JS - RegExp for detecting ".-" , "-." - javascript

I am bit confused with the RegExp I should be using to detect ".-", "-." it indeed passes this combinations as valid but in the same time, "-_","_-" get validated as well. Am I missing something or not escaping something properly?
var reg=new RegExp("(\.\-)|(\-\.)");
Actually seems any combination containing '-' gets passed. it
Got it thank you everyone.

You need to use
"(\\.-)|(-\\.)"

Since you're using a string with the RegExp constructor rather than /, you need to escape twice.
>>> "asd_-ads".search("(\.\-)|(\-\.)")
3
>>> "asd_-ads".search(/(\.\-)|(\-\.)/)
-1
>>> "asd_-ads".search(new RegExp('(\\.\-)|(\-\\.)'))
-1

In notation /(\.\-)|(\-\.)/, the expression would be right.
In the notation you chose, you must double all backslashes, because it still has a special meaning of itself, like \\, \n and so on.
Note there is no need to escape the dash here: var reg = new RegExp("(\\.-)|(-\\.)");
If you don't need to differentiate the matches, you can use a single enclosing capture, or none at all if you only want to check the match: "\\.-|-\\." is still valid.

You are using double quotes so the . doesn't get escaped with one backslash, use this notation:
var reg = /(\.\-)|(\-\.)/;

Related

Search for a string with a point in JavaScript

When I try to seach for a string with a point (i.e. '1.'), js also points to substrings with commas instead of points. It's better to look at the example:
'1,'.search('1.'); // 0
'xxx1,xxx'.search('1.'); // 3
// Normal behaviour
'1.'.search('1,'); // -1
Does anyone know why JavaScript behave itself so?
Is there a way to search for exactly passed string?
Per the docs:
The search() method executes a search for a match between a regular expression and this String object.
. has a special meaning in Regular Expressions. You need to escape the . before matching it. Try the following:
console.log('xxx1,xxx'.search('1\\.'));
Use indexOf().
let str = "abc1,231.4";
console.log(str.indexOf("1."));
indexOf() method should work fine in this case
'1,'.indexOf('1.');
The above code should return -1
String.search is taking regex as parameter.
Regex evaluate . by any character; you gotta escape it using double anti-slash \\.
console.log('1,'.search('1\\.'));
console.log('xxx1,xxx'.search('1\\.'));
console.log('xxx1.xxx'.search('1\\.'));
console.log('1.'.search('1,'));

Regex: get string between last character occurence before a comma

I need some help with Regex.
I have this string: \\lorem\ipsum\dolor,\\sit\amet\conseteteur,\\sadipscing\elitr\sed\diam
and want to get the result: ["dolor", "conseteteur", "diam"]So in words the word between the last backslash and a comma or the end.
I've already figured out a working test, but because of reasons it won't work in neitherChrome (v44.0.2403.130) nor IE (v11.0.9600.17905) console.There i'm getting the result: ["\loremipsumdolor,", "\sitametconseteteur,", "\sadipscingelitrseddiam"]
Can you please tell me, why the online testers aren't working and how i can achieve the right result?
Thanks in advance.
PS: I've tested a few online regex testers with all the same result. (regex101.com, regexpal.com, debuggex.com, scriptular.com)
The string
'\\lorem\ipsum\dolor,\\sit\amet\conseteteur,\\sadipscing\elitr\sed\diam'
is getting escaped, if you try the following in the browser's console you'll see what happens:
var s = '\\lorem\ipsum\dolor,\\sit\amet\conseteteur,\\sadipscing\elitr\sed\diam'
console.log(s);
// prints '\loremipsumdolor,\sitametconseteteur,\sadipscingelitrseddiam'
To use your original string you have to add additional backslashes, otherwise it becomes a different one because it tries to escape anything followed by a single backslash.
The reason why it works in regexp testers is because they probably sanitize the input string to make sure it gets evaluated as-is.
Try this (added an extra \ for each of them):
str = '\\\\lorem\\ipsum\\dolor,\\\\sit\\amet\\conseteteur,\\\\sadipscing\\elitr\\sed\\diam'
re = /\\([^\\]*)(?:,|$)/g
str.match(re)
// should output ["\dolor,", "\conseteteur,", "\diam"]
UPDATE
You can't prevent the interpreter from escaping backslashes in string literals, but this functionality is coming with EcmaScript6 as String.raw
s = String.raw`\\lorem\ipsum\dolor,\\sit\amet\conseteteur,\\sadipscing\elitr\sed\diam`
Remember to use backticks instead of single quotes with String.raw.
It's working in latest Chrome, but I can't say for all other browsers, if they're moderately old, it probably isn't implemented.
Also, if you want to avoid matching the last backslash you need to:
remove the \\ at the start of your regexp
use + instead of * to avoid matching the line end (it will create an extra capture)
use a positive lookahead ?=
like this
s = String.raw`\\lorem\ipsum\dolor,\\sit\amet\conseteteur,\\sadipscing\elitr\sed\diam`;
re = /([^\\]+)(?=,|$)/g;
s.match(re);
// ["dolor", "conseteteur", "diam"]
You may try this,
string.match(/[^\\,]+(?=,|$)/gm);
DEMO

