jq -r '."#graph"[]["rdfs:label"]' 9.0/schemaorg-all-http.jsonld works but jq -r '."#graph"[].["rdfs:label"]' 9.0/schemaorg-all-http.jsonld does not and I don't understand why .["rdfs:label"] does not need the dot. https://stackoverflow.com/a/39798796/308851 suggests it needs .name after [] and https://stedolan.github.io/jq/manual/#Basicfilters says
For example .["foo::bar"] and .["foo.bar"] work while .foo::bar does not,
Where did the dot go?
Using the terminology of the jq manual, jq expressions are
fundamentally composed of pipes and what it calls "basic filters". The
first filter under the heading "Basic Filters" is the identify filter,
.; and .[] is the "Array/Object Value Iterator".
From this perspective, that is, from the perspective of
pipes-and-basic-filters, the expression under consideration
."#graph"[]["rdfs:label"] can be viewed as an abbreviated form of
the pipeline:
.["#graph"] | .[] | .["rdfs:label"]
So from this perspective, the question is what abbreviations are allowed.
One of the most important abbreviation rules is:
E | .[] #=> E[]
Another is:
.["<string>"] #=> ."<string>"
Application of these rules yields the simplified expression.
So perhaps the basic answer to the "why" in this question is: for convenience. :-)
The dot serves two different purposes in jq:
A dot on its own means "the current object". Let's call this the identity dot. It can only appear at the start of an expression or subexpression, for example at the very start, or after a binary operator like the | or + or and, or inside an opening parenthesis (.
A dot followed by a string or an identifier means "retrieve the named field of the current object". Let's call this an indexing dot. Whatever is to the left of it needs to be a complete subexpression, for example a literal value, a parenthesised expression, a function call, etc. It can't appear in any of the places the identity dot can appear.
The thing to understand is that in the square bracket operators, the dot shown in the documentation is an identity dot - it's not actually part of the operator itself. The operator is just the square brackets and their contents, and it needs to be attached to another complete expression.
In general, both square bracket operators (e.g. ["foo"] or [] or [0] or [2:5]) and object identifier indexing operators (e.g. .foo or ."foo") can be appended to another expression. Only the object identifier indexing operators can appear "bare" with no expression on the left. Since the square bracket operators can't appear bare, you will typically see them in the documentation composed after an identity dot.
These are all equivalent:
.foo # indexing dot
."foo" # indexing dot
. .foo # identity dot and indexing dot
. | .foo # identity dot and indexing dot
.["foo"] # identity dot
. | .["foo"] # two identity dots
So the answer to your question is that the last dot in ."#graph"[].["rdfs:label"] isn't allowed because:
It can't be an identity dot because it has an expression on the left.
It can't be an indexing dot because it doesn't have an identifier or a string on the right, it has a square bracket.
All that said, it looks like newer versions of jq are going to extend the syntax to allow square bracket operators immediately after an indexing dot, and having the intuitive meaning of just applying that indexing operation the same as if there had been no dot, so hopefully you won't need to worry about the difference in the future.
For example, this regex
(.*)<FooBar>
will match:
abcde<FooBar>
But how do I get it to match across multiple lines?
abcde
fghij<FooBar>
Try this:
((.|\n)*)<FooBar>
It basically says "any character or a newline" repeated zero or more times.
It depends on the language, but there should be a modifier that you can add to the regex pattern. In PHP it is:
/(.*)<FooBar>/s
The s at the end causes the dot to match all characters including newlines.
The question is, can the . pattern match any character? The answer varies from engine to engine. The main difference is whether the pattern is used by a POSIX or non-POSIX regex library.
A special note about lua-patterns: they are not considered regular expressions, but . matches any character there, the same as POSIX-based engines.
Another note on matlab and octave: the . matches any character by default (demo): str = "abcde\n fghij<Foobar>"; expression = '(.*)<Foobar>*'; [tokens,matches] = regexp(str,expression,'tokens','match'); (tokens contain a abcde\n fghij item).
Also, in all of boost's regex grammars the dot matches line breaks by default. Boost's ECMAScript grammar allows you to turn this off with regex_constants::no_mod_m (source).
As for oracle (it is POSIX based), use the n option (demo): select regexp_substr('abcde' || chr(10) ||' fghij<Foobar>', '(.*)<Foobar>', 1, 1, 'n', 1) as results from dual
POSIX-based engines:
A mere . already matches line breaks, so there isn't a need to use any modifiers, see bash (demo).
