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This is very basic but can't seem to find a answer when searching.
I have a string that's in the format:
[[[0.093493,51.6037],[0.091015,51.5956],[0.088596,51.5857]]]
The doubles inside the brackets are [latitude coordinate,longitude coordinate].
From this I'd like to extract the coordinates.
What should I put inside the search pattern if I use a Pattern/Matcher solution?
Assume that format with brackets is always correct but the doubles can vary in length.
Basicly what I want the code to do is:
Find "[" left of a number, then find this "," and return what's in between
AND another searchpattern that:
find "," and "]" and return what's inbetween.
Keep it simple by using this regex:
\[(\d+\.\d+),(\d+\.\d+)\]
and repeat the matcher.find() till all matches are found.
You matches are in group #1 and group #2
RegEx Demo
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I would like to ask you about regex expression - I need to get all numbers that occur to a certain character. For example:
"$z4~min.~00~s" -> 4
"$z12~min.~00~s" -> 12
I simply need first number in the string, I don't need numbers after dot in the string.
I am using Java for this project.
Do you have any suggestions? Thanks a lot.
java.util.regex.Pattern pattern = java.util.regex.Pattern.compile("^\\D*(\\d+)");
java.util.regex.Matcher matcher = pattern.matcher("$z12~min.~00~s");
if (matcher.find()) {
String firstNumber = matcher.group(1);
System.out.println(firstNumber);
}
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I have the following string:
___abcd-metadata.json
I am trying to do a regex to get ___ and everything after -. So far I have this Regex:
(\-.*?\.json)
To find the last part and (___) to find the first part, but I have not been able to figure how to combine both regexes to make this happen.
My desired result would be ___ -metadata.json = true
"(\-)(.*)" is the pattern, which will give you 2 matches if the string contains them.
Take your string and use String.split('-') to break it into parts.
EDIT
var parts = ('_abcd-metadata.json').split('-'),
start = parts[0].substr(0,3),
rest = parts[1];
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I need to take out all the characters of a string that are not numbers.
You could use a Character Class Intersection like [\p{Punct}\p{Lower}\p{Upper}&&[^.]]
But why not just use
[^\d.]+
As Java String "[^\\d.]+"
This would match one or more characters, that are not \d a digit or the . period.
I'd suggest using \\d+ then (it's consecutive digits), and a capture group. Something like
String str = "";
str = str.replaceAll("(\\d+\\.\\d+)", "$1");
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So I have to use Regex for the name of a character, where the name has to start with a capital and can only exist out of letters, apostrophe's and spaces.
And I dont know how to start, could anyone help me with this?
I suppose your character name is stored in a String variable. For that you can use the String.matches() function, which accepts a regex String parameter.
To create the required regex you will have to combine the folowing:
[A-Z] for the capital letter
[a-z' ]+ for the remaining characters.
Note that when using those in Java you'll need to add some escape characters
You can experiment with regular expressions here: http://www.regexr.com
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I want to use below expression in my program but i don't know what do this regular expression!
please help me.
"(?=(?!^)[,;.:])|(?<=[,;.:])"
in the above expression (?=(?!^)[,;.:]) find any character set that end with [.;,:] or no? what do this (?!^) in this expression?
and this expression find any character set that end with [,;.:] or no?
please help me.
The expression matches 0-length strings that satisfy one of these two conditions:
Ahead of it is one of ,;.:, but not for 0-length strings just before the beginning of the subject string (position 0). So the subject string "." has no match at position 0, only at position 1 because of the following alternative. This is done with positive lookahead (?=) and negative lookahead (?!).
Behind it is one of ,;.:. This is done with positive lookbehind (?<=).
For instance for "aaa,1", you have two matches: at position three (after the last a, because it's followed by ,) and at position 4 (because it's preceded by ,).