Why doesn't this particular regex work in JavaScript?

I have this regex on Javascript :
var myString="aaa#aaa.com";
var mailValidator = new RegExp("\w+([-+.]\w+)*#\w+([-.]\w+)*\.\w+([-.]\w+)*");
if (!mailValidator.test(myString))
{
alert("incorrect");
}
but it shouldn't alert "incorrect" with aaa#aaa.com.
It should return "incorrect" for aaaaaa.com instead (as example).
Where am I wrong?
When you create a regex from a string, you have to take into account the fact that the parser will strip out backslashes from the string before it has a chance to be parsed as a regex.
Thus, by the time the RegExp() constructor gets to work, all the \w tokens have already been changed to just plain "w" in the string constant. You can either double the backslashes so the string parse will leave just one, or you can use the native regex constant syntax instead.
It works if you do this:
var mailValidator = /\w+([-+.]\w+)*#\w+([-.]\w+)*\.\w+([-.]\w+)*/;
What happens in yours is that you need to double escape the backslash because they're inside a string, like "\\w+([-+.]\\w+)*...etc
Here's a link that explains it (in the "How to Use The JavaScript RegExp Object" section).
Try var mailValidator = new RegExp("\\w+([-+.]\\w+)*#\\w+([-.]\\w+)*\\.\\w+([-.]\\w+)*");

Javascript regexp lets undesirable characters

I'm using a regExp in my project but some how I'm getting some undesirable characters
my RegExp looks like this:
new RegExp("[א-ת,A-z,',','(',')','.','-',''']");
which supposed to avoid characters like \ or []
but let my use one and more from (,),-,alphabets etc.
Unfortunately it doesnt happen
Which pattren includes both desirable and undesirable characters??
thanks for your help
Well your regular expression just says to match one "good" character (and incorrectly at that).
I think something closer to this would be what you want, though I'm not sure about the higher-page UTC characters:
var regexp = /^[א-תA-Za-z,()\-']*$/;
If the alefbet part doesn't work (it looks backwards to me, but I guess that's kind of a conundrum :-), try:
var regexp = /^[\u05DA-\05EAA-Za-z,()\-']*$/;
Might be good to tack an "i" (ignore case) modifier on the end too:
var regexp = /^[\u05DA-\05EAA-Za-z,()\-']*$/i;
This also does not handler the various diacritical marks; I don't know if you need those matched or not.
First of all, you don't need all those single quotes and commas. Second, you want A-Za-z, not.A-z. The latter includes ASCII characters between "Z" and "a".
var re = new RegExp("[א-תA-Za-z,()\.'\s-]");

