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Java regex get exact token value

I've string like below , want to get the value of cn=ADMIN , but dont know how to get to using regex efficient way.
group:192.168.133.205:387/cn=ADMIN,cn=groups,dc=mi,dc=com,dc=usa

well ... like this?
package test;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexSample {
public static void main(String[] args) {
String str = "group:192.168.133.205:387/cn=ADMIN,cn=groups,dc=mi,dc=com,dc=usa";
Pattern pattern = Pattern.compile("^.*/(.*)$");
Matcher matcher = pattern.matcher(str);
if (matcher.matches()) {
String right = matcher.group(1);
String[] parts = right.split(",");
for (String part : parts) {
System.err.println("part: " + part);
}
}
}
}
Output is:
part: cn=ADMIN
part: cn=groups
part: dc=mi
part: dc=com
part: dc=usa

String bubba = "group:192.168.133.205:387/cn=ADMIN,cn=groups,dc=mi,dc=com,dc=usa";
String target = "cn=ADMIN";
for(String current: bubba.split("[/,]")){
if(current.equals(target)){
System.out.println("Got it");
}
}

Pattern for regex
cn=([a-zA-Z0-9]+?),
Your name will be in group 1 of matcher. You can extend character classes if you allow spaces etc.

Removing link from Text in Java?

I need to change somethign like this -> Hello, go here http://www.google.com for your ...
grab the link, and change it in a method i made, and replace it back into the string like this
-> Hello, go here http://www.yahoo.com for your...
Here is what i have so far:
if(Text.toLowerCase().contains("http://"))
{
// Do stuff
}
else if(Text.toLowerCase().contains("https://"))
{
// Do stuff
}
All i need to do is change the URL in the String to something different. The Url in the String will not always be http://www.google.com, so i can not just say replace("http://www.google.com","")

Use regex:
String oldUrl = text.replaceAll(".*(https?://)www((\\.\\w+)+).*", "www$2");
text = text.replaceAll("(https?://)www(\\.\\w+)+", "$1" + traslateUrl(oldUrl));
Note: code changed to meet extra requirements in comments below.

you can grab the link from the string using below code. I assumed the string will contain only .com domain
String input = "Hello, go here http://www.google.com";
Pattern pattern = Pattern.compile("http[s]{0,1}://www.[a-z-]*.com");
Matcher m = pattern.matcher(input);
while (m.find()) {
String str = m.group();
}

Have you tried something like:
s= s.replaceFirst("http:.+[ ]", new link);
This will find any word beginning with http up till the first white space and replace it with whatever you want
if you want to keep the link then you can do:
String oldURL;
if (s.contains("http")) {
String[] words = s.split(" ");
for (String word: words) {
if (word.contains("http")) {
oldURL = word;
break;
}
}
//then replace the url or whatever
}

You can try this
private String removeUrl(String commentstr)
{
String urlPattern = "((https?|ftp|gopher|telnet|file|Unsure|http):((//)|(\\\\))+[\\w\\d:##%/;$()~_?\\+-=\\\\\\.&]*)";
Pattern p = Pattern.compile(urlPattern,Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher(commentstr);
int i = 0;
while (m.find()) {
commentstr = commentstr.replaceAll(m.group(i),"").trim();
i++;
}
return commentstr;
}

Matcher can't match

I have the following code. I need to check the text for existing any of the words from some list of banned words. But even if this word exists in the text matcher doesn't see it. here is the code:
final ArrayList<String> regexps = config.getProperty(property);
for (String regexp: regexps){
Pattern pt = Pattern.compile("(" + regexp + ")", Pattern.CASE_INSENSITIVE);
Matcher mt = pt.matcher(plainText);
if (mt.find()){
result = result + "message can't be processed because it doesn't satisfy the rule " + property;
reason = false;
System.out.println("reason" + mt.group() + regexp);
}
}
What is wrong? This code can'f find regexp в[ыy][шs]лит[еe], which is regexp in the plainText = "Вышлите пожалуйста новый счет на оплату на Санг, пока согласовывали, уже
прошли его сроки. Лиценз...". I also tried another variants of the regexp but everything is useless

