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Merge two pattern into one

I need write a pattern to remove currency symbol and comma. eg Fr.-145,000.01
After the pattern matcher should return -145000.01.
The pattern i am using:
^[^0-9\\-]*([0-9\\-\\.\\,]*?)[^0-9\\-]*$
This will return -145,000.01
Then I remove the comma to get -145000.01, I want to ask if that's possible that I change the pattern and directly get -145000.01
String pattern = "^[^0-9\\-]*([0-9\\-\\.\\,]*?)[^0-9\\-]*$";
Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher(str);
if(m.matches()) {
System.out.println(m.group(1));
}
I expect the output could resolve the comma

You can simply it with String.replaceAll() and simpler regex (providing you are expecting the input to be reasonably sane, i.e. without multiple decimal points embedded in the numbers or multiple negative signs)
String str = "Fr.-145,000.01";
str.replaceAll("[^\\d-.]\\.?", "")
If you are going down this route, I would sanity check it by parsing the output with BigDecimal or Double.

One approach would be to just collect our desired digits, ., + and - in a capturing group followed by an optional comma, and then join them:
([+-]?[0-9][0-9.]+),?
Test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "([+-]?[0-9][0-9.]+),?";
final String string = "Fr.-145,000.01\n"
+ "Fr.-145,000\n"
+ "Fr.-145,000,000\n"
+ "Fr.-145\n"
+ "Fr.+145,000.01\n"
+ "Fr.+145,000\n"
+ "Fr.145,000,000\n"
+ "Fr.145\n"
+ "Fr.145,000,000,000.01";
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));
}
}
Demo

String str = "Fr.-145,000.01";
Pattern regex = Pattern.compile("^[^0-9-]*(-?[0-9]+)(?:,([0-9]{3}))?(?:,([0-9]{3}))?(?:,([0-9]{3}))?(\\.[0-9]+)?[^0-9-]*$");
Matcher matcher = regex.matcher(str);
System.out.println(matcher.replaceAll("$1$2$3$4$5"));
Output:
-145000.01
It looks for number with up to 3 commas (Up to 999,999,999,999.99), and replaces it with the digits.

My approach would be to remove all the unnecessary parts using replaceAll.
The unnecessary parts are, apparently:
Any sequence which is not digits or minus at the beginning of the string.
Commas
The first pattern is represented by ^[^\\d-]+. The second is merely ,.
Put them together with an |:
Pattern p = Pattern.compile("(^[^\\d-]+)|,");
Matcher m = p.matcher(str);
String result = m.replaceAll("");

You could 2 capturing groups and make use of repeating matching using the \G anchor to assert the position at the end of the previous match.
(?:^[^0-9+-]+(?=[.+,\d-]*\.\d+$)([+-]?\d{1,3})|\G(?!^)),(\d{3})
In Java
String regex = "(?:^[^0-9+-]+(?=[.+,\\d-]*\\.\\d+$)([+-]?\\d{1,3})|\\G(?!^)),(\\d{3})";
Explanation
(?: Non capturing group
^[^0-9+-]+ Match 1+ times not a digit, + or -
(?= Positive lookahead, assert that what follows is:
[.+,\d-]*\.\d+$ Match 0+ times what is allowed and assert ending on . and 1+ digits
) Close positive lookahead
( Capturing group 1
[+-]?\d{1,3}) Match optional + or - followed by 1-3 digits
| Or
\G(?!^) Assert position at the end of prevous match, not at the start
), Close capturing group 1 and match ,
(\d{3}) Capture in group 2 matching 3 digits
In the replacement use the 2 capturing groups $1$2
See the Regex demo | Java demo

Regex to match strings in-between double quotes that are not containing some other strings

How to match words between double quotes in lines not containing specific words
input:
System.log("error");
new Exception("error");
view.setText("message");
From the above input, I would like to ignore lines with log and Exception words in them(Case sensitive) and match words in between double quotes.
Expected output
message
I have been trying to use look ahead without luck
(?s)^(?!log)".+"
I need this for a search in IntelliJ using regex

In your pattern (?s)^(?!log)".+" the negative lookahead does not contain a quantifier so it will assert that what is directly after the start of the string is not log
What you could do is use a quantifier .* with an alternation to match either log or Exception and add word boundaries \b to prevent them being part of a larger word.
Then you might use negated character classes [^"] to match not a double quote and use a capturing group ([^"]+) for the value between the double quotes.
^(?!.*\b(?:log|Exception)\b)[^"]*"([^"]+)"
In Java:
String regex = "^(?!.*\\b(?:log|Exception)\\b)[^\"]*\"([^\"]+)\"";
Regex demo
If you want to make the dot to match a newline you can prepend (?s) to the pattern.

My guess is that this expression would likely work for capturing the message,
^(?!.*log.*|.*exception.*).*?"(.+?)".*
Demo 1
Example
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "^(?!.*log.*|.*exception.*).*?"(.+?)".*";
final String string = "System.log(\"error\");\n"
+ "new Exception(\"error\");\n"
+ "view.setText(\"message\");";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE | Pattern.CASE_INSENSITIVE);
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 match double quoted substrings

I want to parse the following string:
String text = "\"w1 w\"2\" w3 | w4 w\"5 \"w6 w7\"";
// "w1 w"2" w3 | w4 w"5 "w6 w7"
I'm using Pattern.compile(regex).matcher(text), so what I'm missing here is the proper regex.
The rules are that regex has to:
isolate any single word
any substring surrounded by double quotes is a match
double quotes within a word have to be ignored (I will later replace them with a whitespace).
So the resulting matches should be:
w1 w"2
w3
|
w4
w"5
w6 w7
Whether the double quotes are included or not in the double quotes surrounded substrings is irrelevant (e.g. 1. could be either w1 w"2 or "w1 w"2").
What I came up with is something like this:
"\"(.*)\"|(\\S+)"
I also tried many diffent variants of the above regex (including lookbehind/forward) but none is giving me the expected result.
Any idea on how to improve this?

