Regex Pattern to Match the End of a String

regex pattern to match the end of a string

Use the $ metacharacter to match the end of a string.

In Perl, this looks like:

my $str = 'red/white/blue';
my($last_match) = $str =~ m/.*\/(.*)$/;

Written in JavaScript, this looks like:

var str = 'red/white/blue'.match(/.*\/(.*)$/);

Regular Expression only match if String ends with target

Use an end anchor ($):

.*\.ccf$

This will match any string that ends with .ccf, or in multi-line mode, any line that ends with .ccf.

Regex: match patterns starting from the end of string

You can replace the lookahead with a negated character class:

([^\/\\]+):(\d+):(\d+)\D*$

See the regex demo. Details:

([^\/\\]+) - Group 1: one or more chars other than / and \
: - a colon
(\d+) - Group 2: one or more digits
: - a colon
(\d+) - Group 3: one or more digits
\D*$ - zero or more non-digit chars till end of string.

Regex start searching from the end of the string (reverse)

Have a try with:

((?:_[^_\r\n]*){2})$

It matches an underscore followed by any number of any character that is not underscore or line break, all that occurs twice before the end of lien.

How to match a string's end using a regex pattern in Java?

You need to match "s", but only if it is the last character in a word. This is achieved with the boundary assertion $:

input.replaceAll("s$", " ");

If you enhance the regular expression, you can replace multiple suffixes with one call to replaceAll:

input.replaceAll("(ed|s)$", " ");

In regex, match either the end of the string or a specific character

Use:

/(&|\?)list=.*?(&|$)/

Note that when you use a bracket expression, every character within it (with some exceptions) is going to be interpreted literally. In other words, [&|$] matches the characters &, |, and $.

RegEx match a number at the end of a string

. matches any character. You need to escape it as \.

Try this:

/-?\d+\.?\d*$/

That is:

-?           // optional minus sign
\d+          // one or more digits
\.?          // optional .
\d*          // zero or more digits

As you can see at MDN's regex page, +? is a non-greedy match of 1 or more, not an optional match of 1 or more.

Regex to match the beginning and the end of a string in Java

UPDATE

You want to match a sed, so you can use a\\s+sed if there is only whitespace between a and sed:

String s = "afsgdhgd gfgshfdgadh a sed afdsgdhgdsfgdfagdfhh";
Pattern pattern = Pattern.compile("a\\s+sed");
Matcher matcher = pattern.matcher(s);
while (matcher.find()){
    System.out.println(matcher.group(0)); 
}

See IDEONE demo

Now, if there can be anything between a and sed, use a tempered greedy token:

Pattern pattern = Pattern.compile("(?s)a(?:(?!a|sed).)*sed");
                                         ^^^^^^^^^^^^^

See another IDEONE demo.

ORIGINAL ANSWER

The main problem with your regex is the \n at the end. $ is the end of string, and you try to match one more character after a string end, which is impossible. Also, \\s matches a whitespace symbol, but you need a literal s.

You need to remove \\s and \n and make . match a newline, and also it is advisbale to use * quantifier to allow 0 symbols in-between:

pattern = "(?s)^a.*sed$";

See the regex demo

The regex matches:

^ - start of string
a - a literal a
.* - 0 or more any characters (since (?s) modifier makes a . match any character including a newline)
sed - a literal letter sequence sed
$ - end of string

Regex for string ending with particular pattern

You can just use the $ anchor to match the end of the string;

re.match(r"\.[0-9]$", i)

R regex to match beginning and end of string, ignoring middle

Simply adding a match all pattern .* between the start and end should work:

grep("^\\./xl/worksheets.*\\.xml$", myfiles) 
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Regex Pattern to Match the End of a String