The space character as a punctuator in JavaScript - javascript

In chapter 7.7 (Punctuators) of the ECMAScript spec ( http://www.ecma-international.org/publications/files/ECMA-ST/ECMA-262.pdf ) the grid of punctuators appears to have a gap in row 3 of the last column. This is in fact the space character punctuator, correct?
I understand that space characters may be inserted optionally between tokens in the JavaScript code (in order to improve readability), however, I was wondering where they are actually required...
In order to find this out, I searched for space characters in the minified version of the jQuery library. These are my results:
A space is required... (see Update below)
... between a keyword and an identifier:
function x(){}
var x;
return x;
typeof x;
new X();
... between two keywords:
return false;
if(x){}else if(y){}else{}
These are the two cases that I identified. Are there any other cases?
Note: Space characters inside string literals are not regarded as punctuator tokens (obviously).
Update: As it turns out, a space character is not required in those cases. For example a keyword token and a identifier token have to be seperated by something, but that something does not have to be a space character. It could be any input element which is not a token (WhiteSpace, LineTerminator or Comment).
Also... It seems that the space character is regarded as a WhiteSpace input element, and not a token at all, which would mean that it's not a punctuator.

Update (2021): The spec is much clearer now, and space is definitely not in the list of punctuators. Space is whitespace, which is covered in the White Space section.
Answer from 2010:
I don't think that gap is meant to be a space, no, I think it's just a gap (an unfortunate one). If they really meant to be listing a space, I expect they'd use "Whitespace" as they have elsewhere in the document. But whitespace as a punctuator doesn't really make sense.
I believe spaces (and other forms of whitespace) are delimiters. The spec sort of defines them by omission rather than explicitly. The space is required between function and x because otherwise you have the token functionx, which is not of course a keyword (though it could be a name token — e.g., a variable, property, or function name).
You need delimiters around some tokens (Identifiers and ReservedWords), because that's how we recognize where those tokens begin and end — an IdentifierName starts with an IdentifierStart followed by zero or more IdentifierParts, a class which doesn't include whitespace or any of the characters used for punctuators. Other tokens (Punctuators for instance) we can recognize without delimiters. I think that's about it, and so your two rules are pretty much just two examples of the same rule: IdentifierNames must be delimited (by whitespace, by punctuators, by beginning or end of file, ...).
Somewhat off-topic, but of course not all delimiters are equal. Line-breaking delimiters are sometimes treated specially by the grammar for the horror that is "semicolon insertion".

Whitespaces are not required in any of these cases. You just have to write a syntax that is understandable for the parser. In other words: the machine has to know whether you're using a keyword like function or new or just defining another variable like newFunction.
Each keyword has to be delimited somehow - whitespaces are the most sensible and readable, however they can be replaced:
return/**/false;
return(false);

This is just a guess, but I would say that spaces aren't actually required anywhere. They are used just as one of many alternatives to generate word boundaries between keywords. This means you could just as well replace them with other characters.
If what you want is to remove the unnecessary spaces from some code I would say that spaces (white-space to be more exact, tabs will work just as well) are mandatory only where there are no other means of separating keywords and/or variable identifiers. I.e. where by removing the white-space you no longer have the same keywords and identifiers in the resulting code.
What follows is not exactly relevant to your needs but you may find it interesting. You can write you examples so that they no longer have those spaces. I hope none of the examples are wrong.
x=function(){} instead of function x(){}
this.x=null; instead of var x;
return(x); instead of return x;
typeof(x); instead of typeof x;
y=X(); instead of y = new X();
return(false) instead of return false
if(x){}else{if(y){}else{}} instead of if(x){}else if(y){}else{}

