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]
Related
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\ .
package com.j;
public class Program {
public static void main(String[] args) {
System.out.println(Puzzel.class.getName().replaceAll(".", "/")
+ ".class");
System.out.println(Program.class.getName());
}
}
in the above program i was expecting a output com/j/Program.class
But it is coming //////.class its y?
In the replacement, . is treated as a regular expression, where . means "any character" and here is replaced with / , so the output becomes
////////////.class
For the expected answer, change the expression to escape the .:
Name.class.getName().replaceAll("\\.", "/") + ".class");
Then the output will be what you expected:
com/j/Puzzel.class
Because . is a special char when it comes to regex. You should escape it with backslash.
replaceAll() takes a regular expression for the matcher. Your code says to replace every character (.) with a /. you need replaceAll("\\.") or maybe replaceAll("\\\\."). I can never remember how many escapes to use offhand.
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.
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}+");
I want to add a '\' character to every string in a list of strings... I m doing something like this but it adds 2 backslashes instead.
feedbackMsgs.add(behaviorName+"\\"+fbCode);
result is like: "abc\\def"
how to make sure a single backslash is added??
I've just run a program with the following -
String s = "test" + "\\" + "test2";
System.out.println(s);
And it prints out the following -
test\test2
Are you sure there is no \ in the behaviourName or fbCode variables?
Looks like either your behaviourName ends with a \ or fbCode starts with one.
Try to Log/print behaviorName fbCode and find it yourself !
System.out.println(behaviorName);
System.out.println(fbCode);