Develop Reference

Groovy remove beginning of path

I'm trying to delete the beginning of a path that has '\' and ' ' in it. I seem to be getting the some issues saying escape character issue at character 3.
Example:
SomePath: C:\Users\ADMINISTRATOR\App Play\blah\blah
SomePath.replaceFirst('C:\\Users\\ADMINISTRATOR\\App Play\\', '');
Path should be blah\blah
I've tried:
SomePath.replaceFirst("C:\Users\ADMINISTRATOR\App Play\", "");
SomePath.replaceFirst("C:\\Users\\ADMINISTRATOR\\App Play\\", "");
SomePath.replaceFirst("C:\\\\Users\\\\ADMINISTRATOR\\\\App Play\\\\", "");
SomePath.replaceAll("C:\Users\ADMINISTRATOR\App Play\", "");
SomePath.replaceAll("C:\\Users\\ADMINISTRATOR\\App Play\\", "");
SomePath.replaceAll("C:\\\\Users\\\\ADMINISTRATOR\\\\App Play\\\\", "");

Just gave it a try... the examples with four backslashes work for me:
def somePath = "C:\\Users\\ADMINISTRATOR\\App Play\\blah\\blah"
println somePath
somePath.replaceFirst("C:\\\\Users\\\\ADMINISTRATOR\\\\App Play\\\\", "");
The problem is that the string needs one escaping \ and since the replaceFirst uses a regexp, the regexp-engine needs another \ to escape the \. The result are four backslashes.
Btw: you can use string operations to get your path, but you could also try file operations like this:
def root= new File("C:\\Users\\ADMINISTRATOR\\App Play\\")
def full= new File("C:\\Users\\ADMINISTRATOR\\App Play\\blah\\blah")
def relPath = root.toPath().relativize( full.toPath() ).toFile()
println relPath
(taken from https://gist.github.com/ysb33r/5804364)

You can tackle this problem differently. You could tokenize your input path using \ as a delimiter and then you could pick the last 2 elements (blah and blah) or skip first 4 elements (C:, Users, ADMINISTRATOR, App Play). It depends which assumption is easier to deduct for you. Consider following example:
def somePath = 'C:\\Users\\ADMINISTRATOR\\App Play\\blah\\blah'
// Build a new path by accepting the last 2 parts of the initial path
assert 'blah\\blah' == somePath.tokenize('\\')[-2..-1].join('\\')
// Build a new path by skipping the first 4 parts from initial path
assert 'blah\\blah' == somePath.tokenize('\\').drop(4).join('\\')
First option works better if you want only two last parts from the initial path. Second option works better if you can expect final path like blah\blah\blahhhh because you don't know how many nested children initial path contains and you want to start building a new path right after \App Play\ .

Scala RegEx String extractors behaving inconsistently

I have two regular expression extractors.
One for .java files and the other is for .scala files
val JavaFileRegEx =
"""\S*
\s+
//
\s{1}
([^\.java]+)
\.java
""".replaceAll("(\\s)", "").r
val ScalaFileRegEx =
"""\S*
\s+
//
\s{1}
([^\.scala]+)
\.scala
""".replaceAll("(\\s)", "").r
I want to use these extractors above to extract a java file name and a scala file name from the example code below.
val string1 = " // Tester.java"
val string2 = " // Hello.scala"
string1 match {
case JavaFileRegEx(fileName1) => println(" Java file: " + fileName1)
case other => println(other + "--NO_MATCH")
}
string2 match {
case ScalaFileRegEx(fileName2) => println(" Scala file: " + fileName2)
case other => println(other + "--NO_MATCH")
}
I get this output indicating that the .java file matched but the .scala file did not.
Java file: Tester
// Hello.scala--NO_MATCH
How is it that the Java file matched but the .scala file did not?

NOTE
[] denotes character class. It matches only a single character.
[^] denotes match anything except the characters present in the character class.
In your first regex
\S*\s+//\s{1}([^\.java]+)\.java
\S* matches nothing as there is space in starting
\s+ matches the space which is in starting
// matches // literally
\s{1} matches next space
You are using [^\.java] which says match anything except . or j or a or v or a which can be written as [^.jav].
So, the left string now to be tested is
Tester.java
(Un)luckily any character from Tester does not matches . or j or a or v until we encounter a .. So Tester is matched and then java is also matched.
In your second regex
\S*\s+//\s{1}([^\.scala]+)\.scala
\S* matches nothing as there is space in starting
\s+ matches the space which is in starting
// matches // literally
\s{1} matches next space
Now, you are using [^\.scala] which says that match anything except . or s or c or a or l or a which can be written as [^.scla].
You have now
Hello.scala
but (un)luckily Hello here contains l which is not allowed according to character class and the regex fails.
How to correct it?
I will modify only a bit of your regex
\S*\s+//\s{1}([^.]*)\.java
<-->
This says that match anything except .
You can also use \w here instead if [^.]
Regex Demo
\S*\s+//\s{1}([^.]*)\.scala
Regex Demo
There is no need of {1} in \s{1}. You can simply write it as \s and it will match exactly one space like
\S*\s+//\s([^.]*)\.java

Java Regex - Match line word that not inside string

For my project, I want to get class using regex but not inside string
for e.g
class fo{
void foo{
System.out.println("example writing of class");
System.out.println("class cls{");
}
}
so, I hope result like this :
class fo{
I have try create a pattern but not working, Here.
Pattern.compile("\\b(?!(\"))class\\s+\\w+\\{\\b(?!(\"))")

Try this regex:
(?:^|\\n)\\s*class.*
Explaining:
(?:^|\\n) # from start or new line
\\s* # as many as possible spaces
class # the 'class' text
.* # all characters till the end of line
Hope it helps.

