Sorry, but once again I need help to understand rather complicated snippet from the "Programming Perl" book. Here it is (what is obscure to me marked as bold):
patterns are parsed like double-quoted strings, all the normal double-quote conventions will work, including variable interpolation (unless you use single quotes
as the delimiter) and special characters indicated with backslash escapes. These are applied before the string is interpreted as a regular expression (This is one of the
few places in the Perl language where a string undergoes more than one pass of
processing). ...
Another consequence of this two-pass parsing is that the ordinary Perl tokener
finds the end of the regular expression first, just as if it were looking for the
terminating delimiter of an ordinary string. Only after it has found the end of the
string (and done any variable interpolation) is the pattern treated as a regular
expression. Among other things, this means you can’t “hide” the terminating
delimiter of a pattern inside a regex construct (such as a bracketed character class
or a regex comment, which we haven’t covered yet). Perl will see the delimiter
wherever it is and terminate the pattern at that point.
First, why it is said that Only after it has found the end of the string not the end of the regular expression which it was looking, as stated before?
Second, what does it mean you can’t “hide” the terminating delimiter of a pattern inside a regex construct? Why I can't hide the terminating delimiter /, whereas I can place it wherever I want either in the regexp directly /A\/C/ or in a interpolated variable (even without \):
my $s = 'A/';
my $p = 'A/C';
say $p =~ /$s/;
outputs 1.
While I was writing and re-reading my question I thought that this snippet tells about using a single-quote as a regexp delimiter, then it all seems quite cohesive. Is my assumption correct?
My appreciation.
It says "end of the string" instead of "end of the regular expression" because at that point it's treating the regex as if it were just a string.
It's trying to say that this does not work:
/foo[-/_]/
Even though normal regex metacharacters are not special inside [], Perl will see the regex as /foo[-/ and complain about an unterminated class.
It's trying to say that Perl does not parse the regex as it reads it. First it finds the end of the regex in your source code as if it were a quoted string, so the only special character is \. Then it interpolates any variables. Then it parses the result as a regular expression.
You can hide the terminating delimiter with \ because that works in ordinary strings. You can hide the delimiter inside an interpolated variable, because interpolation happens after the delimiter is found. If you use a bracketing delimiter (e.g. { } or [ ]), you can nest matching pairs of delimiters inside the regex, because q{} works like that too.
But you can't hide it inside any other regex construct.
Say you want to match a *. You would use
m/\*/
But what if you were using you used * as your delimiter? The following doesn't work:
m*\**
because it's interpreted as
m/*/
as seen in the following:
$ perl -e'm*\**'
Quantifier follows nothing in regex; marked by <-- HERE in m/* <-- HERE / at -e line 1.
Take the string literal
"a\"b"
It produces the string
a"b
Similarly, the match operator
m*a\*b*
produces the regex pattern
a*b
If you want to match a literal *, you have to use other means. In other words.
m*a\*b* === m/a*b/ matches pattern a*b
m*a\x{2A}b* === m/a\*b/ matches pattern a\*b
Please do pardon me if my question sounds a bit awkward. I am looking for a regex which will replace line numbers in perl source file without affecting values assigned to scalars.
I think below will make my question a little bit clearer. Say I have a perl source which looks like this:
1. $foo = 2.4;
2. print $foo;
I would like a regular expression to replace those line numbers (1. 2. etc..) without affecting value assigned to scalars, and so in this case $foo.
Thanks
anchor your regexp to the start of the line:
to remove the numbers:
perl -p -i.bak -e's{^\d+\. }{}' myperl
Within a perl regex you can use the caret symbol ^ to represent the start of a line. $ represents the end of a line. These are known as anchors.
So to find a number \d at the beginning of a line (only) you can search for
/^\d+/
If you wanted to remove those numbers you can "replace" them with nothing, as in
s/^\d+//g
You also want to include the dot after the number, so you might try
;
/^\d+./
But in regex a dot represents "any character" so you will need to escape the dot to have it interpreted literally
/^\d+\./
The caret symbol ^ also serves double-duty in character sets (it negates them), I only mention this as it is a common source of confusion when learning regex.
/[^\d]/ # Match characters that are not digits
I have a tag that is like
tag="text textwithdot. text text"
followed by a further tag that would resemble
tag="text text text"
I wanted to use the following regular expression
tag="\w+"
but that only finds one word, how do I find the whole string within the quotes, what wildcard does that?
This should work for you:
tag="([^"]*)"
That basically means tag=" followed by zero or more characters that are not a double quote, followed by a double quote.
BTW: I'm assuming that there is no such thing as a tag that contains the double quote character. If there is such a thing, it would need some escaping rule applied to it and the regular expression would be more complicated.
