Develop Reference

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
...
}

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

Getting more characters inside matcher group than expected

I'm trying to match the following string against the pattern:
String s = "AAA|VY~1055~ ~~BCN~09/24/2012~";
Matcher m = Pattern.compile("(.*)\\|VY\\~(.*)\\~").matcher(s);
if (m.find())
{
String value = m.group(2);
System.out.print("value = " + value);
}
The output is:
value = 1055~ ~~BCN~09/24/2012
But I want this:
value = 1055
Why is it getting all the characters until the end of string?
I've read something about to consume to the end of string, and I've tried:
Matcher m = Pattern.compile("(.*)\\|VY\\~(.*)\\~(.*)").matcher(s);
Matcher m = Pattern.compile("(.*)\\|VY\\~(.*)\\~.*").matcher(s);
But it doesn't work.
Can anybody help me?

Use the *? (reluctant) quantifier, which is lazy (stops matching as soon as possible).
Matcher m = Pattern.compile("(.*)\\|VY\\~(.*?)\\~").matcher(s);

You want to read about the gready, reluctant and possessive quantifiers (need to scroll down a bit).

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
}