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Regex to grab validate email address in complete XML string or normal string

Need to grab string text of email value in big XML/normal string.
Been working with Regex for it and as of now below Regex is working correctly for normal String
Regex : ^[\\w!#$%&'*+/=?`{|}~^-]+(?:\\.[\\w!#$%&'*+/=?`{|}~^-]+)*#(?:[a-zA-Z0-9-]+\\.)+[a-zA-Z]{1,6}$
Text : paris#france.c
but in case when above text is enclosed in XML tag it fails to return.
<email>paris#france.c</email>
I am trying to amend some change to this regex so that it will work for both of the scenarios

You have put ^ at the beginning which means the "Start of the string", and $ at the end which means the "End of the string". Now, look at your string:
<email>paris#france.c</email>
Do you think, it starts and ends with an email address?
I have removed them and also escaped the - in your regex. Here you can check the following auto-generated Java code with the updated regex.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Example {
public static void main(String[] args) {
final String regex = "[\\w!#$%&'*+/=?`\\{|\\}~^\\-]+(?:\\\\.[\\w!#$%&'*+/=?`\\{|\\}~^\\-]+)*#(?:[a-zA-Z0-9-]+\\.)+[a-zA-Z]{1,6}";
final String string = "paris#france.c\n"
+ "<email>paris#france.c</email>";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}
}
}
Output:
Full match: paris#france.c
Full match: paris#france.c

Problems defining the pattern to extract multiple dates from a json string in java

I have the following code:
public static void main(String[] args) {
String str = "{\"$and\":[{\\\"$or\\\":[{\\\"origen\\\":{\\\"$eq\\\":\\\"LEMD\\\"}},{\\\"origen\\\":{\\\"$eq\\\":\\\"LEBL\\\"}}]},{\"horasacta\":{\"$gte\":\"28/02/2015 00:00:00\"}},{\"horasacta\":{\"$lte\":\"28/02/2015 23:59:59\"}}]}";
Pattern pattern = Pattern.compile("\\{\"(.*?)\":\\{\"\\$(.*?)\":\"[0-9]+/[0-9]+/[0-9]+ [0-9]+:[0-9]+:[0-9]+\"}}");
Matcher matcher = pattern.matcher(str);
while (matcher.find()) {
System.out.println(matcher.group(0));
}
I want to get the substrings:
{\"departure\":{\"$gte\":\"28/02/2015 00:00:00\"}}
{\"departure\":{\"$lte\":\"28/02/2015 23:59:59\"}}
but the program give me:
{"$and":[{\"$or\":[{\"origin\":{\"$eq\":\"LEMD\"}},{\"origin\":{\"$eq\":\"LEBL\"}}]},{"departure":{"$gte":"28/02/2015 00:00:00"}}
{"departure":{"$lte":"28/02/2015 23:59:59"}}
the 2nd time the find() matches the pattern but the 1st time it doesn't do the job.
Any help?
thanks

It's frowned upon, yet if you have to, I'm guessing that you may be trying to write an expression looking somewhat like:
{\\"([^\\]+)\\":{\\"\$([^\\]+)\\":\\"[0-9]+\/[0-9]+\/[0-9]+\s+[0-9]+:[0-9]+:[0-9]+\\"}}
not so sure though.
The expression is explained on the top right panel of regex101.com, if you wish to explore/simplify/modify it, and in this link, you can watch how it would match against some sample inputs, if you like.
Test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "\\{\\\\\"([^\\\\]+)\\\\\":\\{\\\\\"\\$([^\\\\]+)\\\\\":\\\\\"[0-9]+\\/[0-9]+\\/[0-9]+\\s+[0-9]+:[0-9]+:[0-9]+\\\\\"\\}}";
final String string = "{\\\"departure\\\":{\\\"$gte\\\":\\\"28/02/2015 00:00:00\\\"}}\n"
+ "{\\\"departure\\\":{\\\"$lte\\\":\\\"28/02/2015 23:59:59\\\"}}";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}

JAVA regex to find string

i have a string like this:
font-size:36pt;color:#ffffff;background-color:#ff0000;font-family:Times New Roman;
How can I get the value of the color and the value of background-color?
color:#ffffff;
background-color:#ff0000;
i have tried the following code but the result is not my expected.
Pattern pattern = Pattern.compile("^.*(color:|background-color:).*;$");
The result will display:
font-size:36pt; color:#ffffff; background-color:#ff0000; font-family:Times New Roman;

If you want to have multiple matches in a string, don't assert ^ and $ because if those matches, then the whole string matches, which means that you can't match it again.
Also, use a lazy quantifier like *?. This will stop matching as soon as it finds some string that matches the pattern after it.
This is the regex you should use:
(color:|background-color:)(.*?);
Group 1 is either color: or background-color:, group 2 is the color code.
Demo

