java Regex Question

Question

In Java - I need to search/validate an input string for a number or for a specific string. it must be one of them. For example - If I have this input lines:

cccccccc 123 vvvvvvvvv jhdakfksah
cccccccc ABC vvvvvvnhj  yroijpotpo
cccdcdcd 234 vcbvbvbvbv lkjd dfdggf
ccccbvff ABC jflkjlkjlj fgdgfg

I need to find 123, ABC, 234, ABC

to look for the number I could use regex: "\d+" but to look for either of them How can I combine them?

Answer 1

Something like this: [0-9]+|[a-zA-Z]+ or maybe [0-9]+|(ABC)+; not sure what your rule for the characters is.

Answer 2

Try this regex: "(\d+|ABC)"

Answer 3

You can specify alternatives in a regular expression by using the | character:

\d+|[ABC]+

In your specific example, the string you want is seemingly always the second "word" (delimited by a space), so it can be beneficial to include the space in the regular expression to look for it. Either by using a capturing group to capture the part you actually want to extract:

" (\d+|[ABC]+)"

(I used quotation marks to delimit the regex). Or by using a lookbehind assertion:

(?<= )(\d+|[ABC]+)

which will only match the desired portion, but still requires the space before it.

Answer 4

Try this regular expression:

^\w+\W+(\w+)

And make your expression to match the begin of a line with ^.

Answer 5

As at least one other answer has mentioned, it looks like you're picking out the second word in a space-delimited string, and regular expressions are more work than is necessary for that task. String.indexOf would be enough:

String line = ...;
int start = line.indexOf(" ") + 1;
int end = line.indexOf(" ", end);
String word2 = line.substring(start, end);

or, now that I think about it

String word2 = line.split(" ")[1];