The tcl (demo), postgresql (demo), r (TRE, base R default engine with no perl=TRUE, for base R with perl=TRUE or for stringr/stringi patterns, use the (?s) inline modifier) (demo) also treat . the same way.
However, most POSIX-based tools process input line by line. Hence, . does not match the line breaks just because they are not in scope. Here are some examples how to override this:
sed - There are multiple workarounds. The most precise, but not very safe, is sed 'H;1h;$!d;x; s/\(.*\)><Foobar>/\1/' (H;1h;$!d;x; slurps the file into memory). If whole lines must be included, sed '/start_pattern/,/end_pattern/d' file (removing from start will end with matched lines included) or sed '/start_pattern/,/end_pattern/{{//!d;};}' file (with matching lines excluded) can be considered.
perl - perl -0pe 's/(.*)<FooBar>/$1/gs' <<< "$str" (-0 slurps the whole file into memory, -p prints the file after applying the script given by -e). Note that using -000pe will slurp the file and activate 'paragraph mode' where Perl uses consecutive newlines (\n\n) as the record separator.
gnu-grep - grep -Poz '(?si)abc\K.*?(?=<Foobar>)' file. Here, z enables file slurping, (?s) enables the DOTALL mode for the . pattern, (?i) enables case insensitive mode, \K omits the text matched so far, *? is a lazy quantifier, (?=<Foobar>) matches the location before <Foobar>.
pcregrep - pcregrep -Mi "(?si)abc\K.*?(?=<Foobar>)" file (M enables file slurping here). Note pcregrep is a good solution for macOS grep users.
See demos.
Non-POSIX-based engines:
php - Use the s modifier PCRE_DOTALL modifier: preg_match('~(.*)<Foobar>~s', $s, $m) (demo)
c# - Use RegexOptions.Singleline flag (demo): - var result = Regex.Match(s, #"(.*)<Foobar>", RegexOptions.Singleline).Groups[1].Value;- var result = Regex.Match(s, #"(?s)(.*)<Foobar>").Groups[1].Value;
powershell - Use the (?s) inline option: $s = "abcde`nfghij<FooBar>"; $s -match "(?s)(.*)<Foobar>"; $matches[1]
perl - Use the s modifier (or (?s) inline version at the start) (demo): /(.*)<FooBar>/s
python - Use the re.DOTALL (or re.S) flags or (?s) inline modifier (demo): m = re.search(r"(.*)<FooBar>", s, flags=re.S) (and then if m:, print(m.group(1)))
java - Use Pattern.DOTALL modifier (or inline (?s) flag) (demo): Pattern.compile("(.*)<FooBar>", Pattern.DOTALL)
kotlin - Use RegexOption.DOT_MATCHES_ALL : "(.*)<FooBar>".toRegex(RegexOption.DOT_MATCHES_ALL)
groovy - Use (?s) in-pattern modifier (demo): regex = /(?s)(.*)<FooBar>/
scala - Use (?s) modifier (demo): "(?s)(.*)<Foobar>".r.findAllIn("abcde\n fghij<Foobar>").matchData foreach { m => println(m.group(1)) }
javascript - Use [^] or workarounds [\d\D] / [\w\W] / [\s\S] (demo): s.match(/([\s\S]*)<FooBar>/)[1]
c++ (std::regex) Use [\s\S] or the JavaScript workarounds (demo): regex rex(R"(([\s\S]*)<FooBar>)");
vba vbscript - Use the same approach as in JavaScript, ([\s\S]*)<Foobar>. (NOTE: The MultiLine property of the RegExp object is sometimes erroneously thought to be the option to allow . match across line breaks, while, in fact, it only changes the ^ and $ behavior to match start/end of lines rather than strings, the same as in JavaScript regex)
behavior.)
ruby - Use the /m MULTILINE modifier (demo): s[/(.*)<Foobar>/m, 1]
rtrebase-r - Base R PCRE regexps - use (?s): regmatches(x, regexec("(?s)(.*)<FooBar>",x, perl=TRUE))[[1]][2] (demo)
ricustringrstringi - in stringr/stringi regex funtions that are powered with the ICU regex engine. Also use (?s): stringr::str_match(x, "(?s)(.*)<FooBar>")[,2] (demo)
go - Use the inline modifier (?s) at the start (demo): re: = regexp.MustCompile(`(?s)(.*)<FooBar>`)
swift - Use dotMatchesLineSeparators or (easier) pass the (?s) inline modifier to the pattern: let rx = "(?s)(.*)<Foobar>"
objective-c - The same as Swift. (?s) works the easiest, but here is how the option can be used: NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:pattern options:NSRegularExpressionDotMatchesLineSeparators error:®exError];
re2, google-apps-script - Use the (?s) modifier (demo): "(?s)(.*)<Foobar>" (in Google Spreadsheets, =REGEXEXTRACT(A2,"(?s)(.*)<Foobar>"))
NOTES ON (?s):
In most non-POSIX engines, the (?s) inline modifier (or embedded flag option) can be used to enforce . to match line breaks.
