This is related to: RegEx: Grabbing values between quotation marks.
If there is a String like this:
HYPERLINK "hyperlink_funda.docx" \l "Sales"
The regex given on the link
(["'])(?:(?=(\\?))\2.)*?\1
is giving me
[" HYPERLINK ", " \l ", " "]
What regex will return values enclosed in quotation mark (specifically between the \" marks) ?
["hyperlink_funda.docx", "Sales"]
Using Java, String.split(String regex) way.
You're not supposed to use that with .split() method. Instead use a Pattern with capturing groups:
{
Pattern pattern = Pattern.compile("([\"'])((?:(?=(\\\\?))\\3.)*?)\\1");
Matcher matcher = pattern.matcher(" HYPERLINK \"hyperlink_funda.docx\" \\l \"Sales\" ");
while (matcher.find())
System.out.println(matcher.group(2));
}
Output:
hyperlink_funda.docx
Sales
Here is a regex demo, and here is an online code demo.
I think you are misunderstanding the nature of the String.split method. Its job is to find a way of splitting a string by matching the features of the separator, not by matching features of the strings you want returned.
Instead you should use a Pattern and a Matcher:
String txt = " HYPERLINK \"hyperlink_funda.docx\" \\l \"Sales\" ";
String re = "\"([^\"]*)\"";
Pattern p = Pattern.compile(re);
Matcher m = p.matcher(txt);
ArrayList<String> matches = new ArrayList<String>();
while (m.find()) {
String match = m.group(1);
matches.add(match);
}
System.out.println(matches);
Related
so I have this string like
"#tag1 #tag2 #tag3 not_tag1 not_tag2 #tag4" (the space between tag2 and tag4 is to indicate there can be many spaces). From this string I want to parse just a tag1, tag2 and so on. They are similar to #tags we see on LinkedIn or any other social media. Is there any easy way to do this using regex or any other function in Java. Or should I do it hard way(i.e. using loops and conditions).
Tag format should be "#" (to indicate tag is starting) and space " "(to indicate end of tag). In between there can be character or numbers but start should be a character only.
example,
input : "#tag1 #tag2 #tag3 not_tag1 not_tag2 #12tag #tag4"
output : ["tag1", "tag2", "tag3", "tag4"]
split by regex: "#\w+"
EDIT: this is the correct regex, but split is not the right method.
same solution as javadev suggested, but use instead:
String input = "#tag1 #tag2 #tag3 not_tag1 not_tag2 #12tag #tag4";
Matcher matcher = Pattern.compile("#\\w+").matcher(input);
while (matcher.find()) {
System.out.println(matcher.group(0));
}
output with # as expected.
Maybe something like:
public static void main(String[] args ) {
String input = "#tag1 #tag2 #tag3 not_tag1 not_tag2 #12tag #tag4";
Pattern pattern = Pattern.compile("#([A-z][A-z0-9]*) *");
Matcher matcher = pattern.matcher(input);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
}
worked for me :)
Output:
tag1
tag2
tag3
tag4
I have one input String like this:
"I am Duc/N Ta/N Van/N"
String "/N" present it is the Name of one person.
The expected output is:
Name: Duc Ta Van
How can I do it by using regular expression?
You can use Pattern and Matcher like this :
String input = "I am Duc/N Ta/N Van/N";
Pattern pattern = Pattern.compile("([^\\s]+)/N");
Matcher matcher = pattern.matcher(input);
String result = "";
while (matcher.find()) {
result+= matcher.group(1) + " ";
}
System.out.println("Name: " + result.trim());
Output
Name: Duc Ta Van
Another Solution using Java 9+
From Java9+ you can use Matcher::results like this :
String input = "I am Duc/N Ta/N Van/N";
String regex = "([^\\s]+)/N";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(input);
String result = matcher.results().map(s -> s.group(1)).collect(Collectors.joining(" "));
System.out.println("Name: " + result); // Name: Duc Ta Van
Here is the regex to use to capture every "name" preceded by a /N
(\w+)\/N
Validate with Regex101
Now, you just need to loop on every match in that String and concatenate the to get the result :
String pattern = "(\\w+)\\/N";
String test = "I am Duc/N Ta/N Van/N";
Matcher m = Pattern.compile(pattern).matcher(test);
StringBuilder sbNames = new StringBuilder();
while(m.find()){
sbNames.append(m.group(1)).append(" ");
}
System.out.println(sbNames.toString());
Duc Ta Van
It is giving you the hardest part. I let you adapt this to match your need.