The trouble is elsewhere.
import java.util.regex.*;
public class HelloWorld {
public static void main(String []args) {
Pattern pt = Pattern.compile("(qwer)");
Matcher mt = pt.matcher("asdf qwer zxcv");
System.out.println(mt.find());
}
}
This prints out true. You may want to use word boundary as delimiter, though:
import java.util.regex.*;
public class HelloWorld {
public static void main(String []args) {
Pattern pt = Pattern.compile("\\bqwer\\b");
Matcher mt = pt.matcher("asdf qwer zxcv");
System.out.println(mt.find());
mt = pt.matcher("asdfqwer zxcv");
System.out.println(mt.find());
}
}
The parenthesis are useless unless you need to capture the keyword in a group. But you already have it to begin with.

Use ArrayList's built in functions indexOf(Object o) and contains(Object o) to check if a String exists anywhere in the Array and where.
e.g.
ArrayList<String> keywords = new ArrayList<String>();
keywords.add("hello");
System.out.println(keywords.contains("hello"));
System.out.println(keywords.indexOf("hello"));
outputs:
true
0

Try this to filter out messages which contain banned words using the following regex which uses OR operator.
private static void findBannedWords() {
final ArrayList<String> keywords = new ArrayList<String>();
keywords.add("f$%k");
keywords.add("s!#t");
keywords.add("a$s");
String input = "what the f$%k";
String bannedRegex = "";
for (String keyword: keywords){
bannedRegex = bannedRegex + ".*" + keyword + ".*" + "|";
}
Pattern pt = Pattern.compile(bannedRegex.substring(0, bannedRegex.length()-1));
Matcher mt = pt.matcher(input);
if (mt.matches()) {
System.out.println("message can't be processed because it doesn't satisfy the rule ");
}
}

How to replace a word within a square bracket based a certain condition

I've a tricky condition which does not seem to work. For a given string, "Hi [HandleKey], you have [Action]", and a map which contains, map<"HandleKey","Peter"> I want to replace the square bracket and the word within if the key is found in the map. In this case, the map does not contain the key Action. The string should return "Hi Peter, you have [Action]".
Here is the code that I'm working on:
private String messageFormatter(String tMessage, Map<String, String> messageMap)
{
String formattedMsg = null;
Set<String> keyset = messageMap.keySet();
Iterator<String> keySetItr = keyset.iterator();
String msgkey = null;
boolean isFormatted = false;
while (keySetItr.hasNext())
{
msgkey = keySetItr.next();
if(t.contains(msgkey))
{
if(!isFormatted)
{
formattedMsg = tMessage.replaceAll("\\[", "").replaceAll("\\]", "");
formattedMsg = formattedMsg.replaceAll(msgkey, messageMap.get(msgkey));
isFormatted= true;
}else
{
formattedMsg = formattedMsg.replaceAll(msgkey, messageMap.get(msgkey));;
}
}else
{
formattedMsg=tMessage;
}
}
return formattedMsg;
}
The last else part is not right. Can anyone please help me with this. This code works fine for all the cases except when a matching key is not found in the map

is this idea ok for you?
instead of applying regex or extracting the stuff between [..], you could do some trick on your map side. e.g.
String s = "Hi [HandleKey], you have [Action]";
for(String k: yourMap.keySet()){
s=s.replaceAll("\\["+k+"\\]",yourMap.get(k));
}

You can do this with regex, here is a complete example code
public static void main(String[] args) {
String str = "Hi [HandleKey], you have [Action] ";
Hashtable<String, String> table = new Hashtable<String, String>();
table.put("HandleKey", "Peter");
Pattern pattern = Pattern.compile("\\[(\\w+)\\]");
Matcher matcher = pattern.matcher(str);
while (matcher.find()) {
String key = matcher.group(1);
if (table.containsKey(key)) {
str = str.replaceFirst("\\[" + key + "\\]", table.get(key));
}
}
System.out.println(str);
}
Output:
Hi Peter, you have [Action]
Note that this is more efficient than looping over the Map if the map size is already large or growing.

