match(r);
var part1 = match[1];
var part2 = match[2];
I am complete novice at regex and Javascript. I have the following problem: need to check into a textfield the existence of one (1) or many (n) consecutive * (asterisk) character/characters eg. * or ** or *** or infinite (n) *. Strings allowed eg. *tomato or tomato* or **tomato or tomato** or as many(n)*tomato many(n)*. So, far I had tried the following:
var str = 'a string'
var value = encodeURIComponent(str);
var reg = /([^\s]\*)|(\*[^\s])/;
if (reg.test(value) == true ) {
alert ('Watch out your asterisks!!!')
}
By your question it's hard to decipher what you're after... But let me try:
Only allow asterisks at beginning or at end
If you only allow an arbitrary number (at least one) of asterisks either at the beginning or at the end (but not on both sides) like:
*****tomato
tomato******
but not **tomato*****
Then use this regular expression:
reg = /^(?:\*+[^*]+|[^*]+\*+)$/;
Match front and back number of asterisks
If you require that the number of asterisks at the biginning matches number of asterisks at the end like
*****tomato*****
*tomato*
but not **tomato*****
then use this regular expression:
reg = /^(\*+)[^*]+\1$/;
Results?
It's unclear from your question what the results should be when each of these regular expressions match? Are strings that test positive to above regular expressions fine or wrong is on you and your requirements. As long as you have correct regular expressions you're good to go and provide the functionality you require.
I've also written my regular expressions to just exclude asterisks within the string. If you also need to reject spaces or anything else simply adjust the [^...] parts of above expressions.
Note: both regular expressions are untested but should get you started to build the one you actually need and require in your code.
If I understand correctly you're looking for a pattern like this:
var pattern = /\**[^\s*]+\**/;
this won't match strings like ***** or ** ***, but will match ***d*** *d or all of your examples that you say are valid (***tomatos etc).If I misunderstood, let me know and I'll see what I can do to help. PS: we all started out as newbies at some point, nothing to be ashamed of, let alone apologize for :)
After the edit to your question I gather the use of an asterisk is required, either at the beginning or end of the input, but the string must also contain at least 1 other character, so I propose the following solution:
var pattern = /^\*+[^\s*]+|[^\s*]+\*+$/;
'****'.match(pattern);//false
' ***tomato**'.match(pattern);//true
If, however *tomato* is not allowed, you'll have to change the regex to:
var pattern = /^\*+[^\s*]+$|^[^\s*]+\*+$/;
Here's a handy site to help you find your way in the magical world of regular expressions.