This question already has an answer here:
Regex - two specific digits followed by optional digits
(1 answer)
Closed 3 years ago.
I want to match this strings value, value1, value2.
I got the number so far, but I need to match the word with no numbers after, also.
sed -e 's/value[0-9]//g'
You can combine a multiple expressions into one by separating with semicolons. Hope this helps.
sed 's/value[0-9]//g;s/value//g' inputfile
The following statement is from a homework question which I tested out and answered, but I'm just not understanding how come this line behaves the way it does and I want to understand why. I realize why this expression is flawed to find an IP address but I don't fully understand why it behaves the way it does since it seems as if the question mark doesn't actually behave as 0 or 1 times in like it's supposed to.
"user#machine:~$ grep -E '[01]?[0-9][0-9]?' "
To my understanding "[01]?" should look for any number 0-1 as indicated by the brackets while the question mark tells grep to look for zero or one instance only and similar with "[0-9]?". Thing is this line will print an unlimited number of digits far exceeding 3 digits. I ruled out that it was due to the 3rd bracket that didn't have a proceeding question mark since it would still print an unlimited amount of digits if I piped an echo or used a testing .txt file full of numbers.
This above example made me than wonder how to find IP's with grep the correct way. So I found countless examples like the following expression for IPv4 octets:
\.(25[0-5]\|2[0-4][0-9]\|[01][0-9][0-9]\|[0-9][0-9]).\
Is this telling me to look for any number 2-5 anywhere from 0-5 times? 0-5 is too many digits for an octet. Is it telling me to look for any number 0-5 up to 25 times? Again that's way too many digits for an octet. What does \2[0-4][0-9]\ mean in this case? I'm confused about how this expression finds numbers strictly between 1-255?
Look at it this way: x?[0-9]x? matches anything which contains a digit because both the x:es are optional. You might as well leave them out because they do not constrain the match at all.
25[0-5] looks for 25 followed by a digit in the range 0-5. In other words, the expression matches a number in the range 250-255.
The full expression in your example looks for a number in the range 00-255 by enumerating strings beginning with 25, 20-24, etc; though it's incomplete in that it doesn't permit single-digit numbers.
The expression matches a single octet (incompletely), not an entire IP address. Here is a common way to match an IPv4 address:
([3-9][0-9]?|2([0-4][0-9]?|5[0-9]?|[6-9])?|1([0-9][0-9]?)?)(\.([3-9][0-9]?|2([0-4][0-9]?|5[0-9]?|[6-9])?|1([0-9][0-9]?)?){3}
where the square brackets express character classes which match a single character out of a set, and the final curly braces {3} express a repetition.
Some regex dialects (e.g. POSIX grep) require backslashes before | and \( but I have used the extended notation (a la grep -E and most online regex exploration tools) which doesn't want backslashes.
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Closed 11 years ago.
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I don't really understand regular expressions. Can you explain them to me in an easy-to-follow manner? If there are any online tools or books, could you also link to them?
The most important part is the concepts. Once you understand how the building blocks work, differences in syntax amount to little more than mild dialects. A layer on top of your regular expression engine's syntax is the syntax of the programming language you're using. Languages such as Perl remove most of this complication, but you'll have to keep in mind other considerations if you're using regular expressions in a C program.
If you think of regular expressions as building blocks that you can mix and match as you please, it helps you learn how to write and debug your own patterns but also how to understand patterns written by others.
Start simple
Conceptually, the simplest regular expressions are literal characters. The pattern N matches the character 'N'.
Regular expressions next to each other match sequences. For example, the pattern Nick matches the sequence 'N' followed by 'i' followed by 'c' followed by 'k'.
If you've ever used grep on Unix—even if only to search for ordinary looking strings—you've already been using regular expressions! (The re in grep refers to regular expressions.)
Order from the menu
Adding just a little complexity, you can match either 'Nick' or 'nick' with the pattern [Nn]ick. The part in square brackets is a character class, which means it matches exactly one of the enclosed characters. You can also use ranges in character classes, so [a-c] matches either 'a' or 'b' or 'c'.
The pattern . is special: rather than matching a literal dot only, it matches any character†. It's the same conceptually as the really big character class [-.?+%$A-Za-z0-9...].
Think of character classes as menus: pick just one.
Helpful shortcuts
Using . can save you lots of typing, and there are other shortcuts for common patterns. Say you want to match a digit: one way to write that is [0-9]. Digits are a frequent match target, so you could instead use the shortcut \d. Others are \s (whitespace) and \w (word characters: alphanumerics or underscore).
The uppercased variants are their complements, so \S matches any non-whitespace character, for example.
Once is not enough
From there, you can repeat parts of your pattern with quantifiers. For example, the pattern ab?c matches 'abc' or 'ac' because the ? quantifier makes the subpattern it modifies optional. Other quantifiers are
* (zero or more times)
+ (one or more times)
{n} (exactly n times)
{n,} (at least n times)
{n,m} (at least n times but no more than m times)
Putting some of these blocks together, the pattern [Nn]*ick matches all of
ick
Nick
nick
Nnick
nNick
nnick
(and so on)
The first match demonstrates an important lesson: * always succeeds! Any pattern can match zero times.