Try this Regex:
(?:(?<=^")|(?<=\s")).*?(?="(?:\s|$))|(?![\s"])\S+
Click for Demo
EXPLANATION:
(?:(?<=^")|(?<=\s")) - Positive Lookbehind to find the position which is preceeded by a ". This " either needs to be at the start of the string or after a whitespace
.*? - matches 0+ occurrences of any character other than a newline character lazily
(?="(?:\s|$)) - Positive lookahead to validate that whatever is matched so far is followed by either a whitespace or there is nothing after the match($).
| - OR (either the above match or the following)
(?![\s"]) - Negative lookahead to validate that the position in not followed by either a whitespace or a "
\S+ - matches 1+ occurrences of a non-whitespace character
Java Code(Generated from here):
Run code here to see the output
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class MyClass {
public static void main(String args[]) {
final String regex = "(?:(?<=^\")|(?<=\\s\")).*?(?=\"(?:\\s|$))|(?![\\s\"])\\S+";
final String string = "\"w1 w\"2\" w3 | w4 w\"5 \"w6 w7\"";
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));
}
}
}
}
OUTPUT:

This seems to do the job:
"(?:[^"]|\b"\b)+"|\S+
Debuggex Demo
Regex101 Demo
Note that in Java, because we're using string literals for regexes, a backslash needs to be preceded by another backslash:
String regex = "\"(?:[^\"]|\\b\"\\b)+\"|\\S+";

Extracting a string using Regex

I have the following code to extract the string within double quotes using Regex.
String str ="\"Java\",\"programming\"";
final Pattern pattern = Pattern.compile("\"([^\"]*)\"");
final Matcher matcher = pattern.matcher(str);
while(matcher.find()){
System.out.println(matcher.group(1));
}
The output I get now is java programming.But from the String str I want the content in the second double quotes which is programming. Can any one tell me how to do that using Regex.

If you take your example, and change it slightly to:
String str ="\"Java\",\"programming\"";
final Pattern pattern = Pattern.compile("\"([^\"]*)\"");
final Matcher matcher = pattern.matcher(str);
int i = 0
while(matcher.find()){
System.out.println("match " + ++i + ": " + matcher.group(1) + "\n");
}
You should find that it prints:
match 1: Java
match 2: programming
This shows that you are able to loop over all of the matches. If you only want the last match, then you have a number of options:
Store the match in the loop, and when the loop is finished, you have the last match.
Change the regex to ignore everything until your pattern, with something like: Pattern.compile(".*\"([^\"]*)\"")
If you really want explicitly the second match, then the simplest solution is something like Pattern.compile("\"([^\"]*)\"[^\"]*\"([^\"]*)\""). This gives two matching groups.

If you want the last token inside double quotes, add an end-of-line archor ($):
final Pattern pattern = Pattern.compile("\"([^\"]*)\"$");
In this case, you can replace while with if if your input is a single line.

Great answer from Paul. Well,You can also try this pattern
final Pattern pattern = Pattern.compile(",\"(\\w+)\"");
Java program
String str ="\"Java\",\"programming\"";
final Pattern pattern = Pattern.compile(",\"(\\w+)\"");
final Matcher matcher = pattern.matcher(str);
while(matcher.find()){
System.out.println(matcher.group(1));
}
Explanation
,\": matches a comma, followed by a quotation mark "
(\\w+): matches one or more words
\": matches the last quotation mark "
Then the group(\\w+) is captured (group 1 precisely)
Output
programming

How to parse a range input in java

I want to parse a range of data (e.g. 100-2000) in Java. Is this code correct:
String patternStr = "^(\\\\d+)-(\\\\d+)$";
Pattern pattern = Pattern.compile(patternStr);
Matcher matcher = pattern.matcher(inputStr);
if(matcher.find()){
// Doing some parser
}

Too many backslashes, and you can use matches() without anchors (^$).
String inputStr = "100-2000";
String patternStr = "(\\d+)-(\\d+)";
Pattern pattern = Pattern.compile(patternStr);
Matcher matcher = pattern.matcher(inputStr);
if (matcher.matches()) {
System.out.println(matcher.group(1) + " - " + matcher.group(2));
}
As for your question "Is this code correct", all you had to do was wrap the code in a class with a main method and run it, and you'd get the answer: No.

No, you're double (well, quadruple)-escaping the digits.
It should be: "^(\\d+)-(\\d+)$".
Meaning:
Start of input: ^
Group 1: 1+ digit(s): (\\d+)
Hyphen literal: -
Group 2: 1+ digit(s): (\\d+)
End of input: $
Notes
The groups are useful for back-references. Here you're using none, so you can ditch the parenthesis around the \\d+ expressions.
You are parsing the representation of a range in this example.
If you want an actual range class, you can use the [min-max] idiom, where "min" and "max" are numbers, for instance [0-9].
As mentioned by Andreas, you can use String.matches without the Pattern-Matcher idiom and the ^ and $, if you want to match the whole input.