To handle when key not in map with minimal changes to what you have above try
formattedMsg.replaceAll(msgkey,
(messageMap.containsKey(msgKey) ? messageMap.get(msgkey) : "[" + msgKey + "]"));
but looking again I can see that you're iterating the set of keys from the messageMap so the issue of a key not appearing in the map doesn't arise?
There's also a reference to if(t.contains(msgKey))... but not sure what t is
if you want the text to contain the formatted [msgKey] when its no found then replacing all "[" & "]" seems the wrong way to start if you want to put them back in in some cases.
I'd look at #iTech's suggestion and get regex doing more for you

Tokenize a string with a space in java

I want to tokenize a string like this
String line = "a=b c='123 456' d=777 e='uij yyy'";
I cannot split based like this
String [] words = line.split(" ");
Any idea how can I split so that I get tokens like
a=b
c='123 456'
d=777
e='uij yyy';

The simplest way to do this is by hand implementing a simple finite state machine. In other words, process the string a character at a time:
When you hit a space, break off a token;
When you hit a quote keep getting characters until you hit another quote.

Depending on the formatting of your original string, you should be able to use a regular expression as a parameter to the java "split" method: Click here for an example.
The example doesn't use the regular expression that you would need for this task though.
You can also use this SO thread as a guideline (although it's in PHP) which does something very close to what you need. Manipulating that slightly might do the trick (although having quotes be part of the output or not may cause some issues). Keep in mind that regex is very similar in most languages.
Edit: going too much further into this type of task may be ahead of the capabilities of regex, so you may need to create a simple parser.

line.split(" (?=[a-z+]=)")
correctly gives:
a=b
c='123 456'
d=777
e='uij yyy'
Make sure you adapt the [a-z+] part in case your keys structure changes.
Edit: this solution can fail miserably if there is a "=" character in the value part of the pair.

StreamTokenizer can help, although it is easiest to set up to break on '=', as it will always break at the start of a quoted string:
String s = "Ta=b c='123 456' d=777 e='uij yyy'";
StreamTokenizer st = new StreamTokenizer(new StringReader(s));
st.ordinaryChars('0', '9');
st.wordChars('0', '9');
while (st.nextToken() != StreamTokenizer.TT_EOF) {
switch (st.ttype) {
case StreamTokenizer.TT_NUMBER:
System.out.println(st.nval);
break;
case StreamTokenizer.TT_WORD:
System.out.println(st.sval);
break;
case '=':
System.out.println("=");
break;
default:
System.out.println(st.sval);
}
}
outputs
Ta
=
b
c
=
123 456
d
=
777
e
=
uij yyy
If you leave out the two lines that convert numeric characters to alpha, then you get d=777.0, which might be useful to you.

Assumptions:
Your variable name ('a' in the assignment 'a=b') can be of length 1 or more
Your variable name ('a' in the assignment 'a=b') can not contain the space character, anything else is fine.
Validation of your input is not required (input assumed to be in valid a=b format)
This works fine for me.
Input:
a=b abc='123 456' &=777 #='uij yyy' ABC='slk slk' 123sdkljhSDFjflsakd#*#&=456sldSLKD)#(
Output:
a=b
abc='123 456'
&=777
#='uij yyy'
ABC='slk slk'
123sdkljhSDFjflsakd#*#&=456sldSLKD)#(
Code:
import java.util.ArrayList;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexTest {
// SPACE CHARACTER followed by
// sequence of non-space characters of 1 or more followed by
// first occuring EQUALS CHARACTER
final static String regex = " [^ ]+?=";
// static pattern defined outside so that you don't have to compile it
// for each method call
static final Pattern p = Pattern.compile(regex);
public static List<String> tokenize(String input, Pattern p){
input = input.trim(); // this is important for "last token case"
// see end of method
Matcher m = p.matcher(input);
ArrayList<String> tokens = new ArrayList<String>();
int beginIndex=0;
while(m.find()){
int endIndex = m.start();
tokens.add(input.substring(beginIndex, endIndex));
beginIndex = endIndex+1;
}
// LAST TOKEN CASE
//add last token
tokens.add(input.substring(beginIndex));
return tokens;
}
private static void println(List<String> tokens) {
for(String token:tokens){
System.out.println(token);
}
}
public static void main(String args[]){
String test = "a=b " +
"abc='123 456' " +
"&=777 " +
"#='uij yyy' " +
"ABC='slk slk' " +
"123sdkljhSDFjflsakd#*#&=456sldSLKD)#(";
List<String> tokens = RegexTest.tokenize(test, p);
println(tokens);
}
}