Regex to match all instances not inside quotes

From this q/a, I deduced that matching all instances of a given regex not inside quotes, is impossible. That is, it can't match escaped quotes (ex: "this whole \"match\" should be taken"). If there is a way to do it that I don't know about, that would solve my problem.
If not, however, I'd like to know if there is any efficient alternative that could be used in JavaScript. I've thought about it a bit, but can't come with any elegant solutions that would work in most, if not all, cases.
Specifically, I just need the alternative to work with .split() and .replace() methods, but if it could be more generalized, that would be the best.
For Example:
An input string of: +bar+baz"not+or\"+or+\"this+"foo+bar+
replacing + with #, not inside quotes, would return: #bar#baz"not+or\"+or+\"this+"foo#bar#
Actually, you can match all instances of a regex not inside quotes for any string, where each opening quote is closed again. Say, as in you example above, you want to match \+.
The key observation here is, that a word is outside quotes if there are an even number of quotes following it. This can be modeled as a look-ahead assertion:
\+(?=([^"]*"[^"]*")*[^"]*$)
Now, you'd like to not count escaped quotes. This gets a little more complicated. Instead of [^"]* , which advanced to the next quote, you need to consider backslashes as well and use [^"\\]*. After you arrive at either a backslash or a quote, you need to ignore the next character if you encounter a backslash, or else advance to the next unescaped quote. That looks like (\\.|"([^"\\]*\\.)*[^"\\]*"). Combined, you arrive at
\+(?=([^"\\]*(\\.|"([^"\\]*\\.)*[^"\\]*"))*[^"]*$)
I admit it is a little cryptic. =)
Azmisov, resurrecting this question because you said you were looking for any efficient alternative that could be used in JavaScript and any elegant solutions that would work in most, if not all, cases.
There happens to be a simple, general solution that wasn't mentioned.
Compared with alternatives, the regex for this solution is amazingly simple:
"[^"]+"|(\+)
The idea is that we match but ignore anything within quotes to neutralize that content (on the left side of the alternation). On the right side, we capture all the + that were not neutralized into Group 1, and the replace function examines Group 1. Here is full working code:
<script>
var subject = '+bar+baz"not+these+"foo+bar+';
var regex = /"[^"]+"|(\+)/g;
replaced = subject.replace(regex, function(m, group1) {
if (!group1) return m;
else return "#";
});
document.write(replaced);
Online demo
You can use the same principle to match or split. See the question and article in the reference, which will also point you code samples.
Hope this gives you a different idea of a very general way to do this. :)
What about Empty Strings?
The above is a general answer to showcase the technique. It can be tweaked depending on your exact needs. If you worry that your text might contain empty strings, just change the quantifier inside the string-capture expression from + to *:
"[^"]*"|(\+)
See demo.
What about Escaped Quotes?
Again, the above is a general answer to showcase the technique. Not only can the "ignore this match" regex can be refined to your needs, you can add multiple expressions to ignore. For instance, if you want to make sure escaped quotes are adequately ignored, you can start by adding an alternation \\"| in front of the other two in order to match (and ignore) straggling escaped double quotes.
Next, within the section "[^"]*" that captures the content of double-quoted strings, you can add an alternation to ensure escaped double quotes are matched before their " has a chance to turn into a closing sentinel, turning it into "(?:\\"|[^"])*"
The resulting expression has three branches:
\\" to match and ignore
"(?:\\"|[^"])*" to match and ignore
(\+) to match, capture and handle
Note that in other regex flavors, we could do this job more easily with lookbehind, but JS doesn't support it.
The full regex becomes:
\\"|"(?:\\"|[^"])*"|(\+)
See regex demo and full script.
Reference
How to match pattern except in situations s1, s2, s3
How to match a pattern unless...
You can do it in three steps.
Use a regex global replace to extract all string body contents into a side-table.
Do your comma translation
Use a regex global replace to swap the string bodies back
Code below
// Step 1
var sideTable = [];
myString = myString.replace(
/"(?:[^"\\]|\\.)*"/g,
function (_) {
var index = sideTable.length;
sideTable[index] = _;
return '"' + index + '"';
});
// Step 2, replace commas with newlines
myString = myString.replace(/,/g, "\n");
// Step 3, swap the string bodies back
myString = myString.replace(/"(\d+)"/g,
function (_, index) {
return sideTable[index];
});
If you run that after setting
myString = '{:a "ab,cd, efg", :b "ab,def, egf,", :c "Conjecture"}';
you should get
{:a "ab,cd, efg"
:b "ab,def, egf,"
:c "Conjecture"}
It works, because after step 1,
myString = '{:a "0", :b "1", :c "2"}'
sideTable = ["ab,cd, efg", "ab,def, egf,", "Conjecture"];
so the only commas in myString are outside strings. Step 2, then turns commas into newlines:
myString = '{:a "0"\n :b "1"\n :c "2"}'
Finally we replace the strings that only contain numbers with their original content.
Although the answer by zx81 seems to be the best performing and clean one, it needes these fixes to correctly catch the escaped quotes:
var subject = '+bar+baz"not+or\\"+or+\\"this+"foo+bar+';
and
var regex = /"(?:[^"\\]|\\.)*"|(\+)/g;
Also the already mentioned "group1 === undefined" or "!group1".
Especially 2. seems important to actually take everything asked in the original question into account.
It should be mentioned though that this method implicitly requires the string to not have escaped quotes outside of unescaped quote pairs.

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