Note :
In java, it is not required to escape a forward slash, but to use the same regex in the entire answer, I will keep "(\\w+)\\/N", but "(\\w+)/N" will work as well.
I've used "[/N]+" as the regular expression.
Regex101
[] = Matches characters inside the set
\/ = Matches the character / literally (case sensitive)
+ = Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
I have a string email = John.Mcgee.r2d2#hitachi.com
How can I write a java code using regex to bring just the r2d2?
I used this but got an error on eclipse
String email = John.Mcgee.r2d2#hitachi.com
Pattern pattern = Pattern.compile(".(.*)\#");
Matcher matcher = patter.matcher
for (Strimatcher.find()){
System.out.println(matcher.group(1));
}
To match after the last dot in a potential sequence of multiple dots request that the sequence that you capture does not contain a dot:
(?<=[.])([^.]*)(?=#)
(?<=[.]) means "preceded by a single dot"
(?=#) means "followed by # sign"
Note that since dot . is a metacharacter, it needs to be escaped either with \ (doubled for Java string literal) or with square brackets around it.
Demo.
Not sure if your posting the right code. I'll rewrite it based on what it should look like though:
String email = John.Mcgee.r2d2#hitachi.com
Pattern pattern = Pattern.compile(".(.*)\#");
Matcher matcher = pattern.matcher(email);
int count = 0;
while(matcher.find()) {
count++;
System.out.println(matcher.group(count));
}
but I think you just want something like this:
String email = John.Mcgee.r2d2#hitachi.com
Pattern pattern = Pattern.compile(".(.*)\#");
Matcher matcher = pattern.matcher(email);
if(matcher.find()){
System.out.println(matcher.group(1));
}
No need to Pattern you just need replaceAll with this regex .*\.([^\.]+)#.* which mean get the group ([^\.]+) (match one or more character except a dot) which is between dot \. and #
email = email.replaceAll(".*\\.([^\\.]+)#.*", "$1");
Output
r2d2
regex demo
If you want to go with Pattern then you have to use this regex \\.([^\\.]+)# :
String email = "John.Mcgee.r2d2#hitachi.com";
Pattern pattern = Pattern.compile("\\.([^\\.]+)#");
Matcher matcher = pattern.matcher(email);
if (matcher.find()) {
System.out.println(matcher.group(1));// Output : r2d2
}
Another solution you can use split :
String[] split = email.replaceAll("#.*", "").split("\\.");
email = split[split.length - 1];// Output : r2d2
Note :
Strings in java should be between double quotes "John.Mcgee.r2d2#hitachi.com"
You don't need to escape # in Java, but you have to escape the dot with double slash \\.
There are no syntax for a for loop like you do for (Strimatcher.find()){, maybe you mean while
I want to match every file name which ends with .js and is stored in a directory called lib.
Therefore I created the following regular expression: (lib/)(.*?).js$.
I tested the expression (lib/)(.*?).js$ in a Regex Tester and matched this filename: src/main/lib/abc/DocumentHandler.js.
To use my expression in Java, I escaped it to: (lib/)(.*?)\\.js$.
Nevertheless, Java tells me that my expression does not match.
Here is my code:
String regEx = "(lib/)(.*?).js$";
String escapedRegEx = "(lib/)(.*?)\\.js$";
Pattern pattern = Pattern.compile(escapedRegEx);
Matcher matcher = pattern.matcher("src/main/lib/abc/DocumentHandler.js");
System.out.println("Matches: " + matcher.matches()); // false :-(
Did I forgot to escape something?
Use Matcher.find() instead of Matcher.matches() to check for subset of any string.
As per Java Doc:
Matcher#matches()
Attempts to match the entire region against the pattern.