Related

Javascript - how to use regex process the following complicated string

I have the following string that will occur repeatedly in a larger string:
[SM_g]word[SM_h].[SM_l] "
Notice in this string after the phrase "[SM_g]word[Sm_h]" there are three components:
A period (.) This could also be a comma (,)
[SM_l]
"
Zero to all three of these components will always appear after "[SM_g]word[SM_h]". However, they can also appear in any order after "[SM_g]word[SM_h]". For example, the string could also be:
[SM_g]word[SM_h][SM_l]"
or
[SM_g]word[SM_h]"[SM_l].
or
[SM_g]word[SM_h]".
or
[SM_g]word[SM_h][SM_1].
or
[SM_g]word[SM_h].
or simply just
[SM_g]word[SM_h]
These are just some of the examples. The point is that there are three different components (more if you consider the period can also be a comma) that can appear after "[SM_h]word[SM_g]" where these three components can be in any order and sometimes one, two, or all three of the components will be missing.
Not only that, sometimes there will be up to one space before " and the previous component/[SM_g]word[SM_h].
For example:
[SM_g]word[SM_h] ".
or
[SM_g]word[SM_h][SM_l] ".
etc. etc.
I am trying to process this string by moving each of the three components inside of the core string (and preserving the space, in case there is a space before &\quot; and the previous component/[SM_g]word[SM_h]).
For example, [SM_g]word[SM_h].[SM_l]" would turn into
[SM_g]word.[SM_l]"[SM_h]
or
[SM_g]word[SM_h]"[SM_l]. would turn into
[SM_g]word"[SM_l].[SM_h]
or, to simulate having a space before "
[SM_g]word[SM_h] ".
would turn into
[SM_g]word ".[SM_h]
and so on.
I've tried several combinations of regex expressions, and none of them have worked.
Does anyone have advice?
You need to put each component within an alternation in a grouping construct with maximum match try of 3 if it is necessary:
\[SM_g]word(\[SM_h])((?:\.|\[SM_l]| ?"){0,3})
You may replace word with .*? if it is not a constant or specific keyword.
Then in replacement string you should do:
$1$3$2
var re = /(\[SM_g]word)(\[SM_h])((?:\.|\[SM_l]| ?"){0,3})/g;
var str = `[SM_g]word[SM_h][SM_l] ".`;
console.log(str.replace(re, `$1$3$2`));
This seems applicable for your process, in other word, changing sub-string position.
(\[SM_g])([^[]*)(\[SM_h])((?=([,\.])|(\[SM_l])|( ?&\\?quot;)).*)?
Demo,,, in which all sub-strings are captured to each capture group respectively for your post processing.
[SM_g] is captured to group1, word to group2, [SM_h] to group3, and string of all trailing part is to group4, [,\.] to group5, [SM_l] to group6, " ?&\\?quot;" to group7.
Thus, group1~3 are core part, group4 is trailing part for checking if trailing part exists, and group5~7 are sub-parts of group4 for your post processing.
Therefore, you can get easily matched string's position changed output string in the order of what you want by replacing with captured groups like follows.
\1\2\7\3 or $1$2$7$3 etc..
For replacing in Javascript, please refer to this post. JS Regex, how to replace the captured groups only?
But above regex is not sufficiently precise because it may allow any repeatitions of the sub-part of the trailing string, for example, \1\2\3\5\5\5\5 or \1\2\3\6\7\7\7\7\5\5\5, etc..
To avoid this situation, it needs to adopt condition which accepts only the possible combinations of the sub-parts of the trailing string. Please refer to this example. https://regex101.com/r/6aM4Pv/1/ for the possible combinations in the order.
But if the regex adopts the condition of allowing only possible combinations, the regex will be more complicated so I leave the above simplified regex to help you understand about it. Thank you:-)

javascript regex replace multiline strings [duplicate]

