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Match pattern which is not wrapped by some character

I have an input string like this:
one `two three` four five `six` seven
where some parts can be wrapped by grave accent character (`).
I want to match only these parts which are not wrapped by it, it is one, four five and seven in example (skip two three and six).
I tryied to do it using lookaheads ((?<=) and (?=)) but it recognised four five group like two three and six. Is it possible to solve this problem using regex only, or I have to do it programmatically? (I'm using java 1.8)

If you are sure that there are no unclosed backticks, you could do this:
((?:\w| )+)(?=(?:[^`]*`[^`]*`)*[^`]*$)
This will match:
"one "
" four five "
" seven"
But it's a little bit expensive, because the lookahead that checks whether the number of backtics in the remaining part of line is divisible by 2 takes O(n^2) time to scan through the entire string.
Note that this works regardless of where the whitespace is, it really counts the backticks, it does not care about the relative position of the backticks. If you don't need this kind of robustness, #anubhava's answer is certainly more performant.
Demo: regex101.

You may use this regex using a lookahead and lookbehind:
(?<!`)\b\w+(?:\s+\w+)*\b(?!`)
RegEx Demo
Explanation:
- (?<!`): Negative Lookbehind to assert that we don't have ` at previous position
- \b\w+(?:\s+\w+)*\b: Match our text surrounded by word boundaries
- (?!`): Negative Lookahead to assert that we don't have ` at next position

I solve issues like this by specifying to exclude closing characters (in your case whitespace) like so:
`[^\s]+`

Using regex to match beginning and end of string [Java]

I have a list of files in a folder:
maze1.in.txt
maze2.in.txt
maze3.in.txt
I've used substring to remove the .txt extensions.
How do I use regex to match the front and the back of the file name?
I need it to match "maze" at the front and ".in" at the back, and the middle must be a digit (can be single or double digit).
I've tried the following
if (name.matches("name\\din")) {
//dosomething
}
It doesn't match anything. What is the correct regex expression to use?

I'm a little confused what you are asking for in particular
^(maze[0-9]*\.in)$
This will match maze(any number).in
^(maze[0-9]*\.in)\.txt$
this will match maze(any number).in.txt -- excludes the .txt NO NEED FOR USING SUB STRING!
Edit live on Debuggex
The think i would be wary about as of right now is the capture groups... I'm not particularly sure what you are doing with this regex. However, I believe explaining capture groups could benefit you.
A capture group for instance is denoted by () this is basically store them in the pattern array and is a way to parse stuff.
example maze1.in.txt
So if you want to capture the entire line minus .txt i would use this ^(maze[0-9]*\.in\.txt)$
However, if I wanted to capture things separately I would do this ^(maze)([0-9]*)(\.in)\.txt$ this will exclude .txt but include maze, the number, and .in IN separate indexes of the pattern array.

Your original solution doesn't work because string "name" is not in your text. It is "maze".
You can try this
name.matches("maze\\d{1,2}\\.in")
d{1,2} is used to match a digit(can be single or double digit).

You need regex anchors that tell the regex to
start at the beginning: ^
and signal the end of the string: $
^maze[\d]{0,2}\.in$
or in Java:
name.matches("^maze[\\d]{0,2}\\.in$");
Also, your regex wasn't matching strings with a dot (.) which would not accept your examples given. You need to add \. to the regex to accept dots because . is a special character.

It is always good to think of what you are trying to do in english, before you create regular expressions.
You want to match a word maze followed by a digit, followed by a literal period . followed by another word.
word `\w` matches a word character
digit `\d` matches a single digit
period `\.` matches a literal period
word `\w` matches a word character
putting it all together into a single string you get (keep in mind the double backslash for the Java escape and the pluses to repeat the previous match one or more times):
"\\w+\\d\\.\\w+"
The above is the generic case for any file name in the format xxx1.yyy, if you wanted to match maze and in specifically, you can just add those in as literal strings.
"maze\\d+\\.in"
example: http://ideone.com/rS7tw1

name.matches("^maze[0-9]+\\.in\\.txt$")

Regex for phone numbers with * in it

I have created a regex for phone numbers as follows \\d+ ?\\w{0,9} ?\\d+ . Now i have a problem that this only accepts numbers. Sometimes i receive the phone number as starified
so it can be 011***1334. How can i incorporate the stars portion into the above regex expression.

