We Keep Coding

Regular expression get the third element from a string

Hello Im having trouble getting the third element of a string (F604080)
<sourceDocumentId>AX02_APF604_F604080</sourceDocumentId>
I have tried with this regular expression and variations, but i can manage to get
F604080.
(?<=\w+_)\w+(?=\<)
(?<=\w+_\w+_)\w+(?=\<)
....
Any help will be appreciated.
Thanks.

You don't need look behind or look ahead, instead just use this simple regex,
.*_(\w+)
and capture group 1.
Java codes,
public static void main(String[] args) {
String s = "<sourceDocumentId>AX02_APF604_F604080</sourceDocumentId>";
Pattern p = Pattern.compile(".*_(\\w+)");
Matcher m = p.matcher(s);
if (m.find()) {
System.out.println(m.group(1));
} else {
System.out.println("Didn't match");
}
}
Prints this like you wanted.
F604080

Using regex you can use something like >\w+_\w+_(\w+)<\/
String str = "<sourceDocumentId>AX02_APF604_F604080</sourceDocumentId>";
String code = null;
Matcher m = Pattern.compile(">\\w+_\\w+_(\\w+)</").matcher(str);
if (m.find()) {
code = m.group(1);
}
Simply use substring() operation
String code = str.substring(str.lastIndexOf('_') + 1, str.lastIndexOf('<'));
If later you parse XML with more element, you may use something like Java DOM Parser XML, but here this is not the best option as you have only one element

Can you just parse the string using "_" as separator and take the 3rd element ?

Both of your regular expressions seems to be matching the given string.
Anyway you could be a little bit more specific with this one:
^(?:<\w+>)(?:\w+)_(?:\w+)_(\w+)(?:<\/\w+>)$
Be sure that the input is the string you think it is and no additional text is given after that.

replacing <img> by <img></img> in a string

I want to replace every <img> tag with closing <img></img> tags in a string. The string is actually an html document where the img tag are generated by me and always look like this :
<img src="some_source.jpg" style="some style attributes and values">
Src is user input so it can be anything.
I made a regex expression, not sure if correct because it's my first time using it but upon testing it was working. The problem is that I don't know how to keep the content of the src.
/<img\ssrc=".+?"\sstyle=".+?">/g
But I have difficulties replacing the tags in the string.
and all I got is this:
Pattern p = Pattern.compile("/<img\\ssrc=\".+?\"\\sstyle=\".+?\">/g");
Matcher m = p.matcher(str);
List<String> imgStrArr = new ArrayList<String>();
while (m.find()) {
imgStrArr.add(m.group(0));
}
Matcher m2 = p.matcher(str);

You can use the following regex to match:
(<img[^>]+>)
And replace with $1</img>
Code:
str = str.replaceAll("(<img[^>]+>)", "$1</img>");
Edit: Considering #MarcusMüller's advice you can do the following:
Regex: (<img[^>]+)>
Replace with $1/>
Code:
str = str.replaceAll("(<img[^>]+)>", "$1/>");

You don't have to use Pattern and Matcher classes, you can use the regular replace method like this:
str = str.replaceAll("(<img.*?>)", "$1</img>");
IdeOne working demo

java regular expression word without ending with dot

I need to print the simple bind variable names in the SQL query.
I need to print the words starting with : character But NOT ending with dot . character.
in this sample I need to print pOrg, pBusinessId but NOT the parameter.
The regular expression ="(:)(\\w+)^\\." is not working.
Could you help in correcting the regular expression.
Thanks
Peddi
public void testMethod(){
String regEx="(:)(\\w+)([^\\.])";
String input= "(origin_table like 'I%' or (origin_table like 'S%' and process_status =5))and header_id = NVL( :parameter.number1:NULL, header_id) and (orginization = :pOrg) and (businsess_unit = :pBusinessId";
Pattern pattern;
Matcher matcher;
pattern = Pattern.compile(regEx);
matcher = pattern.matcher(input);
String grp = null;
while(matcher.find()){
grp = matcher.group(2);
System.out.println(grp);
}
}

You can try with something like
String regEx = "(:)(\\w+)\\b(?![.])";
(:)(\\w+)\\b will make sure that you are matching only entire words starting with :
(?![.]) is look behind mechanism which makes sure that after found word there is no .
This regex will also allow :NULL so if there is some reason why it shouldn't be matched share it with us.
Anyway to exclude NULL from results you can use
String regEx = "(:)(\\w+)\\b(?![.])(?<!:NULL)";
To make regex case insensitive so NULL could also match null compile this pattern with Pattern.CASE_INSENSITIVE flag like
Pattern pattern = Pattern.compile(regEx,Pattern.CASE_INSENSITIVE);