var ss= "<pre>aaaa\nbbb\nccc</pre>ddd";
var arr= ss.match( /<pre.*?<\/pre>/gm );
alert(arr); // null
I'd want the PRE block be picked up, even though it spans over newline characters. I thought the 'm' flag does it. Does not.
Found the answer here before posting. SInce I thought I knew JavaScript (read three books, worked hours) and there wasn't an existing solution at SO, I'll dare to post anyways. throw stones here
So the solution is:
var ss= "<pre>aaaa\nbbb\nccc</pre>ddd";
var arr= ss.match( /<pre[\s\S]*?<\/pre>/gm );
alert(arr); // <pre>...</pre> :)
Does anyone have a less cryptic way?
Edit: this is a duplicate but since it's harder to find than mine, I don't remove.
It proposes [^] as a "multiline dot". What I still don't understand is why [.\n] does not work. Guess this is one of the sad parts of JavaScript..
DON'T use (.|[\r\n]) instead of . for multiline matching.
DO use [\s\S] instead of . for multiline matching
Also, avoid greediness where not needed by using *? or +? quantifier instead of * or +. This can have a huge performance impact.
See the benchmark I have made: https://jsben.ch/R4Hxu
Using [^]: fastest
Using [\s\S]: 0.83% slower
Using (.|\r|\n): 96% slower
Using (.|[\r\n]): 96% slower
NB: You can also use [^] but it is deprecated in the below comment.
[.\n] does not work because . has no special meaning inside of [], it just means a literal .. (.|\n) would be a way to specify "any character, including a newline". If you want to match all newlines, you would need to add \r as well to include Windows and classic Mac OS style line endings: (.|[\r\n]).
That turns out to be somewhat cumbersome, as well as slow, (see KrisWebDev's answer for details), so a better approach would be to match all whitespace characters and all non-whitespace characters, with [\s\S], which will match everything, and is faster and simpler.
In general, you shouldn't try to use a regexp to match the actual HTML tags. See, for instance, these questions for more information on why.
Instead, try actually searching the DOM for the tag you need (using jQuery makes this easier, but you can always do document.getElementsByTagName("pre") with the standard DOM), and then search the text content of those results with a regexp if you need to match against the contents.
You do not specify your environment and version of JavaScript (ECMAScript), and I realise this post was from 2009, but just for completeness:
With the release of ECMA2018 we can now use the s flag to cause . to match \n (see https://stackoverflow.com/a/36006948/141801).
Thus:
let s = 'I am a string\nover several\nlines.';
console.log('String: "' + s + '".');
let r = /string.*several.*lines/s; // Note 's' modifier
console.log('Match? ' + r.test(s)); // 'test' returns true
This is a recent addition and will not work in many current environments, for example Node v8.7.0 does not seem to recognise it, but it works in Chromium, and I'm using it in a Typescript test I'm writing and presumably it will become more mainstream as time goes by.
Now there's the s (single line) modifier, that lets the dot matches new lines as well :)
\s will also match new lines :D
Just add the s behind the slash
/<pre>.*?<\/pre>/gms
[.\n] doesn't work, because dot in [] (by regex definition; not javascript only) means the dot-character. You can use (.|\n) (or (.|[\n\r])) instead.
I have tested it (Chrome) and it's working for me (both [^] and [^\0]), by changing the dot (.) with either [^\0] or [^] , because dot doesn't match line break (See here: http://www.regular-expressions.info/dot.html).
var ss= "<pre>aaaa\nbbb\nccc</pre>ddd";
var arr= ss.match( /<pre[^\0]*?<\/pre>/gm );
alert(arr); //Working
In addition to above-said examples, it is an alternate.
^[\\w\\s]*$
Where \w is for words and \s is for white spaces
[\\w\\s]*
This one was beyond helpful for me, especially for matching multiple things that include new lines, every single other answer ended up just grouping all of the matches together.

Greedy Regex for varying number of new line [duplicate]

