开发者

What does the "Nothing to repeat" error mean when using a regex in javascript?

开发者 https://www.devze.com 2023-03-11 05:34 出处:网络
I\'m new to Regex and I\'m trying to work it into one of my new projects to see if I can learn it and add it to my repitoire of skills. However, I\'m hitting a roadblock here.

I'm new to Regex and I'm trying to work it into one of my new projects to see if I can learn it and add it to my repitoire of skills. However, I'm hitting a roadblock here.

I'm trying to see if the user's input has illegal characters in it by using the .search function as so:

if (name.search("[\[\]\?\*\+\|\{\}\\\(\)\@\.\n\r]") != -1) {
    ...
}

However, when I try to execute the function this line is contained it, it throws the following error for that specific line:

Uncaught SyntaxError: Invalid regular expression: /[[]?*+|{}\()@.

]/: Nothing to repeat

I can't for the life of me see what's wr开发者_StackOverflow中文版ong with my code. Can anyone point me in the right direction?


You need to double the backslashes used to escape the regular expression special characters. However, as @Bohemian points out, most of those backslashes aren't needed. Unfortunately, his answer suffers from the same problem as yours. What you actually want is:

The backslash is being interpreted by the code that reads the string, rather than passed to the regular expression parser. You want:

"[\\[\\]?*+|{}\\\\()@.\n\r]"

Note the quadrupled backslash. That is definitely needed. The string passed to the regular expression compiler is then identical to @Bohemian's string, and works correctly.


Building off of @Bohemian, I think the easiest approach would be to just use a regex literal, e.g.:

if (name.search(/[\[\]?*+|{}\\()@.\n\r]/) != -1) {
    // ... stuff ...
}

Regex literals are nice because you don't have to escape the escape character, and some IDE's will highlight invalid regex (very helpful for me as I constantly screw them up).


For Google travelers: this stupidly unhelpful error message is also presented when you make a typo and double up the + regex operator:

Okay:

\w+

Not okay:

\w++


Firstly, in a character class [...] most characters don't need escaping - they are just literals.

So, your regex should be:

"[\[\]?*+|{}\\()@.\n\r]"

This compiles for me.


Well, in my case I had to test a Phone Number with the help of regex, and I was getting the same error,

Invalid regular expression: /+923[0-9]{2}-(?!1234567)(?!1111111)(?!7654321)[0-9]{7}/: Nothing to repeat'

So, what was the error in my case was that + operator after the / in the start of the regex. So enclosing the + operator with square brackets [+], and again sending the request, worked like a charm.

Following will work:

/[+]923[0-9]{2}-(?!1234567)(?!1111111)(?!7654321)[0-9]{7}/

This answer may be helpful for those, who got the same type of error, but their chances of getting the error from this point of view, as mine! Cheers :)


for example I faced this in express node.js when trying to create route for paths not starting with /internal

app.get(`\/(?!internal).*`, (req, res)=>{

and after long trying it just worked when passing it as a RegExp Object using new RegExp()

app.get(new RegExp("\/(?!internal).*"), (req, res)=>{

this may help if you are getting this common issue in routing


This can also happen if you begin a regex with ?.

? may function as a quantifier -- so ? may expect something else to come before it, thus the "nothing to repeat" error. Nothing preceded it in the regex string so it didn't get to quantify anything; there was nothing to repeat / nothing to quantify.

? also has another role -- if the ? is preceded by ( it may indicate the beginning of a lookaround assertion or some other special construct. See example below.

If one forgets to write the () parentheses around the following lookbehind assertion ?<=x, this will cause the OP's error:

Incorrect: const xThenFive = /?<=x5/;

Correct: const xThenFive = /(?<=x)5/;

This /(?<=x)5/ is a positive lookbehind: we're looking for a 5 that is preceded by an x e.g. it would match the 5 in x563 but not the 5 in x652.

0

精彩评论

暂无评论...
验证码 换一张
取 消