Scala RegEx String extractors behaving inconsistently - java

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

Related

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\ .

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.

A custom tokenizer for Java

I am developing an application in which I need to process text files containing emails. I need all the tokens from the text and the following is the definition of token:
Alphanumeric
Case-sensitive (case to be preserved)
'!' and '$' are to be considered as constituent characters. Ex: FREE!!, $50 are tokens
'.' (dot) and ',' comma are to be considered as constituent characters if they occur between numbers. For ex:
192.168.1.1, $24,500
are tokens.
and so on..
Please suggest me some open-source tokenizers for Java which are easy to customize to suit my needs. Will simply using StringTokenizer and regex be enough? I have to perform stopping also and that's why I was looking for an open source tokenizer which will also perform some extra things like stopping, stemming.
A few comments up front:
From StringTokenizer javadoc:
StringTokenizer is a legacy class that is retained for compatibility
reasons although its use is discouraged in new code. It is recommended
that anyone seeking this functionality use the split method of String
or the java.util.regex package instead.
Always use Google first - the first result as of now is JTopas. I did not use it, but it looks it could work for this
As for regex, it really depends on your requirements. Given the above, this might work:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Mkt {
public static void main(String[] args) {
Pattern p = Pattern.compile("([$\\d.,]+)|([\\w\\d!$]+)");
String str = "--- FREE!! $50 192.168.1.1 $24,500";
System.out.println("input: " + str);
Matcher m = p.matcher(str);
while(m.find()) {
System.out.println("token: " + m.group());
}
}
}
Here's a sample run:
$ javac Mkt.java && java Mkt
input: --- FREE!! $50 192.168.1.1 $24,500
token: FREE!!
token: $50
token: 192.168.1.1
token: $24,500
Now, you might need to tweak the regex, for example:
You gave $24,500 as an example. Should this work for $24,500abc or $24,500EUR?
You mentioned 192.168.1.1 should be included. Should it also include 192,168.1,1 (given . and , are to be included)?
and I guess there are other things to consider.
Hope this helps to get you started.

Use java regex to find all strings that start with '#' and end with ' ' , and not include ' ' and '#'

I need to get all strings(not empty) starts with # and end with ' '(space) in String below:
String s = "#test1 #test2 #test3 #test4 ## #test5";
I hope I can get all "test1", "test2", "test3", "test4", "test5" strings.
How to do it with java regx? thanks a lot!
You can use the following regex
#\w+
\w is similar to [a-zA-Z\d_]
\w+ matches 1 to many characters which are from [a-zA-Z\d_]
The Java regex (?<=#)[^# ]+(?= ) should do the trick. According to Regex Planet's Java regex page that regex matches test1, test2, test3 and test4. (#test5 does not end with a space, so test5 is not matched.)
If you're OK with matching the leading #s and trailing s as well, you can get away with the simpler Java regex #[^# ]+.
Finally I solved it with code below:
Pattern pattern = Pattern.compile("#\\p{L}+");

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

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