Matcher#find()
Attempts to find the next subsequence of the input sequence that matches the pattern.
sample code:
String regEx = "(lib/)(.*)\\.js$";
String str = "src/main/lib/abc/DocumentHandler.js";
Pattern pattern = Pattern.compile(regEx);
Matcher matcher = pattern.matcher(str);
if (matcher.find()) { // <== returns true if found
System.out.println("Matches: " + matcher.group());
System.out.println("Path: " + matcher.group(2));
}
output:
Matches: lib/abc/DocumentHandler.js
Path: abc/DocumentHandler
Use Matcher#group(index) to get the matched group that is grouped by enclosing inside parenthesis (...) in the regex pattern.
You can use String#matches() method to match the whole string.
String regEx = "(.*)(/lib/)(.*?)\\.js$";
String str = "src/main/lib/abc/DocumentHandler.js";
System.out.println("Matched :" + str.matches(regEx)); // Matched : true
Note: Don't forget to escape dot . that has special meaning in regex pattern to match any thing other than new line.
Try this RegEx pattern
String regEx = "(.*)(lib\\/)(.*)(\\.js$)";
Pattern pattern = Pattern.compile(regEx);
Matcher matcher = pattern.matcher("src/main/lib/abc/DocumentHandler.js");
It's working for me:
Firstly you don't need to escape it, and secondly you are not matching the first part of the string.
String regEx = "(.*)(lib/)(.*?).js$";
Pattern pattern = Pattern.compile(regEx);
Matcher matcher = pattern.matcher("src/main/lib/abc/DocumentHandler.js");
I have a long string let's say
I like this #computer and I want to buy it from #XXXMall.
I know the regular expression pattern is
Pattern tagMatcher = Pattern.compile("[#]+[A-Za-z0-9-_]+\\b");
Now i want to get all the hashtags in an array. How can i use this expression to get array of all hash tags from string something like
ArrayList hashtags = getArray(pattern, str)
You can write like?
private static List<String> getArray(Pattern tagMatcher, String str) {
Matcher m = tagMatcher.matcher(str);
List<String> l = new ArrayList<String>();
while(m.find()) {
String s = m.group(); //will give you "#computer"
s = s.substring(1); // will give you just "computer"
l.add(s);
}
return l;
}
Also you can use \\w- instead of A-Za-z0-9-_ making the regex [#]+[\\w]+\\b
This link would surely be helpful for achieving what you want.
It says:
The find() method searches for occurrences of the regular expressions
in the text passed to the Pattern.matcher(text) method, when the
Matcher was created. If multiple matches can be found in the text, the
find() method will find the first, and then for each subsequent call
to find() it will move to the next match.
The methods start() and end() will give the indexes into the text
where the found match starts and ends.
Example:
String text =
"This is the text which is to be searched " +
"for occurrences of the word 'is'.";
String patternString = "is";
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(text);
int count = 0;
while(matcher.find()) {
count++;
System.out.println("found: " + count + " : "
+ matcher.start() + " - " + matcher.end());
}
You got the hint now.
Here is one way, using Matcher
Pattern tagMatcher = Pattern.compile("#+[-\\w]+\\b");
Matcher m = tagMatcher.matcher(stringToMatch);
ArrayList<String> hashtags = new ArrayList<>();
while (m.find()) {
hashtags.add(m.group());
}
I took the liberty of simplifying your regex. # does not need to be in a character class. [A-Za-z0-9_] is the same as \w, so [A-Za-z0-9-_] is the same as [-\w]
You can use :
String val="I like this #computer and I want to buy it from #XXXMall.";
String REGEX = "(?<=#)[A-Za-z0-9-_]+";
List<String> list = new ArrayList<String>();
Pattern pattern = Pattern.compile(REGEX);
Matcher matcher = pattern.matcher(val);
while(matcher.find()){
list.add(matcher.group());
}
(?<=#) Positive Lookbehind - Assert that the character # literally be matched.
you can use the following code for getting the names
String saa = "#{akka}nikhil#{kumar}aaaaa";
Pattern regex = Pattern.compile("#\\{(.*?)\\}");
Matcher m = regex.matcher(saa);
while(m.find()) {
String s = m.group(1);
System.out.println(s);
}
It will print
akka
kumar