javascript regex saying illegal

Question

I was using the following code to detect all the javadoc comments in the pre elements to replace them with a color of my choice.

<script type="text/javascript">
    var arr = document.getElementsByTagName("pre");
    for (var i=0; i<arr.length; i++){
        arr[i].innerHTML = arr[i].innerHTML.replace(/\t/g, '&nbsp;&nbsp;&nbsp;&nbsp;');
        arr[i].innerHTML = arr[i].innerHTML.replace(/</g, '&lt;');
        arr[i].innerHTML = arr[i].innerHTML.replace(/>/g, '&gt;');
        arr[i].innerHTML = arr[i].innerHTML.replace(/\\/*\*\*([^\*]|\*(?!/))*\\*/g, '<font color="#3F5FBF">' + $& + '</font>');
    }
</script>

but it gives me the following error.

Uncaught SyntaxError: Unexpected token ILLEGAL

EDIT: changed the code to

<script type="text/javascript">
var arr = document.getElementsByTagName("pre");
for (var i=0; i<arr.length; i++){
    arr[i].innerHTML = arr[i].innerHTML.replace(/\t/g, '&nbsp;&nbsp;&nbsp;&nbsp;');
    arr[i].innerHTML = arr[i].innerHTML.replace(/</g, '&lt;');
    arr[i].innerHTML = arr[i].innerHTML.replace(/>/g, '&gt;');
    var reg = /\\\/*\*\*([^\*]|\*(?!\/))*\\*/;
    var cds = arr[i].innerHTML.match(reg);
    for (var j=0; j<cds.length; j++){
        arr[i].innerHTML = arr[i].innerHTML.replace(cds[j], '<font color="#3F5FBF">' + cds[j] + '</font>'); 
    }
}
prettyPrint();
</script>

But the error changes to

Uncaught TypeError: Cannot read property 'length' of null start-ahead.html:234

(anonymous function)

The unexpected token is probably \ after /\\/*. Look at the syntax highlighter, your escape is wrong. — Felix Kling
– Felix Kling, Commented May 30, 2012 at 1:15
@FelixKling is right, and there's another error in (?!/). Replace it with (?!\/) — Danilo Valente
– Danilo Valente, Commented May 30, 2012 at 1:19
@FelixKling is right, OP is escaping the escape character and not escaping the scope character. Just as Mark explained in his answer. +1 for Mark. — Fabrício Matté
– Fabrício Matté, Commented May 30, 2012 at 1:19
Syntax errors are something you should be able to fix by yourself. SO is not your personal debugger. Your expression is still incorrect. — Felix Kling
– Felix Kling, Commented May 30, 2012 at 1:24
Why don't you take a look at the 234 line of your code start-ahead.html:234 — Derek 朕會功夫
– Derek 朕會功夫, Commented May 30, 2012 at 1:27

Danilo Valente · Accepted Answer · 2012-05-30 01:16:45Z

4

The problem is here /\\/*\*\*([^\*]|\*(?!/))*\\*/g, when you used the \\ and then used /. The program understands that your regex is like this /\\/ and what comes after results in an error. Try this regex:

/\/*\*\*([^\*]|\*(?!\/))*\\*/g

Or, depending on what you want:

/\\\/*\*\*([^\*]|\*(?!\/))*\\*/g

answered May 30, 2012 at 1:16

Danilo Valente

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Comments

Bergi · Accepted Answer · 2012-05-30 01:25:55Z

1

The replace() method needs a string as a second parameter. The patterns in it to be replaced with captures need to be in that string.

You try to build that string out of two string literals and a variable named "$&". Yet, that is an illegal name - the "&" is a bitwise AND operator and ~~lacks a second operand, because it is followed by "+" - SyntaxError.~~ will produce an horrible runtime error.

Also, as @FelixKling mentioned, your regexp is badly escaped. It might should be something like new Regexp("\\/*\*\*([^\*]|\*(?!/))*\\*", "g"), yet then there were some useless backslashes. What do you want to approach? Is it /\*{3}([^*]|\*(?!\/))\**/g?

edited May 30, 2012 at 1:25

answered May 30, 2012 at 1:16

Bergi

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2 Comments

Felix Kling Over a year ago

@Mark: It won't produce this error, but it will nevertheless be a problem once the expression is fixed.

Felix Kling Over a year ago

This operation does not create an error though. Assuming $ is defined, 'a' + $& + 'b' just results in 0. I assume the whole expression is evaluated as 'a' + $ & (+'b') (the second + being an unary plus).

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