Also,
tag=['"]([^"]*)['"]
if that tags could change between ' and "
You could use an ungreedy match everything.
tag="[\s\S]*?"
Or use the . with dot matches newlines flag (assuming \n is a possibility).
Currently I use this reg ex:
"\bI([ ]{1,2})([a-zA-Z]|\d){2,13}\b"
It was just brought to my attention that the text that I use this against could contain a "\" (backslash). How do I add this to the expression?
Add |\\ inside the group, after the \d for instance.
This expression could be simplified if you're also allowing the underscore character in the second capture register, and you are willing to use metacharacters. That changes this:
([a-zA-Z]|\d){2,13}
into this ...
([\w]{2,13})
and you can also add a test for the backslash character with this ...
([\w\x5c]{2,13})
which makes the regex just a tad easier to eyeball, depending on your personal preference.
"\bI([\x20]{1,2})([\w\x5c]{2,13})\b"
See also:
WP Metacharacter
Metacharacters
Shorthand character class
Both #slavy13 and #dreftymac give you the basic solution with pointers, but...
You can use \d inside a character class to mean a digit.
You don't need to put blank into a character class to match it (except, perhaps, for clarity, though that is debatable).
You can use [:alpha:] inside a character class to mean an alpha character, [:digit:] to mean a digit, and [:alnum:] to mean an alphanumeric (specifically not including underscore, unlike \w). Note that these character classes might mean more characters than you expect; think of accented characters and non-arabic digits, especially in Unicode.
If you want to capture the whole of the information after the space, you need the repetition inside the capturing parentheses.
Contrast the behaviour of these two one-liners:
perl -n -e 'print "$2\n" if m/\bI( {1,2})([a-zA-Z\d\\]){2,13}\b/'
perl -n -e 'print "$2\n" if m/\bI( {1,2})([a-zA-Z\d\\]{2,13})\b/'
Given the input line "I a123", the first prints "3" and the second prints "a123". Obviously, if all you wanted was the last character of the second part of the string, then the original expression is fine. However, that is unlikely to be the requirement. (Obviously, if you're only interested in the whole lot, then using '$&' gives you the matched text, but it has negative efficiency implications.)
I'd probably use this regex as it seems clearest to me:
m/\bI( {1,2})([[:alnum:]\\]{2,13})\b/
Time for the obligatory plug: read Jeff Friedl's "Mastering Regular Expressions".
As I pointed out in my comment to slavy's post, \\ -> \b as a backslash is not a word character. So my suggestion is
/\bI([ ]{1,2})([\p{IsAlnum}\\]{2,13})(?:[^\w\\]|$)/
I assumed that you wanted to capture the whole 2-13 characters, not just the first one that applies, so I adjusted my RE.
You can make the last capture a lookahead if the engine supports it and you don't want to consume it. That would look like:
/\bI([ ]{1,2})([\p{IsAlnum}\\]{2,13})(?=[^\w\\]|$)/
I have a value like this:
"Foo Bar" "Another Value" something else
What regex will return the values enclosed in the quotation marks (e.g. Foo Bar and Another Value)?
In general, the following regular expression fragment is what you are looking for:
"(.*?)"
This uses the non-greedy *? operator to capture everything up to but not including the next double quote. Then, you use a language-specific mechanism to extract the matched text.
In Python, you could do:
>>> import re
>>> string = '"Foo Bar" "Another Value"'
>>> print re.findall(r'"(.*?)"', string)
['Foo Bar', 'Another Value']
I've been using the following with great success:
(["'])(?:(?=(\\?))\2.)*?\1
It supports nested quotes as well.
For those who want a deeper explanation of how this works, here's an explanation from user ephemient:
([""']) match a quote; ((?=(\\?))\2.) if backslash exists, gobble it, and whether or not that happens, match a character; *? match many times (non-greedily, as to not eat the closing quote); \1 match the same quote that was use for opening.
I would go for:
"([^"]*)"
The [^"] is regex for any character except '"'
The reason I use this over the non greedy many operator is that I have to keep looking that up just to make sure I get it correct.
Lets see two efficient ways that deal with escaped quotes. These patterns are not designed to be concise nor aesthetic, but to be efficient.
These ways use the first character discrimination to quickly find quotes in the string without the cost of an alternation. (The idea is to discard quickly characters that are not quotes without to test the two branches of the alternation.)
