I'm trying to get the functions from an input string by using regular expressions.
So far, I managed to get the JavaScript kind of function declarations with the following regex:
/(\b(f|F)unction(.*?)\((.*?)\)\s*\{)/g
Which when applied like this, will return me whole declaration on the first index:
var re = /(\b(f|F)unction(.*?)\((.*?)\)\s*\{)/g;
while ((m = re.exec(text)) !== null) {
//m[0] contains the function declaration
declarations.push(m[0]);
}
Now, I would like to get in the returned match, the whole content of each of the functions so I can work with it later on (removed it, wrap it...)
I haven't managed to find a regext to do so, so far, I got this:
(\b(f|F)unction(.*?)\((.*?)\)\s*\{)(.*?|\n)*\}
But of course, it catches the first closing bracket } instead of the one at the end of each of the functions.
Any idea of how to get the closing } of each function?
Any idea of how to get the closing } of each function?
This will be very hard with a regex. Because the function body can include any number of possibly nested brace pairs. And then consider strings containing unmatched braces in the function body.
To parse a non-regular language you need something more powerful than regular expressions: a parser for that language.
(Some regex variants have some ability to matched paired characters, but firstly JavaScript's regex engine isn't one; and secondly then there are those strings….)
Related
This question already has answers here:
RegEx for a^b instead of pow(a,b)
(6 answers)
Closed 4 years ago.
I'd like to apply a regexp code to the part of a string that is before and/or after a specific character, and that character must be outside parenthesis.
To be more specific, I am coding a website (in React.JS) showing logical calculation, and I want to remove the first and last parenthesis before and after the main logical operator.
For example, in the string:
"((p∧r)∧(q∧r))∧(p∧q)"
I want to get only: "(p∧r)∧(q∧r)" and "p∧q".
That means I want to get all the character before and after the only "∧" outside of any parenthesis, and I want to remove the first opening parenthesis and the last closing parenthesis of the two string. (The result could be an array with the part before and after for example.)
I am already able to remove the first parenthis with this code :
str.replace(/(\()(.*)(\))/, "$2");
But that code is applied to the whole string right now.
So how do I apply this code to the two parts before and after the only "∧" outside parenthesis ? If possible I'd prefer a code only in regexp, but a JavaScript part would do. Thanks by advance.
You will not be able to do this in pure regex, this is similar of trying to parse HTML with regex. in short, it is not possible (and have been answered many time over in all possible way)
So you will need javascript. To do this in javascript will not be simple, as you will need to parse the entire string to find "^-who-are-not-in-parenthese".
I would first check if there is some parser for mathematical operation or other who already exist and could be adapted for your need.
If you do not find something, you will need to create yourself.
You could do it with a loop passing character by character, having a counter for the indentation level of parenthesis, and a tree-like data structure as an output.
you could also use regex to find the last level of indentation and creating a bottom-up parser. you find all the innerest parenthesis group, saved them, and replace them with a special identifier character. you can then redo the operation to find the now innerest parenthesis group, check if there is special character in them and place them in the tree (as parent of the element identified by the character.)
once you have your tree structure (in one way or another), the root and the first level of element should be what you want.
You could do it this way:
// your string
const fixMe = "((p∧r)∧(q∧r))^(p∧q)";
// Separate double paren items from single paren items
const parts = fixMe.match(/\(\(.*\)\)|\(.*\)/g).map(
// Get rid of leading and following parens
item => item.replace(/^\(|\)$/g, '')
);
However, note: this solution is not very flexible. Better might some sort of recursive parenthesis lookup where each loop keeps an accounting of nesting level, etc...
I ran into the below monster of a regex in the wild today. The regex is meant to validate a url.
function superUrlValidation(url) {
return new RegExp(/^/.source + "((.+):\/\/)?" + /(((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:)*#)?(((\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5]))|((([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.)+(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.?)(:\d*)?)(\/((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)+(\/(([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)*)*)?)?(\?((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|[\uE000-\uF8FF]|\/|\?)*)?(\#((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|\/|\?)*)?$/.source, "i")
.test(url);
}
I've never seen .source used in a regex like this so I looked it up.
The MDN docs for RegExp.prototype.source states:
The source property returns a String containing the source text of the regexp object, and it doesn't contain the two forward slashes on both sides and any flags.
... and gives this example:
var regex = /fooBar/ig;
console.log(regex.source); // "fooBar", doesn't contain /.../ and "ig".
I understand the MDN example (you're getting the source text of the regex object after it is created, makes sense), but I dont understand how this is being used in the superUrlValidation regex above.
