Say I have a string:
/first/second/third
And I want to remove everything after the last instance of / so I would end up with:
/first/second
What regular expression would I use? I've tried:
String path = "/first/second/third";
String pattern = "$(.*?)/";
Pattern r = Pattern.compile(pattern2);
Matcher m = r.matcher(path);
if(m.find()) path = m.replaceAll("");
Why use a regex at all here? Look for the last / character with lastIndexOf. If it's found, then use substring to extract everything before it.
Do you mean like this
s = s.replaceAll("/[^/]*$", "");
Or better if you are using paths
File f = new File(s);
File dir = f.getParent(); // works for \ as well.
replaceAll I think.
If you have a string that contains your character (whether a supplemental code-point or not), then you can use Pattern.quote and match the inverse charset up to the end thus:
String myCharEscaped = Pattern.quote(myCharacter);
Pattern pattern = Pattern.compile("[^" + myCharEscaped + "]*\\z");
should do it, but really you can just use lastIndexOf as in
myString.substring(0, s.lastIndexOf(myCharacter) + 1)
To get a code-point as a string just do
new StringBuilder().appendCodePoint(myCodePoint).toString()
Despite the answers avoiding regex Pattern and Matcher, it's useful for performance (compiled patterns) and it'still pretty straightforward and worth mastering. :)
Not sure why you have "$" up front. Try either:
Matching starting group
String path = "/first/second/third";
String pattern = "^(.*)/"; // * = "greedy": maximum string from start to last "/"
Pattern r = Pattern.compile(pattern2);
Matcher m = r.matcher(path);
if (m.find()) path = m.group();
Stripping tail match:
String path = "/first/second/third";
String pattern = "/(.*?)$)/"; // *? = "reluctant": minimum string from last "/" to end
Pattern r = Pattern.compile(pattern2);
Matcher m = r.matcher(path);
if (m.find()) path = m.replace("");
