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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: Need to extract a number from a string

I have a string containing a number. Something like "Incident #492 - The Title Description".
I need to extract the number from this string.
Tried
Pattern p = Pattern.compile("\\d+");
Matcher m = p.matcher(theString);
String substring =m.group();
By getting an error
java.lang.IllegalStateException: No match found
What am I doing wrong?
What is the correct expression?
I'm sorry for such a simple question, but I searched a lot and still not found how to do this (maybe because it's too late here...)

You are getting this exception because you need to call find() on the matcher before accessing groups:
Matcher m = p.matcher(theString);
while (m.find()) {
String substring =m.group();
System.out.println(substring);
}
Demo.

There are two things wrong here:
The pattern you're using is not the most ideal for your scenario, it's only checking if a string only contains numbers. Also, since it doesn't contain a group expression, a call to group() is equivalent to calling group(0), which returns the entire string.
You need to be certain that the matcher has a match before you go calling a group.
Let's start with the regex. Here's what it looks like now.
Debuggex Demo
That will only ever match a string that contains all numbers in it. What you care about is specifically the number in that string, so you want an expression that:
Doesn't care about what's in front of it
Doesn't care about what's after it
Only matches on one occurrence of numbers, and captures it in a group
To that, you'd use this expression:
.*?(\\d+).*
Debuggex Demo
The last part is to ensure that the matcher can find a match, and that it gets the correct group. That's accomplished by this:
if (m.matches()) {
String substring = m.group(1);
System.out.println(substring);
}
All together now:
Pattern p = Pattern.compile(".*?(\\d+).*");
final String theString = "Incident #492 - The Title Description";
Matcher m = p.matcher(theString);
if (m.matches()) {
String substring = m.group(1);
System.out.println(substring);
}

You need to invoke one of the Matcher methods, like find, matches or lookingAt to actually run the match.

regular expression text between two sign

I have a text and I want to replace variables in it with proper values and my variables located between two #. When I use [/(?m)#.*?#/] to get these texts it also returns texts before and after first and last #. how could I get texts only between these two # sign. thanks in advance.
I use String.split("") method in Java.
for example I want use on the following String:
this is #the best# possible way #t#o do result!!!
and I wanna get these two results:
the best
t

In Java you can use this regex to grab value between first and second #:
String repl = input.replaceFirst("(?m)^[^#]*#([^#]*)#.*$" "$1");
To grab value between first and last #:
String repl = input.replaceFirst("(?m)^[^#]*#(.*?)#[^#]*$" "$1");
To find multiple matches use Pattern, Matcher:
Pattern p = Pattern.compile("#([^#]*)#"):
Matcher m = p.matcher(p);
while (m.find()) {
System.out.prinln(m.group(1));
}
RegEx Demo

Split() is the wrong tool to use here, use the Matcher() method to do this instead.
String s = "this is #the best# possible way #t#o do result!!!";
Pattern p = Pattern.compile("#([^#]*)#");
Matcher m = p.matcher(s);
while (m.find()) {
System.out.println(m.group(1));
}
Output
the best
t

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

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));
...