How is the source being used before the regex object is completed and what does this accomplish? I cant find any documentation showing .source being used in this way.
Note that .source is used twice in the regex, at the beginning and the end
Use of .source everywhere in your regex seems totally unnecessary, may be just a trick to avoid double escaping. In fact even use of new RegExp is not needed and you can get away with just the regex literal as this:
var re = /^((.+):\/\/)?(((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:)*#)?(((\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5]))|((([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.)+(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.?)(:\d*)?)(\/((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)+(\/(([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)*)*)?)?(\?((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|[\uE000-\uF8FF]|\/|\?)*)?(\#((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|\/|\?)*)?$/i;
/^/ is a regex literal, meaning it's a valid regex object in it's own right. This means that /^/.source === "^".
This seems like an arbitrary example of using the source property as this means the author could have just placed a "^" in it's place, or even just put a ^ at the beginning of the next string, and it would have the same effect.
The .source property returns the content of the regex between the forward slashes as you say. so the result of the above is equivalent to this string:
/^((.+):\/\/)?(((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:)*#)?(((\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5]))|((([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.)+(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.?)(:\d*)?)(\/((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)+(\/(([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)*)*)?)?(\?((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|[\uE000-\uF8FF]|\/|\?)*)?(\#((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|\/|\?)*)?$/i
In JavaScript you can write regexes like this: /matchsomething/ or using the RegExp function/constructor above. It looks like the code you found is the result of someone not know what they were doing. They seem to have taken a few regexes using the literal syntax (i.e /match_here/) and plugged it into the constructor version and stuck them all together.
I can't see any benefit in using the source property this way. I would just use the string version or the constructor version. Or better, find out what the original author intended and write it again or find a respected regex library with the criteria you need.
And, yeah, wow. It's massive.
I have a string. The data is from a text area and contains occurrences of (varying) content enclosed in square brackets.
I would like to run functions (e.g. replace) on the string content that is enclosed in square brackets, then update the original string to reflect the changes made.
Example
not this [potentially act on this token] not this [potentially act on this token] not this
I think some of the following may be involved but I can’t figure out how to implement a working solution. I wouldn’t like to use jQuery or external libraries.
Match + RegEx Something like this.
indexOf. Possibly to get the content of the match if it can’t be done with RegEx alone.
For each. As there are numerous instances of content enclosed in square brackets.
Here’s a starting point JSFiddle.
If anyone could be of any assistance, I’d really appreciate it.
Thank you very much.
So your jsfiidle has pretty much all you need, except that the replace function can take a callback. This callback receives one parameter which is the matched string and whatever you return is substituted for the match
user_input = user_input.replace(/\[.*\]/gi,function(match){
return match + "foo";
});
http://jsfiddle.net/fdtxvyo7/6/
Try something like this ...
var startEnd = user_input.match( /(\[.*?\])/g );
startend is now an array of the bracketed values that you can now use for search/replace.
I'm trying to improve my understanding of Regex, but this one has me quite mystified.
I started with some text defined as:
var txt = "{\"columns\":[{\"text\":\"A\",\"value\":80},{\"text\":\"B\",\"renderer\":\"gbpFormat\",\"value\":80},{\"text\":\"C\",\"value\":80}]}";
and do a replace as follows:
txt.replace(/\"renderer\"\:(.*)(?:,)/g,"\"renderer\"\:gbpFormat\,");
which results in:
"{"columns":[{"text":"A","value":80},{"text":"B","renderer":gbpFormat,"value":80}]}"
What I expected was for the renderer attribute value to have it's quotes removed; which has happened, but also the C column is completely missing! I'd really love for someone to explain how my Regex has removed column C?
As an extra bonus, if you could explain how to remove the quotes around any value for renderer (i.e. so I don't have to hard-code the value gbpFormat in the regex) that'd be fantastic.
You are using a greedy operator while you need a lazy one. Change this:
"renderer":(.*)(?:,)
^---- add here the '?' to make it lazy
To
"renderer":(.*?)(?:,)
Working demo
Your code should be:
txt.replace(/\"renderer\"\:(.*?)(?:,)/g,"\"renderer\"\:gbpFormat\,");
If you are learning regex, take a look at this documentation to know more about greedyness. A nice extract to understand this is:
Watch Out for The Greediness!
Suppose you want to use a regex to match an HTML tag. You know that
the input will be a valid HTML file, so the regular expression does
not need to exclude any invalid use of sharp brackets. If it sits
between sharp brackets, it is an HTML tag.
Most people new to regular expressions will attempt to use <.+>. They
will be surprised when they test it on a string like This is a
first test. You might expect the regex to match and when
continuing after that match, .