var ss= "<pre>aaaa\nbbb\nccc</pre>ddd";
var arr= ss.match( /<pre.*?<\/pre>/gm );
alert(arr); // null
I'd want the PRE block be picked up, even though it spans over newline characters. I thought the 'm' flag does it. Does not.
Found the answer here before posting. SInce I thought I knew JavaScript (read three books, worked hours) and there wasn't an existing solution at SO, I'll dare to post anyways. throw stones here
So the solution is:
var ss= "<pre>aaaa\nbbb\nccc</pre>ddd";
var arr= ss.match( /<pre[\s\S]*?<\/pre>/gm );
alert(arr); // <pre>...</pre> :)
Does anyone have a less cryptic way?
Edit: this is a duplicate but since it's harder to find than mine, I don't remove.
It proposes [^] as a "multiline dot". What I still don't understand is why [.\n] does not work. Guess this is one of the sad parts of JavaScript..
DON'T use (.|[\r\n]) instead of . for multiline matching.
DO use [\s\S] instead of . for multiline matching
Also, avoid greediness where not needed by using *? or +? quantifier instead of * or +. This can have a huge performance impact.
See the benchmark I have made: https://jsben.ch/R4Hxu
Using [^]: fastest
Using [\s\S]: 0.83% slower
Using (.|\r|\n): 96% slower
Using (.|[\r\n]): 96% slower
NB: You can also use [^] but it is deprecated in the below comment.
[.\n] does not work because . has no special meaning inside of [], it just means a literal .. (.|\n) would be a way to specify "any character, including a newline". If you want to match all newlines, you would need to add \r as well to include Windows and classic Mac OS style line endings: (.|[\r\n]).
That turns out to be somewhat cumbersome, as well as slow, (see KrisWebDev's answer for details), so a better approach would be to match all whitespace characters and all non-whitespace characters, with [\s\S], which will match everything, and is faster and simpler.
In general, you shouldn't try to use a regexp to match the actual HTML tags. See, for instance, these questions for more information on why.
Instead, try actually searching the DOM for the tag you need (using jQuery makes this easier, but you can always do document.getElementsByTagName("pre") with the standard DOM), and then search the text content of those results with a regexp if you need to match against the contents.
You do not specify your environment and version of JavaScript (ECMAScript), and I realise this post was from 2009, but just for completeness:
With the release of ECMA2018 we can now use the s flag to cause . to match \n (see https://stackoverflow.com/a/36006948/141801).
Thus:
let s = 'I am a string\nover several\nlines.';
console.log('String: "' + s + '".');
let r = /string.*several.*lines/s; // Note 's' modifier
console.log('Match? ' + r.test(s)); // 'test' returns true
This is a recent addition and will not work in many current environments, for example Node v8.7.0 does not seem to recognise it, but it works in Chromium, and I'm using it in a Typescript test I'm writing and presumably it will become more mainstream as time goes by.
Now there's the s (single line) modifier, that lets the dot matches new lines as well :)
\s will also match new lines :D
Just add the s behind the slash
/<pre>.*?<\/pre>/gms
[.\n] doesn't work, because dot in [] (by regex definition; not javascript only) means the dot-character. You can use (.|\n) (or (.|[\n\r])) instead.
I have tested it (Chrome) and it's working for me (both [^] and [^\0]), by changing the dot (.) with either [^\0] or [^] , because dot doesn't match line break (See here: http://www.regular-expressions.info/dot.html).
var ss= "<pre>aaaa\nbbb\nccc</pre>ddd";
var arr= ss.match( /<pre[^\0]*?<\/pre>/gm );
alert(arr); //Working
In addition to above-said examples, it is an alternate.
^[\\w\\s]*$
Where \w is for words and \s is for white spaces
[\\w\\s]*
This one was beyond helpful for me, especially for matching multiple things that include new lines, every single other answer ended up just grouping all of the matches together.

RegExp sub-pattern reuse for different open-close conditions?