A few other useful examples:
[0-9]+ (and its equivalent \d+) matches any non-negative integer
\d{4}-\d{2}-\d{2} matches dates formatted like 2019-01-01
Grouping
A quantifier modifies the pattern to its immediate left. You might expect 0abc+0 to match '0abc0', '0abcabc0', and so forth, but the pattern immediately to the left of the plus quantifier is c. This means 0abc+0 matches '0abc0', '0abcc0', '0abccc0', and so on.
To match one or more sequences of 'abc' with zeros on the ends, use 0(abc)+0. The parentheses denote a subpattern that can be quantified as a unit. It's also common for regular expression engines to save or "capture" the portion of the input text that matches a parenthesized group. Extracting bits this way is much more flexible and less error-prone than counting indices and substr.
Alternation
Earlier, we saw one way to match either 'Nick' or 'nick'. Another is with alternation as in Nick|nick. Remember that alternation includes everything to its left and everything to its right. Use grouping parentheses to limit the scope of |, e.g., (Nick|nick).
For another example, you could equivalently write [a-c] as a|b|c, but this is likely to be suboptimal because many implementations assume alternatives will have lengths greater than 1.
Escaping
Although some characters match themselves, others have special meanings. The pattern \d+ doesn't match backslash followed by lowercase D followed by a plus sign: to get that, we'd use \\d\+. A backslash removes the special meaning from the following character.
Greediness
Regular expression quantifiers are greedy. This means they match as much text as they possibly can while allowing the entire pattern to match successfully.
For example, say the input is
"Hello," she said, "How are you?"
You might expect ".+" to match only 'Hello,' and will then be surprised when you see that it matched from 'Hello' all the way through 'you?'.
To switch from greedy to what you might think of as cautious, add an extra ? to the quantifier. Now you understand how \((.+?)\), the example from your question works. It matches the sequence of a literal left-parenthesis, followed by one or more characters, and terminated by a right-parenthesis.
If your input is '(123) (456)', then the first capture will be '123'. Non-greedy quantifiers want to allow the rest of the pattern to start matching as soon as possible.
(As to your confusion, I don't know of any regular-expression dialect where ((.+?)) would do the same thing. I suspect something got lost in transmission somewhere along the way.)
Anchors
Use the special pattern ^ to match only at the beginning of your input and $ to match only at the end. Making "bookends" with your patterns where you say, "I know what's at the front and back, but give me everything between" is a useful technique.
Say you want to match comments of the form
-- This is a comment --
you'd write ^--\s+(.+)\s+--$.
Build your own
Regular expressions are recursive, so now that you understand these basic rules, you can combine them however you like.
Tools for writing and debugging regexes:
RegExr (for JavaScript)
Perl: YAPE: Regex Explain
Regex Coach (engine backed by CL-PPCRE)
RegexPal (for JavaScript)
Regular Expressions Online Tester
Regex Buddy
Regex 101 (for PCRE, JavaScript, Python, Golang, Java 8)
I Hate Regex
Visual RegExp
Expresso (for .NET)
Rubular (for Ruby)
Regular Expression Library (Predefined Regexes for common scenarios)
Txt2RE
Regex Tester (for JavaScript)
Regex Storm (for .NET)
Debuggex (visual regex tester and helper)
Books
Mastering Regular Expressions, the 2nd Edition, and the 3rd edition.
Regular Expressions Cheat Sheet
Regex Cookbook
Teach Yourself Regular Expressions
Free resources
RegexOne - Learn with simple, interactive exercises.
Regular Expressions - Everything you should know (PDF Series)
Regex Syntax Summary
How Regexes Work
JavaScript Regular Expressions
Footnote
†: The statement above that . matches any character is a simplification for pedagogical purposes that is not strictly true. Dot matches any character except newline, "\n", but in practice you rarely expect a pattern such as .+ to cross a newline boundary. Perl regexes have a /s switch and Java Pattern.DOTALL, for example, to make . match any character at all. For languages that don't have such a feature, you can use something like [\s\S] to match "any whitespace or any non-whitespace", in other words anything.
This question already has answers here:
Detect most likely words from text without spaces / combined words
(5 answers)
Closed 8 years ago.
Lets suppose a string with no spaces:
Input : "putreturnsbetwenparagaphs"
Output : put returns between paragraphs
This could get more complex as more words overlap. How to achieve this really fast. If required does spell corrections and splits the word. Think about it.
One problem could be the plural or case of the word. In your example it could be difficult to make a difference between paragraph and paragraphs.
Do you have more information? Are some words in a explicit grammatical form, or could any word of a common dictionary including case, numerus etc. occour?
This question already has answers here:
Regex: Determine if two regular expressions could match for the same input?
(5 answers)
Closed 10 years ago.
I have two regexps. I need to determine if it is possible to build string of given length that matches these two regexps simultaneously. I need algorithm to do that.
String's length wouldn't exceed 20 characters.
It depends. For perl compatible regular expressions (pcre), this is not generally possible, as they are turing complete: you cannot even be sure that matching always terminates.
For the original, "clean" form of reguler languages as defined in the Chomsky-hierarchy, it is known that they are closed under intersection, this is already discussed in this thread.
As soon as you have the NFA for the intersection, it is easy to check whether any string matches it - if thera is a path from the start to the end of your NFA, then the string for this path is the string you are searching for, for DFAs, an algorithm is given here, it should be simple to adapt it to NFAs.