Content between quotes is described with an unrolled loop (instead of a repeated alternation) to be more efficient too: [^"\\]*(?:\\.[^"\\]*)*
Obviously to deal with strings that haven't balanced quotes, you can use possessive quantifiers instead: [^"\\]*+(?:\\.[^"\\]*)*+ or a workaround to emulate them, to prevent too much backtracking. You can choose too that a quoted part can be an opening quote until the next (non-escaped) quote or the end of the string. In this case there is no need to use possessive quantifiers, you only need to make the last quote optional.
Notice: sometimes quotes are not escaped with a backslash but by repeating the quote. In this case the content subpattern looks like this: [^"]*(?:""[^"]*)*
The patterns avoid the use of a capture group and a backreference (I mean something like (["']).....\1) and use a simple alternation but with ["'] at the beginning, in factor.
Perl like:
["'](?:(?<=")[^"\\]*(?s:\\.[^"\\]*)*"|(?<=')[^'\\]*(?s:\\.[^'\\]*)*')
(note that (?s:...) is a syntactic sugar to switch on the dotall/singleline mode inside the non-capturing group. If this syntax is not supported you can easily switch this mode on for all the pattern or replace the dot with [\s\S])
(The way this pattern is written is totally "hand-driven" and doesn't take account of eventual engine internal optimizations)
ECMA script:
(?=["'])(?:"[^"\\]*(?:\\[\s\S][^"\\]*)*"|'[^'\\]*(?:\\[\s\S][^'\\]*)*')
POSIX extended:
"[^"\\]*(\\(.|\n)[^"\\]*)*"|'[^'\\]*(\\(.|\n)[^'\\]*)*'
or simply:
"([^"\\]|\\.|\\\n)*"|'([^'\\]|\\.|\\\n)*'
Peculiarly, none of these answers produce a regex where the returned match is the text inside the quotes, which is what is asked for. MA-Madden tries but only gets the inside match as a captured group rather than the whole match. One way to actually do it would be :
(?<=(["']\b))(?:(?=(\\?))\2.)*?(?=\1)
Examples for this can be seen in this demo https://regex101.com/r/Hbj8aP/1
The key here is the the positive lookbehind at the start (the ?<= ) and the positive lookahead at the end (the ?=). The lookbehind is looking behind the current character to check for a quote, if found then start from there and then the lookahead is checking the character ahead for a quote and if found stop on that character. The lookbehind group (the ["']) is wrapped in brackets to create a group for whichever quote was found at the start, this is then used at the end lookahead (?=\1) to make sure it only stops when it finds the corresponding quote.
The only other complication is that because the lookahead doesn't actually consume the end quote, it will be found again by the starting lookbehind which causes text between ending and starting quotes on the same line to be matched. Putting a word boundary on the opening quote (["']\b) helps with this, though ideally I'd like to move past the lookahead but I don't think that is possible. The bit allowing escaped characters in the middle I've taken directly from Adam's answer.
The RegEx of accepted answer returns the values including their sourrounding quotation marks: "Foo Bar" and "Another Value" as matches.
Here are RegEx which return only the values between quotation marks (as the questioner was asking for):
Double quotes only (use value of capture group #1):
"(.*?[^\\])"
Single quotes only (use value of capture group #1):
'(.*?[^\\])'
Both (use value of capture group #2):
(["'])(.*?[^\\])\1
-
All support escaped and nested quotes.
I liked Eugen Mihailescu's solution to match the content between quotes whilst allowing to escape quotes. However, I discovered some problems with escaping and came up with the following regex to fix them:
(['"])(?:(?!\1|\\).|\\.)*\1
It does the trick and is still pretty simple and easy to maintain.
Demo (with some more test-cases; feel free to use it and expand on it).
PS: If you just want the content between quotes in the full match ($0), and are not afraid of the performance penalty use:
(?<=(['"])\b)(?:(?!\1|\\).|\\.)*(?=\1)
Unfortunately, without the quotes as anchors, I had to add a boundary \b which does not play well with spaces and non-word boundary characters after the starting quote.
Alternatively, modify the initial version by simply adding a group and extract the string form $2:
(['"])((?:(?!\1|\\).|\\.)*)\1
PPS: If your focus is solely on efficiency, go with Casimir et Hippolyte's solution; it's a good one.
A very late answer, but like to answer
(\"[\w\s]+\")
http://regex101.com/r/cB0kB8/1
The pattern (["'])(?:(?=(\\?))\2.)*?\1 above does the job but I am concerned of its performances (it's not bad but could be better). Mine below it's ~20% faster.
The pattern "(.*?)" is just incomplete. My advice for everyone reading this is just DON'T USE IT!!!
For instance it cannot capture many strings (if needed I can provide an exhaustive test-case) like the one below:
$string = 'How are you? I\'m fine, thank you';
The rest of them are just as "good" as the one above.