Is it possible to reuse a matching RegExp sub-pattern for a variety of opening and closing conditions of the containing pattern?
I have a complex/long RegExp sub-pattern for a certain expression X, which I expect to reside within any of the open-close statements, defined as: ${...}, $(...), $[...], $/.../, etc., which in combination makes the whole pattern (mixing open-close conditions is not accepted, or it would have been trivial).
What I want is to avoid repeating the same long X sub-pattern for each of the open-close conditions (using |) when defining the whole pattern, as it becomes too long and unreadable, even though it is mostly just repeated X sub-pattern.
My question - is this achievable within the RegExp syntax? And if yes, then how?
Environments: Node 0.12 for ES5 and IO.js 2.0 for ES6.
P.S. Strictly speaking, we are talking RegExp optimization here, for better code readability, and, possibly, performance.
You can use an extremely hacky way of matching specific opening and closing braces when used together:
\$(?:(\[)|(\()|({)|(\/)).*?(?:(?=\2)(?=\3)(?=\4)\]|(?=\1)(?=\3)(?=\4)\)|(?=\1)(?=\2)(?=\4)}|(?=\1)(?=\2)(?=\3)\/)
^^^ Inner Match Here
It basically looks for all groups except one specific one to be empty and happens to only work in JavaScript regex. The .*? section pointed out in the above code just needs to be replaced with the regex to be matched inside of the braces to match an arbitrary pattern.
Demo: https://regex101.com/r/aX7rH1/1
// Matches
${...}
$(...)
$[...]
$/.../
// Does Not Match
${...)
${...]
${.../
$(...}
$(...]
$(.../
$[...}
$[...)
$[.../
$/...}
$/...)
$/...]

Why is my RegExp construction not accepted by JavaScript?