To do this you should use the (?!abc) expression in regex. This finds a match but doesn't select it. After that you can simply select the hexcode, like this:
String s = "font-size:36pt;color:#ffffff;background-color:#ff0000;font-family:Times New Roman";
Pattern pattern = Pattern.compile("(?!color:)#.{6}");
Matcher matcher = pattern.matcher(s);
while (matcher.find()) {
System.out.println(matcher.group());
}

Pattern pattern = Pattern.compile("color\\s*:\\s*([^;]+)\\s*;\\s*background-color\\s*:\\s*([^;]+)\\s*;");
Matcher matcher = pattern.matcher("font-size:36pt; color:#ffffff; background-color:#ff0000; font-family:Times New Roman;");
if (matcher.find()) {
System.out.println("color:" + matcher.group(1));
System.out.println("background-color:" + matcher.group(2));
}

No need to describe the whole input, only the relevant part(s) that you're looking to extract.
The regex color:(#[\\w\\d]+); does the trick for me:
String input = "font-size:36pt;color:#ffffff;background-color:#ff0000;font-family:Times New Roman;";
String regex = "color:(#[\\w\\d]+);";
Matcher m = Pattern.compile(regex).matcher(input);
while (m.find()) {
System.out.println(m.group(1));
}
Notice that m.group(1) returns the matching group which is inside the parenthesis in the regex. So the regex actually matches the whole color:#ffffff; and color:#ff0000; parts, but the print only handles the number itself.

Use a CSS parser like ph-css
String input = "font-size:36pt; color:#ffffff; background-color:#ff0000; font-family:Times New Roman;";
final CSSDeclarationList cssPropertyList =
CSSReaderDeclarationList.readFromString(input, ECSSVersion.CSS30);
System.out.println(cssPropertyList.get(1).getProperty() + " , "
+ cssPropertyList.get(1).getExpressionAsCSSString());
System.out.println(cssPropertyList.get(2).getProperty() + " , "
+ cssPropertyList.get(2).getExpressionAsCSSString());
Prints:
color , #ffffff
background-color , #ff0000
Find more about ph-css on github

java regex take variable between two tag

I am very new in regex and need your help. I wanna take numbers and letters between two span.
<span>454.000 $</span>
I wanna take 454.000 $. There are 12 space before . Please help me.

This Should Work.
Regexp:
\s+<.+>(.+)<.+>
Input:
<span>454.000 $</span>
Output:
454.000 $
JAVA CODE:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "\\s+<.+>(.+)<.+>";
final String string = " <span>454.000 $</span>";
final Pattern pattern = Pattern.compile(regex);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}
See: https://regex101.com/r/2zg5Ws/1

Capturing group using pattern matching is something like below
String x = " <span>454.000 $</span> ";
Pattern p = Pattern.compile("<span>(.*?)</span>");
Matcher m = p.matcher(x);
if (m.find()) {
System.out.println(">> "+ m.group(1)); // output 454.000 $
}
But for such cases I always prefer to use the replaceAll() as it is shorter version of code:
String num = x.replaceAll(".*<span>(.*?)</span>.*", "$1");
// num has 454.000 $
For the replace it is actually capturing the group from the text and replacing the whole text with that group ($1). This solution depends upon how your input string is.

Replace different Regex-Matches with Match-based results in Java

One common usage for regex is the replacement of the matches with something that is based on the matches.
For example a commit-text with ticket numbers ABC-1234: some text (ABC-1234) has to be replaced with <ABC-1234>: some text (<ABC-1234>) (<> as example for some surroundings.)
This is very simple in Java
String message = "ABC-9913 - Bugfix: Some text. (ABC-9913)";
String finalMessage = message;
Matcher matcher = Pattern.compile("ABC-\\d+").matcher(message);
if (matcher.find()) {
String ticket = matcher.group();
finalMessage = finalMessage.replace(ticket, "<" + ticket + ">");
}
System.out.println(finalMessage);
results in<ABC-9913> - Bugfix: Some text. (<ABC-9913>).
But if there are different matches in the input String, this is different. I tried a slightly different code replacing if (matcher.find()) { with while (matcher.find()) {. The result is messed up with doubled replacements (<<ABC-9913>>).
How can I replace all matching values in an elegant way?

You can simply use replaceAll:
String input = "ABC-1234: some text (ABC-1234)";
System.out.println(input.replaceAll("ABC-\\d+", "<$0>"));
prints:
<ABC-1234>: some text (<ABC-1234>)
$0 is a reference to the matched string.
Java regex reference (see "Groups and capturing").

The problem is that the replace() method transforms the string over and over again.
A better way is to replace one match at a time. The matcher class has an appendReplacement-method for this.
String message = "ABC-9913, ABC-9915 - Bugfix: Some text. (ABC-9913,ABC-9915)";
Matcher matcher = Pattern.compile("ABC-\\d+").matcher(message);
StringBuffer sb = new StringBuffer();
while (matcher.find()) {
String ticket = matcher.group();
matcher.appendReplacement(sb, "<" + ticket + ">");
}
matcher.appendTail(sb);
System.out.println(sb);