But it does not. The regex will match first. Obviously not
what we wanted. The reason is that the plus is greedy. That is, the
plus causes the regex engine to repeat the preceding token as often as
possible. Only if that causes the entire regex to fail, will the regex
engine backtrack. That is, it will go back to the plus, make it give
up the last iteration, and proceed with the remainder of the regex.
Like the plus, the star and the repetition using curly braces are
greedy.
Try like this:
txt = txt.replace(/"renderer":"(.*?)"/g,'"renderer":$1');
The issue in the expression you were using was this part:
(.*)(?:,)
By default, the * quantifier is greedy by default, which means that it gobbles up as much as it can, so it will run up to the last comma in your string. The easiest solution would be to turn that in to a non-greedy quantifier, by adding a question mark after the asterisk and change that part of your expression to look like this
(.*?)(?:,)
For the solution I proposed at the top of this answer, I also removed the part matching the comma, because I think it's easier just to match everything between quotes. As for your bonus question, to replace the matched value instead of having to hardcode gbpFormat, I used a backreference ($1), which will insert the first matched group into the replacement string.
Don't manipulate JSON with regexp. It's too likely that you will break it, as you have found, and more importantly there's no need to.
In addition, once you have changed
'{"columns": [..."renderer": "gbpFormat", ...]}'
into
'{"columns": [..."renderer": gbpFormat, ...]}' // remove quotes from gbpFormat
then this is no longer valid JSON. (JSON requires that property values be numbers, quoted strings, objects, or arrays.) So you will not be able to parse it, or send it anywhere and have it interpreted correctly.
Therefore you should parse it to start with, then manipulate the resulting actual JS object:
var object = JSON.parse(txt);
object.columns.forEach(function(column) {
column.renderer = ghpFormat;
});
If you want to replace any quoted value of the renderer property with the value itself, then you could try
column.renderer = window[column.renderer];
Assuming that the value is available in the global namespace.
This question falls into the category of "I need a regexp, or I wrote one and it's not working, and I'm not really sure why it has to be a regexp, but I heard they can do all kinds of things, so that's just what I imagined I must need." People use regexps to try to do far too many complex matching, splitting, scanning, replacement, and validation tasks, including on complex languages such as HTML, or in this case JSON. There is almost always a better way.
The only time I can imagine wanting to manipulate JSON with regexps is if the JSON is broken somehow, perhaps due to a bug in server code, and it needs to be fixed up in order to be parseable.
I'm trying to remove a rectangular brackets(bbcode style) using javascript, this is for removing unwanted bbcode.
I try with this.
theString .replace(/\[quote[^\/]+\]*\[\/quote\]/, "")
it works with this string sample:
theString = "[quote=MyName;225]Test 123[/quote]";
it will fail within this sample:
theString = "[quote=MyName;225]Test [quote]inside quotes[/quote]123[/quote]";
if there any solution beside regex no problem
The other 2 solutions simply do not work (see my comments). To solve this problem you first need to craft a regex which matches the innermost matching quote elements (which contain neither [QUOTE..] nor [/QUOTE]). Next, you need to iterate, applying this regex over and over until there are no more QUOTE elements left. This tested function does what you want:
function filterQuotes(text)
{ // Regex matches inner [QUOTE]non-quote-stuff[/quote] tag.
var re = /\[quote[^\[]+(?:(?!\[\/?quote\b)\[[^\[]*)*\[\/quote\]/ig;
while (text.search(re) !== -1)
{ // Need to iterate removing QUOTEs from inside out.
text = text.replace(re, "");
}
return text;
}
Note that this regex employs Jeffrey Friedl's "Unrolling the loop" efficiency technique and is not only accurate, but is quite fast to boot.
See: Mastering Regular Expressions (3rd Edition) (highly recommended).
Try this one:
/\[quote[^\/]+\].*\[\/quote\]$/
The $ sign indicates that only the closing quote element at the end of the string should be used to determine the ending of the quote you're trying to remove.
And i added a "." before the asterisk so that this will match any sign in between. I tested this with your two strings and it worked.
edit: I don't exactly know how you are using that. But just as an addition. If you want the pattern also to match to a string where no attributes are added for example:
[quote]Hello[/quote]
You should change the "+" sign into an asterisk as well like this:
/\[quote[^\/]*\].*\[\/quote\]$/
This answer has flaws, see Ridgerunner's answer for a more correct one.
Here's my crack at it.
function filterQuotes(text)
{
return text.replace(/\[(\/)?quote([^\/]*)?\]/g,"");
}