Develop Reference

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.

IndexOutOfBoundsException when using Matcher.find()

This Java program showing me IndexOutOfBoundsException when it tries to invoke group(1). If I replace 1 with 0 then the whole line is printed.. What do I have to do?
Pattern pattern = Pattern.compile("<abhi> abhinesh </abhi>");
Matcher matcher = pattern.matcher("<abhi> abhinesh </abhi>");
if (matcher.find())
System.out.println(matcher.group(1));
else
System.out.println("Not found");

index starts at 0 so use matcher.group(0)
Edit : To match the text between tag use this regex <abhi>(.*)<\\/abhi>

This post may shed more light on your question.
Confused about Matcher Group.
In short you haven't defined any regular expression grouping to reference an alternate group. You only have the full matching string.
Below if you try adding a grouped regular expression to parse the xml you'll notice 0 has the full string, 1 has the begin tag, 2 has the value, and 3 has the end tag.
Pattern pattern = Pattern.compile("<([a-z]+)>([a-z ]+)</([a-z]+)>");
Matcher matcher = pattern.matcher("<abhi> abhinesh </abhi>");
if (matcher.find()){
System.out.println(matcher.group(0));//<abhi> abhinesh </abhi>
System.out.println(matcher.group(1));//abhi
System.out.println(matcher.group(2));// abhinesh
System.out.println(matcher.group(3));//abhi
}else{
System.out.println("Not found");
}

Try this this regex:
<abhi>(.*)<\\/abhi>
The text you're after will be stored in the first capture group.
Example:
String regex = "<abhi>(.*)<\\/abhi>";
String input = "<abhi>foo</abhi>";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(input);
if (m.find()) {
System.out.println(m.group(1));
}

Pattern Matching with dynamic matcher

Sample input string : Customer ${/xml:Name} has Ordered Product ${/xml:product} of ${/xml:unit} units.
i able to find get strings that match ${ ...... } using "\\$\\{.*?\\}"
I resolve the value for string from xml and now i have to replace the value back in input string.
i am using this method,
Pattern MY_PATTERN = Pattern.compile("\\$\\{.*?\\}");
Matcher m = MY_PATTERN.matcher(inputstring);
while (m.find()) {
String s = m.group(0); // s is ${/xml:Name}
// escaping wild characters
s = s.replaceAll("${", "\\$\\{"); // s is \$\{/xml:Name}
s = s.replaceAll("}", "\\}"); // s is \$\{/xml:Name\}
Pattern inner_pattern = Pattern.compile(s);
Matcher m1 = inner_pattern.matcher(inputstring);
name = m1.replaceAll(xPathValues.get(s));
}
but i get error at s = s.replaceAll("${", "\\$\\{"); i get Pattern Syntax Exception

You must escape the { too, try $\\{

Instead of:
s = s.replaceAll("${", "\\$\\{"); // s is \$\{/xml:Name}
s = s.replaceAll("}", "\\}"); // s is \$\{/xml:Name\}
You can use it without regex method String#replace(string):
s = s.replace("${", "\\$\\{").replace("}", "\\}"); // s is \$\{/xml:Name\}

It's because you could have a regexp likea{1,4} means to match a,aa,aaa,aaaa so a times 1 to 4, java tries to interpret your regexp like this, therefore try escaping the {

Yes, you must escape the {, but I would rather capture what's inside the braces:
Pattern MY_PATTERN = Pattern.compile("\\$\\{/xml:(.*?)\\}");
Matcher m = MY_PATTERN.matcher(inputstring);
while (m.find()) {
name = m.group(1); // s is Name
...
}

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

rules for use reg exp java

i have pattern:
host=([a-z0-9./:]*)
it's find for me host address. And i have content
host=http//:sdf3452.domain.com/
And my code is:
Matcher m;
Pattern hostP = Pattern.compile("host=([a-z0-9./:]*)");
m=hostP.matcher(content);//string 1
String match = m.group();//string 2
Log.i("host", ""+hostP.matcher(content).find());
if i delete string 1 and 2 i see true in logcat. If left as is I got exception nothing found.
I've tried all kinds of pattern. Through debug looked m variable, finds no match. Please teach me use reg exp!

Before you group() a match, you need to invoke find().
Try it like this:
Pattern hostP = Pattern.compile("host=([a-z0-9./:]*)");
Matcher m = hostP.matcher(content);
if(m.find()) {
String match = m.group();
// ...
}
EDIT
and a little demo that shows what each match-group contains:
Pattern p = Pattern.compile("host=([a-z0-9./:]*)");
Matcher m = p.matcher("host=http://sdf3452.domain.com/");
if (m.find()) {
for(int i = 0; i <= m.groupCount(); i++) {
System.out.printf("m.group(%d) = '%s'\n", i, m.group(i));
}
}
which will print:
m.group(0) = 'host=http://sdf3452.domain.com/'
m.group(1) = 'http://sdf3452.domain.com/'
As you can see, group(0), which is the same as group(), contains what the entire pattern matches.
But realize that a URL can contain much more than what your defined in [a-z0-9./:]*!

String content = "host=http://sdf3452.domain.com/";
Matcher mm;
Pattern hostP = Pattern.compile("host=([a-z0-9./:]*)");
mm=hostP.matcher(content);
String match = "";
if (mm.find()){//use m.find() first
match = mm.group(1);//1 is order number of brackets
}