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REGEX extract two double number separated from hypen

I have strings like:
some foo text
some foo
1-2
1.00-2.00
3.21-1.23
2.12-2.12
I have to check if the string format contains two numbers separated by hyphen.
How can I do it?
Thanks

Regex for float is: ^[1-9]\d*\.\d+$ if decimals are optional : ^[1-9]\d*(?:\.\d+)?$
Repeat it twice with hyphen in between:
`^[1-9]\d*(?:\.\d+)?-[1-9]\d*(?:\.\d+)?$`

You can use the regex:
^\d+(\.\d+)?-\d+(\.\d+)?$
Explanation can be found here.
Using java you can create a method that checks whether your desired pattern exists or not:
public static boolean returnMatch(String input) {
Pattern p1 = Pattern.compile("^\\d+(\\.\\d+)?-\\d+(\\.\\d+)?$");
Matcher m1 = p1.matcher(input);
return m1.find() ? true : false;
}
Now call it using:
System.out.println(returnMatch("some foo text")); // false
System.out.println(returnMatch("1.00-2.00")); // true
System.out.println(returnMatch("2.12-2.12")); // true
System.out.println(returnMatch("10-20")); // true

Use a simple Regex:
(\d+(?:\.\d+)?)-(\d+(?:\.\d+)?)
This solution assumes there is always a decimal part present (at least one digit). Demo at Regex101.
\d is a digit
\d+ is at least one digit
\. matches a dot (.) literally
() is a capturing group
(?:\.\d+)? is a non-capturing group which optionally matches the decimal part
Don't forget the proper escaping in Java String regex = "(\\d+(?:\\.\\d+)?)-(\\d+(?:\\.\\d+)?)";
In case one or more spaced or blank characters appear between the dash and numbers, use:
(\d+(?:\.\d+)?)\s*-\s*(\d+(?:\.\d+)?)

Java. Regular Expressions. How to mix NOT with AND?

I have a string (VIN) like this:
String vin = "XTC53229R71133923";
I can use OR to see if there are characters Q,O,I:
String regExp = ".*[QOI].*";
This works.
However I can not check that any of these 3 letter are NOT in the string.
It means: (NOT Q) AND (NOT O) AND (NOT I).
I tried negative lookahead:
String regExp = "(?!.*[QOI].*)";
This doens't work. In "XTC5Q3229R71133923" it returns true.
The main issue - I have 2 conditions:
Number of characters (A-Z0-9) in the string should be 17.
The string should not have Q,O,I.
I can check this with 2 regexps:
String regExp = "^([A-Z0-9]{17})$"; //should be true
String regExp = ".*[QOI].*"; //should be false
But is there a way to combine these 2 checks in one regular expression?

How about just using a custom range that doesn't include the characters you do not want?
String regexp = "^([A-HJ-NPR-Z0-9]{17})$";

Here you go ^[^QOI]{17}$. Starting a charcter class with ^ means "do not match any of these characters".

Regex to identify strings containing a particular symbol?

I have set of inputs ++++,----,+-+-.Out of these inputs I want the string containing only + symbols.

If you want to see if a String contains nothing but + characters, write a loop to check it:
private static boolean containsOnly(String input, char ch) {
if (input.isEmpty())
return false;
for (int i = 0; i < input.length(); i++)
if (input.charAt(i) != ch)
return false;
return true;
}
Then call it to check:
System.out.println(containsOnly("++++", '+')); // prints: true
System.out.println(containsOnly("----", '+')); // prints: false
System.out.println(containsOnly("+-+-", '+')); // prints: false
UPDATE
If you must do it using regex (worse performance), then you can do any of these:
// escape special character '+'
input.matches("\\++")
// '+' not special in a character class
input.matches("[+]+")
// if "+" is dynamic value at runtime, use quote() to escape for you,
// then use a repeating non-capturing group around that
input.matches("(?:" + Pattern.quote("+") + ")+")
Replace final + with * in each of these, if an empty string should return true.

The regular expression for checking if a string is composed of only one repeated symbol is
^(.)\1*$
If you only want lines composed by '+', then it's
^\++$, or ^++*$ if your regex implementation does not support +(meaning "one or more").

For a sequence of the same symbol, use
(.)\1+
as the regular expression. For example, this will match +++, and --- but not +--.

Regex pattern: ^[^\+]*?\+[^\+]*$
This will only permit one plus sign per string.
Demo Link
Explanation:
^ #From start of string
[^\+]* #Match 0 or more non plus characters
\+ #Match 1 plus character
[^\+]* #Match 0 or more non plus characters
$ #End of string
edit, I just read the comments under the question, I didn't actually steal the commented regex (it just happens to be intellectual convergence):
Whoops, when using matches disregard ^ and $ anchors.
input.matches("[^\\+]*?\+[^\\+]*")

Regex to get first number in string with other characters

I'm new to regular expressions, and was wondering how I could get only the first number in a string like 100 2011-10-20 14:28:55. In this case, I'd want it to return 100, but the number could also be shorter or longer.
I was thinking about something like [0-9]+, but it takes every single number separately (100,2001,10,...)
Thank you.

