Regexes for modifying beginning and end of line or string - java

Given this string 1990January20hello.abc I want to apply regexes to get the final string 1990January-20hello.abc
I thought I could do:
String text = "1990January20hello.abc";
Pattern p = Patter.compile("(.*)(January|Jan)(.*)", Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
Matcher m = p.matcher(text);
while (m.find()){
String val1 = m.group(2);
String val2 = val1.replace("$", "-");
text = text.replace(val1, val2);
}
When I do that seems like in the while loop it does find "January" but val2 and text stay January. what am I doing wrong? Its as if Java doesn't recognize $ for end of line/string. Ideally I want to say val1.replace("(^|$)","-") so I can get 1990-January-20hello.abc as final string. Please help. Thanks for suggestions in advance.


Use this for a pattern:
(\d+)([A-Za-z]+)(\d+)(.*)
The resulting groups and their values are:
Group 1 = 1990,
Group 2 = January,
Group 3 = 20,
Group 4 = the rest of the string.
And then you can append them together and add whatever you want.


String#replace does not use regular expressions so will not match start or end of line characters. You can use one of the replace methods that does use a regular expression such as replaceAll:
String text = "1990January20hello.abc";
Pattern p = Pattern.compile("(.*)(January|Jan)(.*)", Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
Matcher m = p.matcher(text);
while (m.find()) {
String val1 = m.group(2);
String val2 = val1.replaceAll("$|^", "-");
text = text.replace(val1, val2);
}
Output:
1990-January-20hello.abc


Try
String s = "1990January20hello.abc";
String s2 = s.replaceAll("(Jan|January)(\\d+)", "$1-$2");
System.out.println(s2);
Or using Matcher
String s3 = Pattern
.compile("(Jan|January)(\\d+)", Pattern.CASE_INSENSITIVE)
.matcher(s).replaceAll("$1-$2");
System.out.println(s3);


String does not recognize all regex patterns and thats why my replace was failing. Heres how I solved it:
String text = "1990January20hello.abc";
Pattern p = Patter.compile("(.*)(January|Jan)(.*)", Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
Matcher m = p.matcher(text);
while (m.find()){
String val1 = m.group(2); //"January"
Pattern p1 = Pattern.compile(^|$);
Matcher m1 = p1.matcher(val1);
String val2 = m1.replaceAll("-"); //"-January-"
text = text.replace(val1, val2); //"1990-January-20hello.abc"
}
Thank you for feedback!

Related

How to Split with particular condition in Java?

Suppose I have a string like
"resources/json/04-Dec/someName_SomeTeam.json"
In above string I want just "04-Dec" part, this may change to "12-Jan" like this or any date with month with that format. How do I do this?

You can split using / and get the value 2
String text = "resources/json/04-Dec/someName_SomeTeam.json";
String[] split = text.split("\\/");
String result = split[2];//04-Dec
Or you can use patterns with this regex \d{2}\-\[A-Z\]\[a-z\]{2}:
String text = "resources/json/04-Dec/someName_SomeTeam.json";
String regex = "\\d{2}\\-[A-Z][a-z]{2}";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(text);
if (matcher.find()) {
System.out.println(matcher.group());
}

Get a particular string from a data using regular expression

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

Matcher not finding matches

I'm trying to extract the numbers in the following string:
09/29/2014
I am currently using the code:
Pattern p = Pattern.compile("([0-9]{2})/([0-9]{2})/([0-9]{4})");
Matcher m = p.matcher(startDatepicker);
String startYear = m.group(3);
String startMonth = m.group(1);
String startDay = m.group(2);
startDatepicker contains: 09/29/2014
However, I am not receiving any matches.. I also tried escaping the forward slashes with \\ but that also didn't work.
Am I missing something?
Thanks for your help.

Before you could access the matched groups, you need to call find() on the matcher, and check that it has found a match:
Pattern p = Pattern.compile("([0-9]{2})/([0-9]{2})/([0-9]{4})");
Matcher m = p.matcher(startDatepicker);
if (!m.find()) {
return;
}
String startYear = m.group(3);
String startMonth = m.group(1);
String startDay = m.group(2);
The call of m.find() positions the matcher on the first match.
Demo.

You need to call find() to iterate through your match groups.
Pattern p = Pattern.compile("([0-9]{2})/([0-9]{2})/([0-9]{4})");
Matcher m = p.matcher(startDatepicker);
while (m.find()) {
...
}
The find() method searches for occurrences of the regex in the input passed to p.matcher(). If multiple matches can be found, this method will find the first, and then move to the next match for each subsequent call.

First and second tocen regex

How could I get the first and the second text in "" from the string?
I could do it with indexOf but this is really boring ((
For example I have a String for parse like: "aaa":"bbbbb"perhapsSomeOtherText
And I d like to get aaa and bbbbb with the help of Regex pattern - this will help me to use it in switch statement and will greatly simplify my app/

If all that you have is colon delimited string just split it:
String str = ...; // colon delimited
String[] parts = str.split(":");
Note, that split() receives regex and compilies it every time. To improve performance of your code you can use Pattern as following:
private static Pattern pColonSplitter = Pattern.compile(":");
// now somewhere in your code:
String[] parts = pColonSplitter.split(str);
If however you want to use pattern for matching and extraction of string fragments in more complicated cases, do it like following:
Pattert p = Patter.compile("(\\w+):(\\w+):");
Matcher m = p.matcher(str);
if (m.find()) {
String a = m.group(1);
String b = m.group(2);
}
Pay attention on brackets that define captured group.

Something like this?
Pattern pattern = Pattern.compile("\"([^\"]*)\"");
Matcher matcher = pattern.matcher("\"aaa\":\"bbbbb\"perhapsSomeOtherText");
while (matcher.find()) {
System.out.println(matcher.group(1));
}
Output
aaa
bbbbb

String str = "\"aaa\":\"bbbbb\"perhapsSomeOtherText";
Pattern p = Pattern.compile("\"\\w+\""); // word between ""
Matcher m = p.matcher(str);
while(m.find()){
System.out.println(m.group().replace("\"", ""));
}
output:
aaa
bbbbb

there are several ways to do this
Use StringTokenizer or Scanner with UseDelimiter method

Get an array of Strings matching a pattern from a String

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

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