I'm having trouble getting your regex to work in the first place, without stars,
but anyway... you can represent the star by escaping it.
\*
So you should just turn all of your \d into [\d\*] or [\\d\\*] if you have to escape the \ first in your java.
Some regular expression engines don't require you to escape all special characters in [] so I'd watch for that behavior if it doesn't work at first

Anywhere you use \\d, turn it into [\\d\\*]
[\\d\\*]+ ?\\w{0,9} ?[\\d\\*]+
I agree with Sam I am though; the original seems odd. If you just need numbers and asterisks, this should do (escaped Java-style):
[\\d\\*]{7,10}

Regular expression to match a character only once before any whitespace

In Java, what regular expression would I use to match a string that has exactly one colon and makes sure that the colon appears before any whitespace?
For example, it should match these strings:
label: print "Enter input"
But: I still had the money.
ghjkdhfjkgjhalergfyujhrageyjdfghbg:
area:54
But not
label: print "Enter input:"
There was one more thing: I still had the money.
ghfdsjhgakjsdhfkjdsagfjkhadsjkhflgadsjklfglsd
area::54

If you use it with matches (which requires to match the entire string), you could use
[^\\s:]*:[^:]*
Which means: arbitrarily many non-whitespace, non-: characters, then a :, then more arbitrarily many non-: characters.
I've really only used two regex concepts: (negated) character classes and repetition.
If you want to require at least one character before or after :, replace the corresponding * with + (as jlordo pointed out in a comment).

The following should work:
^[^\s:]*:(?!.*:)
If your strings can contain line breaks, use the DOTALL flag or change the regex to the following:
(?s)^[^\s:]*:(?!.*:)

It depends on what we call white space, it could be
[^\\p{Space}:]*:[^:]

The following should get you started:
Matcher MatchedPattern = Pattern.compile("^(\\w+\\:{1}[\"\\w\\s\\.]*)$").matcher("yourstring");

Blank spaces in regular expression

I use this regular to validate many of the input fields of my java web app:
"^[a-zA-Z0-9]+$"
But i need to modify it, because i have a couple of fields that need to allow blank spaces(for example: Address).
How can i modify it to allow blank spaces(if possible not at the start).
I think i need to use some scape character like \
I tried a few different combinations but none of them worked. Can somebody help me with this regex?

I'd suggest using this:
^[a-zA-Z0-9][a-zA-Z0-9 ]+$
It adds two things: first, you're guaranteed not to have a space at the beginning, while allowing characters you need. Afterwards, letters a-z and A-Z are allowed, as well as all digits and spaces (there's a space at the end of my regex).

If you want to use only a whitespace, you can do:
^[a-zA-Z0-9 ]+$
If you want to include tabs \t, new-line \n \r\n characters, you can do:
^[a-zA-Z0-9\s]+$
Also, as you asked, if you don't want the whitespace to be at the begining:
^[a-zA-Z0-9][a-zA-Z0-9 ]+$

Use this: ^[a-zA-Z0-9]+[a-zA-Z0-9 ]+$. This should work. First atom ensures that there must be at least one character at beginning.

try like this ^[a-zA-Z0-9 ]+$ that is, add a space in it

This regex dont allow spaces at the end of string, one downside it accepts underscore character also.
^(\w+ )+\w+|\w+$

Try this one: I assume that any input with a length of at least one character is valid. The previously mentioned answers does not take that into account.
"^[a-zA-Z0-9][a-zA-Z0-9 ]*$"
If you want to allow all whitespace characters, replace the space by "\s"