/^[^\d]*(\d+)/
This will start at the beginning, skip any non-digits, and match the first sequence of digits it finds
EDIT:
this Regex will match the first group of numbers, but, as pointed out in other answers, parseInt is a better solution if you know the number is at the beginning of the string

Try this to match for first number in string (which can be not at the beginning of the string):
String s = "2011-10-20 525 14:28:55 10";
Pattern p = Pattern.compile("(^|\\s)([0-9]+)($|\\s)");
Matcher m = p.matcher(s);
if (m.find()) {
System.out.println(m.group(2));
}

Just
([0-9]+) .*
If you always have the space after the first number, this will work

Assuming there's always a space between the first two numbers, then
preg_match('/^(\d+)/', $number_string, $matches);
$number = $matches[1]; // 100
But for something like this, you'd be better off using simple string operations:
$space_pos = strpos($number_string, ' ');
$number = substr($number_string, 0, $space_pos);
Regexs are computationally expensive, and should be avoided if possible.

the below code would do the trick.
Integer num = Integer.parseInt("100 2011-10-20 14:28:55");

[0-9] means the numbers 0-9 can be used the + means 1 or more times. if you use [0-9]{3} will get you 3 numbers

Try ^(?'num'[0-9]+).*$ which forces it to start at the beginning, read a number, store it to 'num' and consume the remainder without binding.

This string extension works perfectly, even when string not starts with number.
return 1234 in each case - "1234asdfwewf", "%sdfsr1234" "## # 1234"
public static string GetFirstNumber(this string source)
{
if (string.IsNullOrEmpty(source) == false)
{
// take non digits from string start
string notNumber = new string(source.TakeWhile(c => Char.IsDigit(c) == false).ToArray());
if (string.IsNullOrEmpty(notNumber) == false)
{
//replace non digit chars from string start
source = source.Replace(notNumber, string.Empty);
}
//take digits from string start
source = new string(source.TakeWhile(char.IsDigit).ToArray());
}
return source;
}

NOTE: In Java, when you define the patterns as string literals, do not forget to use double backslashes to define a regex escaping backslash (\. = "\\.").
To get the number that appears at the start or beginning of a string you may consider using
^[0-9]*\.?[0-9]+ # Float or integer, leading digit may be missing (e.g, .35)
^-?[0-9]*\.?[0-9]+ # Optional - before number (e.g. -.55, -100)
^[-+]?[0-9]*\.?[0-9]+ # Optional + or - before number (e.g. -3.5, +30)
See this regex demo.
If you want to also match numbers with scientific notation at the start of the string, use
^[0-9]*\.?[0-9]+([eE][+-]?[0-9]+)? # Just number
^-?[0-9]*\.?[0-9]+([eE][+-]?[0-9]+)? # Number with an optional -
^[-+]?[0-9]*\.?[0-9]+([eE][+-]?[0-9]+)? # Number with an optional - or +
See this regex demo.
To make sure there is no other digit on the right, add a \b word boundary, or a (?!\d)
or (?!\.?\d) negative lookahead that will fail the match if there is any digit (or . and a digit) on the right.

public static void main(String []args){
Scanner s=new Scanner(System.in);
String str=s.nextLine();
Pattern p=Pattern.compile("[0-9]+");
Matcher m=p.matcher(str);
while(m.find()){
System.out.println(m.group()+" ");
}

\d+
\d stands for any decimal while + extends it to any other decimal coming directly after, until there is a non number character like a space or letter

Match any word but two in specific

I'm trying to create a regular expresion to match any word ( \w+ ) except true or false.
This is what I got so far is: \w+\s*=\s*[^true|^false]\w+
class Ntnf {
public static void main ( String ... args ) {
System.out.println( args[0].matches("\\w+\\s*=\\s*[^true|^false]\\w+") );
}
}
But is not working for:
a = b
a = true
a = false
It matches always.
How can I match any word ( \w+ ) except true or false?
EDIT
I'm trying to spot this pattern:
a = b
x = y
name = someothername
etc = xyz
x = truea
n = falsea
But avoid matching
a = true
etc = false
name = true

You can use:
^(?!(true|false)$)
^ - beginning of string
?! - negative lookahead
$ - end of string
So it matches as long as the whole string isn't just "true" or "false". Note that it can still start with one of those.
However, it may be more straightforward to use regular string comparisons.
EDIT:
The whole regex (without escaping) for your situation is:
^\w+\s*=\s*(?!(true|false)$)\w+$
It's the same idea, except that we're putting it in the equation form.

[^true] Is a character class. It only matches one character. [^true] means: "Match this character only if it not one of t, r, u or e". This is not what you need, right?
Regex is not a good idea for this task. It will be quite complicated to do it in regex. Just use string comparison.

Square brackets match a list of possible characters, or reject a list of possible characters (not necessarily in the order you specify), so [^true] is not the way to go.
When I'm trying not to match a certain word, I usually do the following:
([^t]|t[^r]|tr[^u]|tru[^e])