Assume that I am given a regex pattern, whole_pattern
Pattern p = Pattern.compile(whole_pattern);
Matcher m = p.matcher(line);
if (m.find()) {
String s1 = m.group(1);
String s2 = m.group(2);
}
Obviously, we can get matched string of each group. But can we get the pattern of each group in the whole_pattern string? For example if whole_pattern = (\\d+)(\\w+), then patterns of group 1 and group 2 are \\d+ and \\w+, respectively.
You can use regexp for regexp:
public void simpleTest() {
String whole_pattern = "(\\d+)(\\w+)";
System.out.println(patternGroups(whole_pattern));
}
private List<String> patternGroups(String patternString) {
List<String> result = new ArrayList<>();
String pattern = "\\(([^()]+)\\)";
Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher(patternString);
while (m.find()) {
result.add(m.group(1));
}
return result;
}
Output of simpleTest() call will be:
[\d+, \w+]
Split your pattern string by ) and remove the first character in each elementh of the resulting array.
I am trying to get particular string from the data below.It is too long am here with sharing sample data. From this I have to get the 'france24Id=7GHYUFGty6fdGFHyy56'
am not that much familier with regex.
How can I retreive the string 'france24Id=7GHYUFGty6fdGFHyy56' from above data?
I tried splitting the data using ',' but it is not an effective way.That's why I choose regex.
2016-07-29 12:08:46,260 s=xGuide, [xre-10-pipe#6da05f7a[,connection=WebSocketConnectionWrapper[/1.8.9]]] INFO c=c.ore., - onConnect event payload={minimumVersion='0', applicationName='shell', fetl='555', authenticationToken='6y777', sessionAuthToken='', sessionGUID='null', connectURL='http://section?ruleName=Default', partnerId='hp', nativeDimensions=null, appParams={heartbeatRequest=1, france24Id=7GHYUFGty6fdGFHyy56, service=false, networkBuffer={min=150, max=150, step=0}}, deviceCaps={platform=Mac, receiverType=Native, revisions={protocol=1, auth=1, video=1}, pixelDimensions=[1280, 720]}, forceSource=null, reconnect=false, currentCommandIndex=0, reconnectReason=7, authService=9}
You can get what you want with (france\d+Id)=([a-zA-Z0-9]+),. This will grab your string and dump the two parts of it into platform-appropriate capture group variables (for instance, in Perl, $1 and $2 respectively).
In Java, your code would look a little like this:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public String matchID(String data) {
Pattern r = new Pattern("(france\\d+Id)=([a-zA-Z0-9]+),");
Matcher m = r.matcher(data);
return m.group(2);
}
public static void main(String[] args) {
String str = "2016-07-29 12:08:46,260 s=xGuide, [xre-10-pipe#6da05f7a[,connection=WebSocketConnectionWrapper[/1.8.9]]] INFO c=c.ore., - onConnect event payload={minimumVersion='0', applicationName='shell', fetl='555', authenticationToken='6y777', sessionAuthToken='', sessionGUID='null', connectURL='http://section?ruleName=Default', partnerId='hp', nativeDimensions=null, appParams={heartbeatRequest=1, france24Id=7GHYUFGty6fdGFHyy56, service=false, networkBuffer={min=150, max=150, step=0}}, deviceCaps={platform=Mac, receiverType=Native, revisions={protocol=1, auth=1, video=1}, pixelDimensions=[1280, 720]}, forceSource=null, reconnect=false, currentCommandIndex=0, reconnectReason=7, authService=9}";
String regex = ".*(france24Id=[\\d|\\w]*),.*";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
if(matcher.matches()){
System.out.println(matcher.group(1));
}
}
You can use Pattern and Matcher classes in Java.
