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In the GWT tutorial where you build a stock watcher there is this regex expression to check if an input is valid:

if (!symbol.matches("^[0-9A-Z\\.]{1,10}$"))

Which allows inputs between 1 and 10 chars that are numbers, letters, or dots.

The part that confuses me is the \\.

I interpret this as escaped backslash \\ and then a . which stands for any character. And I thought the correct expression would be \. to escape the dot but doing this results in a regex error in eclipse Invalid escape sequence.

Am I missing the obvious here?

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    It's a string literal - you want a backslash in the actual string, so you need to escape that for a normal Java string literal. Ignore the regex aspect: just String x = "\."; isn't valid Java code. Commented Nov 14, 2016 at 11:26
  • At least related: stackoverflow.com/questions/18503280/how-to-represent-backslash Commented Nov 14, 2016 at 11:27
  • Oh... OK, kind of obvious. Thanks! Commented Nov 14, 2016 at 11:28
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    You do not need to escape the dot in the character class at all. Escape it or not, it will match a literal dot. There is just no escape sequence as \.. And this thread is about the same issue: Java doesn't work with regex \s, says: invalid escape sequence. Commented Nov 14, 2016 at 11:28
  • \\. is not an escaped backslash followed by a colon. As the others have mentioned, Java needs to escape backslash in Strings, so this is equivalent to \. as Regex. If you want to have an escaped backslash in Regex, you'd have to write it like this: \\\\. where each \\ represents one backslash in the Regex. Commented Nov 14, 2016 at 12:12

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This is one of the hassles of regular expressions in Java. That \\ is not an escaped backslash at the regex level, just at the string level.

This string:

"^[0-9A-Z\\.]{1,10}$"

Defines this regular expression:

^[0-9A-Z\.]{1,10}$

...because the escape is consumed by the string literal.

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\ is the escape symbol in a Java String Literal. For instance the newline character is written as \n. In order to place a normal \ in a Java string, this is done by using \\.

So your Java String literal (string in the code): "^[0-9A-Z\\.]{1,10}$" is the actual string used for the regular expression "^[0-9A-Z\.]{1,10}$" (with a single slash). So as you expected this is \. in the regular expression.

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