正则表达式在应返回true时返回false

Question

I have this regex:

regex = new RegExp("\${")

and this string:

path = "${wf:conf('oozie.wf.application.path')}/workflows/core"

When I test the string against the regexp:

regex .test(path)

it returns false, why?

Answer 1

You regex is equivalent to ${ . Ie you are looking for the end of the string followed by a { which will never match.

Just like in regular expressions, you use \\ in string literals to escape special characters or create special escape sequences. Eg

> 'foo\'bar'
"foo'bar"

However, if you are trying to escape a character that is not special or creating an invalid escape sequence, the backslash is just discarded:

> 'foob\ar'
"foobar"

That means, if you use a string literal to construct the regular expression, and you want to "forward" the backslash to the expression, you have to escape it as well:

new RegExp("\\${")

You can just create the expressions in the console and convince yourself:

> RegExp("\${")
/${/

> /${/
/${/

> RegExp("\\${")
/\${/

> /\${/
/\${/

---

It really is simpler to use a regex literal:

var regex = /\${/;

Answer 2

You need to use double \\ to create \\${ as \\ is the string literal escape character.

You need a string \\${ as the regex, in string literal to represent a \\ , you need to use \\\\ as a single \\ will act like a escape notation.

regex = new RegExp("\\${")

Answer 3

Can you please explain why should I use double slash? I use one to escape $ , why second?

Let us take another example. "\\n" is an escape-sequence in a string, producing a newline character. In order to produce the sequence \\n (backslash, then n ) in a string, you need to escape the backslash: "\\\\n" . To produce a regular expression that matches the string \\n (backslash, n ), you can't use new RegExp("\\\\n") , since that will make a new regexp of \\ and n (result of string escaping); but regexp again interprets \\ and n as newline. So you need double escape: once to tell string that you want a literal backslash, and once more to convince regexp that you want a literal backslash as well. So, to match the backslash followed by n , you need to do this:

new Regexp("\\\\n")

which is equivalent to

/\\n/

which will match the string in foo if

console.log(foo)
// => \n

In your example, it's not the backslash that is escaped, but a dollar; only regexp needs the escape on that. But you need to be sure that the backslash you escape the dollar with survives the string literal's escape mechanism:

"\\$"

will contain two characters: backslash and $ , and regexp constructed form it will be equivalent to /\\$/ . But if you start with the string "\\$" , it is equivalent to "$" , since dollars don't need escaping in regular strings; and a regexp made from that will match the end of the line, and not a dollar sign.