String data = "2016-07-29 12:08:46,260 s=xGuide, [xre-10-pipe#6da05f7a[,connection=WebSocketConnectionWrapper[/1.8.9]]] INFO c=c.ore., - onConnect event payload={minimumVersion='0', applicationName='shell', fetl='555', authenticationToken='6y777', sessionAuthToken='', sessionGUID='null', connectURL='http://section?ruleName=Default', partnerId='hp', nativeDimensions=null, appParams={heartbeatRequest=1, france24Id=7GHYUFGty6fdGFHyy56, service=false, networkBuffer={min=150, max=150, step=0}}, deviceCaps={platform=Mac, receiverType=Native, revisions={protocol=1, auth=1, video=1}, pixelDimensions=[1280, 720]}, forceSource=null, reconnect=false, currentCommandIndex=0, reconnectReason=7, authService=9}";
String regex1 = "france24Id=[a-zA-Z0-9]+"; //this matches france24Id=7GHYUFGty6fdGFHyy56
String regex2 = "(?<=france24Id=)[a-zA-Z0-9]+"; //this matches 7GHYUFGty6fdGFHyy56 or whatever after "france24Id=" and before ','
Pattern pattern1 = Pattern.compile(regex1);
Pattern pattern2 = Pattern.compile(regex2);
Matcher matcher1 = pattern1.matcher(data);
Matcher matcher2 = pattern2.matcher(data);
String result1, result2;
if(matcher1.find())
result1 = matcher1.group(); //if match is found, result1 should contain "france24Id=7GHYUFGty6fdGFHyy56"
if(matcher2.find())
result2 = matcher2.group(); //if match is found, result1 should contain "7GHYUFGty6fdGFHyy56"
You can also try this one:
String str = "france24Id=7GHYUFGty6fdGFHyy56";
Pattern pattern = Pattern.compile("(?<=france24Id=)([a-zA-Z0-9]+)");
Matcher matcher = pattern.matcher(str);
if (matcher.find()) {
System.out.println("ID = " + matcher.group());
}
And the result is:
ID = 7GHYUFGty6fdGFHyy56
Hi i want to increment the integer values in between the string.
for example the initial string is -- m1p1b1.
The below code is working correctly, But it has one problem.
When the string is m10p10b10 it gives the result m21p21b21 not m11p11b11.
Also the integer length between the string dynamic, So i cant do any static code.
Pattern digitPattern = Pattern.compile("(\\d)");
Matcher matcher = digitPattern.matcher("m1p1b1");
StringBuffer result = new StringBuffer();
while (matcher.find()) {
matcher.appendReplacement(result, String.valueOf(Integer.parseInt(matcher.group(1)) + 1));
}
matcher.appendTail(result);
System.out.println(result.toString());
Change \\d to \\d+ to match one or more digits:
Pattern digitPattern = Pattern.compile("\\d+");
Matcher matcher = digitPattern.matcher("m10p10b10");
StringBuffer result = new StringBuffer();
while (matcher.find()) {
matcher.appendReplacement(result, String.valueOf(Integer.parseInt(matcher.group(0)) + 1));
}
matcher.appendTail(result);
System.out.println(result.toString()); // => m11p11b11
See the IDEONE demo
Note you do not have to capture the whole pattern with (...), you can access the value using matcher.group(0).
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
I have a string of text like this:
This is a[WAIT] test.
What I want to do is search the string for a substring that starts with [ and ends with ]
Each one I find I want to add it to an ArrayList and replace substring in original string with a ^
Here is my regex:
String regex_script = "/^\\[\\]$/"; //Match a string which starts with the character [ ending in the character ]
Here is what I have so far:
StringBuffer sb = new StringBuffer();
Pattern p = Pattern.compile(regex_script); // Create a pattern to match
Matcher m = p.matcher(line); // Create a matcher with an input string
boolean result = m.find();
while(result) {
m.appendReplacement(sb, "^");
result = m.find();
}
m.appendTail(sb); // Add the last segment of input to the new String
how would I got about doing this? Thanks
you can do:
String regex_script = "\\[([^\\]]*)\\]";
String line = "This is a[WAIT] testThis is a[WAIT] test";
StringBuffer sb = new StringBuffer();
List<String> list = new ArrayList<String>(); //use to record
Pattern p = Pattern.compile(regex_script); // Create a pattern to match
Matcher m = p.matcher(line); // Create a matcher with an input string
while (m.find()) {
list.add(m.group(1));
m.appendReplacement(sb, "[^]");
}
m.appendTail(sb); // Add the last segment of input to the new String
System.out.println(sb.toString());
If you are searching a substring, don't use ^ and $. Those are for beginning and at end at a string (not a word) Try:
String regex_script = "/\[.*\]/";