Or, with a regex for tokenizing, and a little state machine that just adds the key/val to a map:
String line = "a = b c='123 456' d=777 e = 'uij yyy'";
Map<String,String> keyval = new HashMap<String,String>();
String state = "key";
Matcher m = Pattern.compile("(=|'[^']*?'|[^\\s=]+)").matcher(line);
String key = null;
while (m.find()) {
String found = m.group();
if (state.equals("key")) {
if (found.equals("=") || found.startsWith("'"))
{ System.err.println ("ERROR"); }
else { key = found; state = "equals"; }
} else if (state.equals("equals")) {
if (! found.equals("=")) { System.err.println ("ERROR"); }
else { state = "value"; }
} else if (state.equals("value")) {
if (key == null) { System.err.println ("ERROR"); }
else {
if (found.startsWith("'"))
found = found.substring(1,found.length()-1);
keyval.put (key, found);
key = null;
state = "key";
}
}
}
if (! state.equals("key")) { System.err.println ("ERROR"); }
System.out.println ("map: " + keyval);
prints out
map: {d=777, e=uij yyy, c=123 456, a=b}
It does some basic error checking, and takes the quotes off the values.

This solution is both general and compact (it is effectively the regex version of cletus' answer):
String line = "a=b c='123 456' d=777 e='uij yyy'";
Matcher m = Pattern.compile("('[^']*?'|\\S)+").matcher(line);
while (m.find()) {
System.out.println(m.group()); // or whatever you want to do
}
In other words, find all runs of characters that are combinations of quoted strings or non-space characters; nested quotes are not supported (there is no escape character).

public static void main(String[] args) {
String token;
String value="";
HashMap<String, String> attributes = new HashMap<String, String>();
String line = "a=b c='123 456' d=777 e='uij yyy'";
StringTokenizer tokenizer = new StringTokenizer(line," ");
while(tokenizer.hasMoreTokens()){
token = tokenizer.nextToken();
value = token.contains("'") ? value + " " + token : token ;
if(!value.contains("'") || value.endsWith("'")) {
//Split the strings and get variables into hashmap
attributes.put(value.split("=")[0].trim(),value.split("=")[1]);
value ="";
}
}
System.out.println(attributes);
}
output:
{d=777, a=b, e='uij yyy', c='123 456'}
In this case continuous space will be truncated to single space in the value.
here attributed hashmap contains the values

import java.io.*;
import java.util.Scanner;
public class ScanXan {
public static void main(String[] args) throws IOException {
Scanner s = null;
try {
s = new Scanner(new BufferedReader(new FileReader("<file name>")));
while (s.hasNext()) {
System.out.println(s.next());
<write for output file>
}
} finally {
if (s != null) {
s.close();
}
}
}
}

java.util.StringTokenizer tokenizer = new java.util.StringTokenizer(line, " ");
while (tokenizer.hasMoreTokens()) {
String token = tokenizer.nextToken();
int index = token.indexOf('=');
String key = token.substring(0, index);
String value = token.substring(index + 1);
}

Have you tried splitting by '=' and creating a token out of each pair of the resulting array?