If you really care both about performance and precision then start with the one below:
/(['"])((\\\1|.)*?)\1/gm
In my tests it covered every string I met but if you find something that doesn't work I would gladly update it for you.
Check my pattern in an online regex tester.
This version
accounts for escaped quotes
controls backtracking
/(["'])((?:(?!\1)[^\\]|(?:\\\\)*\\[^\\])*)\1/
MORE ANSWERS! Here is the solution i used
\"([^\"]*?icon[^\"]*?)\"
TLDR;
replace the word icon with what your looking for in said quotes and voila!
The way this works is it looks for the keyword and doesn't care what else in between the quotes.
EG:
id="fb-icon"
id="icon-close"
id="large-icon-close"
the regex looks for a quote mark "
then it looks for any possible group of letters thats not "
until it finds icon
and any possible group of letters that is not "
it then looks for a closing "
I liked Axeman's more expansive version, but had some trouble with it (it didn't match for example
foo "string \\ string" bar
or
foo "string1" bar "string2"
correctly, so I tried to fix it:
# opening quote
(["'])
(
# repeat (non-greedy, so we don't span multiple strings)
(?:
# anything, except not the opening quote, and not
# a backslash, which are handled separately.
(?!\1)[^\\]
|
# consume any double backslash (unnecessary?)
(?:\\\\)*
|
# Allow backslash to escape characters
\\.
)*?
)
# same character as opening quote
\1
string = "\" foo bar\" \"loloo\""
print re.findall(r'"(.*?)"',string)
just try this out , works like a charm !!!
\ indicates skip character
My solution to this is below
(["']).*\1(?![^\s])
Demo link : https://regex101.com/r/jlhQhV/1
Explanation:
(["'])-> Matches to either ' or " and store it in the backreference \1 once the match found
.* -> Greedy approach to continue matching everything zero or more times until it encounters ' or " at end of the string. After encountering such state, regex engine backtrack to previous matching character and here regex is over and will move to next regex.
\1 -> Matches to the character or string that have been matched earlier with the first capture group.
(?![^\s]) -> Negative lookahead to ensure there should not any non space character after the previous match
Unlike Adam's answer, I have a simple but worked one:
(["'])(?:\\\1|.)*?\1
And just add parenthesis if you want to get content in quotes like this:
(["'])((?:\\\1|.)*?)\1
Then $1 matches quote char and $2 matches content string.
All the answer above are good.... except they DOES NOT support all the unicode characters! at ECMA Script (Javascript)
If you are a Node users, you might want the the modified version of accepted answer that support all unicode characters :
/(?<=((?<=[\s,.:;"']|^)["']))(?:(?=(\\?))\2.)*?(?=\1)/gmu
Try here.
echo 'junk "Foo Bar" not empty one "" this "but this" and this neither' | sed 's/[^\"]*\"\([^\"]*\)\"[^\"]*/>\1</g'
This will result in: >Foo Bar<><>but this<
Here I showed the result string between ><'s for clarity, also using the non-greedy version with this sed command we first throw out the junk before and after that ""'s and then replace this with the part between the ""'s and surround this by ><'s.
From Greg H. I was able to create this regex to suit my needs.
I needed to match a specific value that was qualified by being inside quotes. It must be a full match, no partial matching could should trigger a hit
e.g. "test" could not match for "test2".
reg = r"""(['"])(%s)\1"""
if re.search(reg%(needle), haystack, re.IGNORECASE):
print "winning..."
Hunter
If you're trying to find strings that only have a certain suffix, such as dot syntax, you can try this:
\"([^\"]*?[^\"]*?)\".localized
Where .localized is the suffix.
Example:
print("this is something I need to return".localized + "so is this".localized + "but this is not")
It will capture "this is something I need to return".localized and "so is this".localized but not "but this is not".
A supplementary answer for the subset of Microsoft VBA coders only one uses the library Microsoft VBScript Regular Expressions 5.5 and this gives the following code
Sub TestRegularExpression()
Dim oRE As VBScript_RegExp_55.RegExp '* Tools->References: Microsoft VBScript Regular Expressions 5.5
Set oRE = New VBScript_RegExp_55.RegExp
oRE.Pattern = """([^""]*)"""
oRE.Global = True
Dim sTest As String
sTest = """Foo Bar"" ""Another Value"" something else"
Debug.Assert oRE.test(sTest)
Dim oMatchCol As VBScript_RegExp_55.MatchCollection
Set oMatchCol = oRE.Execute(sTest)
Debug.Assert oMatchCol.Count = 2
Dim oMatch As Match
For Each oMatch In oMatchCol
Debug.Print oMatch.SubMatches(0)
Next oMatch
End Sub