Split on Regular Expression per Path

If I have this:
thisisgibberish 1234 /hello/world/
more gibberish 43/7 /good/timing/
just onemore 8888 /thanks/mate
what would the regular expression inside the Java String.split() method be to obtain the paths per line?
ie.
[0]: /hello/world/
[1]: /good/timing/
[2]: /thanks/mate
Doing
myString.split("\/[a-zA-Z]")
causes the splits to occur to every /h, /w, /g, /t, and /m.
How would I go about writing a regular expression to split it only once per line while only capturing the paths?
Thanks in advance.

Why split ? I think running a match here is better, try the following expression:
(?<=\s)(/[a-zA-Z/])+
Regex101 Demo

This uses split() :
String[] split = myString.split(myString.substring(0, myString.lastIndexOf(" ")));
OR
myString.split(myString.substring(0, myString.lastIndexOf(" ")))[1]; //works for current inputs

You must first remove the leading junk, then split on the intervening junk:
String[] paths = str.replaceAll("^.*? (?=/[a-zA-Z])", "")
.split("(?m)((?<=[a-zA-Z]/|[a-zA-Z])\\s|^).*? (?=/[a-zA-Z])");
One important point here is the use of (?m), which is a switch that turns on "dot matches newline", which is required to split across the newlines.
Here's some test code:
String str = "thisisgibberish 1234 /hello/world/\nmore gibberish 43/7 /good/timing/\njust onemore 8888 /thanks/mate";
String[] paths = str.replaceAll("^.*? (?=/[a-zA-Z])", "")
.split("(?m)((?<=[a-zA-Z]/|[a-zA-Z])\\s|^).*? (?=/[a-zA-Z])");
System.out.println( Arrays.toString( paths));
Output (achieving requirements):
[/hello/world/, /good/timing/, /thanks/mate]

Youtube complete Java Regex

I need to parse several pages to get all of their Youtube IDs.
I found many regular expressions on the web, but : the Java ones are not complete (they either give me garbage in addition to the IDs, or they miss some IDs).
The one that I found that seems to be complete is hosted here. But it is written in JavaScript and PHP. Unfortunately I couldn't translate them into JAVA.
Can somebody help me rewrite this PHP regex or the following JavaScript one in Java?
'~
https?:// # Required scheme. Either http or https.
(?:[0-9A-Z-]+\.)? # Optional subdomain.
(?: # Group host alternatives.
youtu\.be/ # Either youtu.be,
| youtube\.com # or youtube.com followed by
\S* # Allow anything up to VIDEO_ID,
[^\w\-\s] # but char before ID is non-ID char.
) # End host alternatives.
([\w\-]{11}) # $1: VIDEO_ID is exactly 11 chars.
(?=[^\w\-]|$) # Assert next char is non-ID or EOS.
(?! # Assert URL is not pre-linked.
[?=&+%\w]* # Allow URL (query) remainder.
(?: # Group pre-linked alternatives.
[\'"][^<>]*> # Either inside a start tag,
| </a> # or inside <a> element text contents.
) # End recognized pre-linked alts.
) # End negative lookahead assertion.
[?=&+%\w]* # Consume any URL (query) remainder.
~ix'
/https?:\/\/(?:[0-9A-Z-]+\.)?(?:youtu\.be\/|youtube\.com\S*[^\w\-\s])([\w\-]{11})(?=[^\w\-]|$)(?![?=&+%\w]*(?:['"][^<>]*>|<\/a>))[?=&+%\w]*/ig;

First of all you need to insert and extra backslash \ foreach backslash in the old regex, else java thinks you escapes some other special characters in the string, which you are not doing.
https?:\\/\\/(?:[0-9A-Z-]+\\.)?(?:youtu\\.be\\/|youtube\\.com\\S*[^\\w\\-\\s])([\\w\\-]{11})(?=[^\\w\\-]|$)(?![?=&+%\\w]*(?:['\"][^<>]*>|<\\/a>))[?=&+%\\w]*
Next when you compile your pattern you need to add the CASE_INSENSITIVE flag. Here's an example:
String pattern = "https?:\\/\\/(?:[0-9A-Z-]+\\.)?(?:youtu\\.be\\/|youtube\\.com\\S*[^\\w\\-\\s])([\\w\\-]{11})(?=[^\\w\\-]|$)(?![?=&+%\\w]*(?:['\"][^<>]*>|<\\/a>))[?=&+%\\w]*";
Pattern compiledPattern = Pattern.compile(pattern, Pattern.CASE_INSENSITIVE);
Matcher matcher = compiledPattern.matcher(link);
while(matcher.find()) {
System.out.println(matcher.group());
}

Marcus above has a good regex, but i found that it doesn't recognize youtube links that have "www" but not "http(s)" in them
for example www.youtube....
i have an update:
^(?:https?:\\/\\/)?(?:[0-9A-Z-]+\\.)?(?:youtu\\.be\\/|youtube\\.com\\S*[^\\w\\-\\s])([\\w\\-]{11})(?=[^\\w\\-]|$)(?![?=&+%\\w]*(?:['\"][^<>]*>|<\\/a>))[?=&+%\\w]*
it's the same except for the start