Since it looks like you're using camelcase, you can actually simplify things a bit when it comes to excluding :NULL:
:([a-z][\\w]+)\\b(?!\\.)
And $1 will return your variable names.
Alternative that doesn't rely on negative lookahead:
:([a-z][\\w]+)\\b(?:[^\\.]|$)

You can try:
Pattern regex = Pattern.compile("^:.*?[^.]$");
Demo

Replace string with part of the matching regex

I have a long string. I want to replace all the matches with part of the matching regex (group).
For example:
String = "This is a great day, is it not? If there is something, THIS IS it. <b>is</b>".
I want to replace all the words "is" by, let's say, "<h1>is</h1>". The case should remain the same as original. So the final string I want is:
This <h1>is</h1> a great day, <h1>is</h1> it not? If there <h1>is</h1> something,
THIS <h1>IS</h1> it. <b><h1>is</h1></b>.
The regex I was trying:
Pattern pattern = Pattern.compile("[.>, ](is)[.<, ]", Pattern.CASE_INSENSITIVE);

The Matcher class is commonly used in conjunction with Pattern. Use the Matcher.replaceAll() method to replace all matches in the string
String str = "This is a great day...";
Pattern p = Pattern.compile("\\bis\\b", Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher(str);
String result = m.replaceAll("<h1>is</h1>");
Note: Using the \b regex command will match on a word boundary (like whitespace). This is helpful to use in order to ensure that only the word "is" is matched and not words that contain the letters "i" and "s" (like "island").

Like this:
str = str.replaceAll(yourRegex, "<h1>$1</h1>");
The $1 refers to the text captured by group #1 in your regex.

Michael's answer is better, but if you happen to specifically only want [.>, ] and [.<, ] as boundaries, you can do it like this:
String input = "This is a great day, is it not? If there is something, THIS IS it. <b>is</b>";
Pattern p = Pattern.compile("(?<=[.>, ])(is)(?=[.<, ])", Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher(input);
String result = m.replaceAll("<h1>$1</h1>");

yourStr.replaceAll("(?i)([.>, ])(is)([.<, ])","$1<h1>$2</h1>$3")
(?i)to indicate ignoring case; wrap everything your want to reuse with brackets, reuse them with $1 $2 and $3, concatenate them into what you want.

Simply use a backreference for that.
"This is a great day, is it not? If there is something, THIS IS it. <b>is</b>".replaceAll("[.>, ](is)[.<, ]", "<h1>$2</h1>"); should do.

It may be a late addition, but if anyone is looking for this like
Searching for 'thing' and also he needs 'Something' too to be taken as result,
Pattern p = Pattern.compile("([^ ])is([^ \.])");
String result = m.replaceAll("<\h1>$1is$2</h1>");
will result <\h1>Something</h1> too

Getting specific portion of string from a Matcher

I'd like to get a portion of a matched string coming from a Matcher, like this:
Pattern pat = Pattern.compile("a.*l.*z");
Matcher match = pat.matcher("abcdlmnoz"); // I'd want to get bcd AND mno
ArrayList<String> values = match.magic(); //here is where your magic happens =)
ArrayList<String> is only for this example, I could be happy to recieve either a List or individual String items. The best would be what.htaccess files and RewriteRule's do:
RewriteRule (.*)/path?(.*) $1/$2/modified-path/
Well, putting those (.*) into $arguments would be as cool as an ArrayList or accessing String separately. I've been looking for something at Java Matcher API, but I didn't happen to see anything useful inside.
Thanks in advance, guys.

You can capture groups in a regexp match using (_):
Pattern pat = Pattern.compile("a(.*)l(.*)z");
boolean b = match.matches(); // don't forget to attempt the match
Then use match.group(n) to get that portion of the capture. The groups are stored in the match object.
Capturing GroupsOracle

Look at the matcher's "group" method and peruse the doc you linked to for references to groups, which is what the parentheses in the regex do :)

...
String testStr = "abcdlmnoz";
String myRE = "a(.*)l(.*)z";
Pattern myRECompiled = Pattern.compile (myRE,
DOTALL);
Matcher myMatcher = myRECompiled.matcher (testStr);
myMatcher.find ();
System.out.println (myMatcher.group (1));
System.out.println (myMatcher.group (2));
...