I'm using a RegExp to validate some user input on an ASP.NET web page. It's meant to enforce the construction of a password (i.e. between 8 and 20 long, at least one upper case character, at least one lower case character, at least one number, at least one of the characters ##!$% and no use of letters L or O (upper or lower) or numbers 0 and 1. This RegExp works fine in my tester (Expresso) and in my C# code.
This is how it looks:
(?-i)^(?=.{8,20})(?=.*[2-9])(?=.*[a-hj-km-np-z])(?=.*[A-HJ-KM-NP-Z])
(?=.*[##!$%])[2-9a-hj-km-np-zA-HJ-KM-NP-Z##!$%]*$
(Line break added for formatting)
However, when I run the code it lives in in IE6 or IE7 (haven't tried other browsers as this is an internal app and we're a Microsoft shop), I get a runtime error saying 'Syntax error in regular expression'. That's it - no further information in the error message aside from the line number.
What is it about this that JavaScript doesn't like?
Well, there are two ways of defining a Regex in Javascript:
a. Through a Regexp object constructor:
var re = new RegExp("pattern","flags");
re.test(myTestString);
b. Using a string literal:
var re = /pattern/flags;
You should also note that JS does not support some of the tenets of Regular Expressions. For a non-comprehensive list of features unsupported in JS, check out the regular-expressions.info site.
Specifically speaking, you appear to be setting some flags on the expression (for example, the case insensitive flag). I would suggest that you use the /i flag (as indicated by the syntax above) instead of using (?-i)
That would make your Regex as follows (Positive Lookahead appears to be supported):
/^(?=.{8,20})(?=.*[2-9])(?=.*[a-hj-km-np-z])(?=.*[A-HJ-KM-NP-Z])(?=.*[##!$%])[2-9a-hj-km-np-zA-HJ-KM-NP-Z##!$%]*$/i;
For a very good article on the subject, check out Regular Expressions in JavaScript.
Edit (after Howard's comment)
If you are simply assigning this Regex pattern to a RegularExpressionValidator control, then you will not have the ability to set Regex options (such as ignore case). Also, you will not be able to use the Regex literal syntax supported by Javascript. Therefore, the only option that remains is to make your pattern intrinsically case insensitive. For example, [a-h] would have to be written as [A-Ha-h]. This would make your Regex quite long-winded, I'm sorry to say.
Here is a solution to this problem, though I cannot vouch for it's legitimacy. Some other options that come to mind may be to turn of Client side validation altogether and validate exclusively on the Server. This will give you access to the full Regex flavour implemented by the System.Text.RegularExpressions.Regex object. Alternatively, use a CustomValidator and create your own JS function which applies the Regex match using the patterns that I (and others) have suggested.
I'm not familiar with C#'s regular expression syntax, but is this (at the start)
(?-i)
meant to turn the case insensitivity pattern modifier on? If so, that's your problem. Javascript doesn't support specifying the pattern modifiers in the expression. There's two ways to do this in javascript
var re = /pattern/i
var re = new RegExp('pattern','i');
Give one of those a try, and your expression should be happy.
As Cerberus mentions, (?-i) is not supported in JavaScript regexps. So, you need to get rid of that and use /i. Something to keep in mind is that there is no standard for regular expression syntax; it is different in each language, so testing in something that uses the .NET regular expression engine is not a valid test of how it will work in JavaScript. Instead, try and look for a reference on JavaScript regular expressions, such as this one.
Your match that looks for 8-20 characters is also invalid. This will ensure that there are at least 8 characters, but it does not limit the string to 20, since the character class with the kleene-closure (* operator) at the end can match as many characters as provided. What you want instead is to replace the * at the end with the {8,20}, and eliminate it from the beginning.
var re = /^(?=.*[2-9])(?=.*[a-hj-km-np-z])(?=.*[A-HJ-KM-NP-Z])(?=.*[##!$%])[2-9a-hj-km-np-zA-HJ-KM-NP-Z##!$%]{8,20}$/i;
On the other hand, I'm not really sure why you would want to restrict the length of passwords, unless there's a hard database limit (which there shouldn't be, since you shouldn't be storing passwords in plain text in the database, but instead hashing them down to something fixed size using a secure hash algorithm with a salt). And as mentioned, I don't see a reason to be so restrictive on the set of characters you allow. I'd recommend something more like this:
var re = /^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##!$%])[a-zA-Z0-9##!$%]{8,}$/i;
Also, why would you forbid 1, 0, L and O from your passwords (and it looks like you're trying to forbid I as well, which you forgot to mention)? This will make it very hard for people to construct good passwords, and since you never see a password as you type it, there's no reason to worry about letters which look confusingly similar. If you want to have a more permissive regexp:
var re = /^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##!$%]).{8,}$/i;
Are you enclosing the regexp in / / characters?
var regexp = /[]/;
return regexp.test();
(?-i)
Doesn't exist in JS Regexp. Flags can be specified as “new RegExp('pattern', 'i')”, or literal syntax “/pattern/i”.
(?=
Exists in modern implementations of JS Regexp, but is dangerously buggy in IE. Lookahead assertions should be avoided in JS for this reason.
between 8 and 20 long, at least one upper case character, at least one lower case character, at least one number, at least one of the characters ##!$% and no use of letters L or O (upper or lower) or numbers 0 and 1.
Do you have to do this in RegExp, and do you have to put all the conditions in one RegExp? Because those are easy conditions to match using multiple RegExps, or even simple string matching:
if (
s.length<8 || s.length>20 ||
s==s.toLowerCase() || s==s.toUpperCase() ||
s.indexOf('0')!=-1 || s.indexOf('1')!=-1 ||
s.toLowerCase().indexOf('l')!=-1 || s.toLowerCase().indexOf('o')!=-1 ||
(s.indexOf('#')==-1 && s.indexOf('#')==-1 && s.indexOf('!')==-1 && s.indexOf('%')==-1 && s.indexOf('%')==-1)
)
alert('Bad password!');
(These are really cruel and unhelpful password rules if meant for end-users BTW!)
I would use this regular expression:
/(?=[^2-9]*[2-9])(?=[^a-hj-km-np-z]*[a-hj-km-np-z])(?=[^A-HJ-KM-NP-Z]*[A-HJ-KM-NP-Z])(?=[^##!$%]*[##!$%])^[2-9a-hj-km-np-zA-HJ-KM-NP-Z##!$%]{8,}$/
The [^a-z]*[a-z] will make sure that the match is made as early as possible instead of expanding the .* and doing backtracking.
(?-i) is supposed to turn case-insensitivity off. Everybody seems to be assuming you're trying to turn it on, but that would be (?i). Anyway, you don't want it to be case-insensitive, since you need to ensure that there are both uppercase and lowercase letters. Since case-sensitive matching is the default, prefacing a regex with (?-i) is pointless even in those flavors (like .NET) that support inline modifiers.

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