If placed at the start of the pattern, (?s) changes the bahavior of all . in the pattern. If the (?s) is placed somewhere after the beginning, only those .s will be affected that are located to the right of it unless this is a pattern passed to Python's re. In Python re, regardless of the (?s) location, the whole pattern . is affected. The (?s) effect is stopped using (?-s). A modified group can be used to only affect a specified range of a regex pattern (e.g., Delim1(?s:.*?)\nDelim2.* will make the first .*? match across newlines and the second .* will only match the rest of the line).
POSIX note:
In non-POSIX regex engines, to match any character, [\s\S] / [\d\D] / [\w\W] constructs can be used.
In POSIX, [\s\S] is not matching any character (as in JavaScript or any non-POSIX engine), because regex escape sequences are not supported inside bracket expressions. [\s\S] is parsed as bracket expressions that match a single character, \ or s or S.
If you're using Eclipse search, you can enable the "DOTALL" option to make '.' match any character including line delimiters: just add "(?s)" at the beginning of your search string. Example:
(?s).*<FooBar>
In many regex dialects, /[\S\s]*<Foobar>/ will do just what you want. Source
([\s\S]*)<FooBar>
The dot matches all except newlines (\r\n). So use \s\S, which will match ALL characters.
We can also use
(.*?\n)*?
to match everything including newline without being greedy.
This will make the new line optional
(.*?|\n)*?
In Ruby you can use the 'm' option (multiline):
/YOUR_REGEXP/m
See the Regexp documentation on ruby-doc.org for more information.
"." normally doesn't match line-breaks. Most regex engines allows you to add the S-flag (also called DOTALL and SINGLELINE) to make "." also match newlines.
If that fails, you could do something like [\S\s].
For Eclipse, the following expression worked:
Foo
jadajada Bar"
Regular expression:
Foo[\S\s]{1,10}.*Bar*
Note that (.|\n)* can be less efficient than (for example) [\s\S]* (if your language's regexes support such escapes) and than finding how to specify the modifier that makes . also match newlines. Or you can go with POSIXy alternatives like [[:space:][:^space:]]*.
Use:
/(.*)<FooBar>/s
The s causes dot (.) to match carriage returns.
Use RegexOptions.Singleline. It changes the meaning of . to include newlines.
Regex.Replace(content, searchText, replaceText, RegexOptions.Singleline);
In notepad++ you can use this
<table (.|\r\n)*</table>
It will match the entire table starting from
rows and columns
You can make it greedy, using the following, that way it will match the first, second and so forth tables and not all at once
<table (.|\r\n)*?</table>
In a Java-based regular expression, you can use [\s\S].
This works for me and is the simplest one:
(\X*)<FooBar>
Generally, . doesn't match newlines, so try ((.|\n)*)<foobar>.
In JavaScript you can use [^]* to search for zero to infinite characters, including line breaks.
$("#find_and_replace").click(function() {
var text = $("#textarea").val();
search_term = new RegExp("[^]*<Foobar>", "gi");;
replace_term = "Replacement term";
var new_text = text.replace(search_term, replace_term);
$("#textarea").val(new_text);
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<button id="find_and_replace">Find and replace</button>
<br>
<textarea ID="textarea">abcde
fghij<Foobar></textarea>
Solution:
Use pattern modifier sU will get the desired matching in PHP.
Example:
preg_match('/(.*)/sU', $content, $match);
Sources:
Pattern Modifiers
In the context of use within languages, regular expressions act on strings, not lines. So you should be able to use the regex normally, assuming that the input string has multiple lines.
In this case, the given regex will match the entire string, since "<FooBar>" is present. Depending on the specifics of the regex implementation, the $1 value (obtained from the "(.*)") will either be "fghij" or "abcde\nfghij". As others have said, some implementations allow you to control whether the "." will match the newline, giving you the choice.
Line-based regular expression use is usually for command line things like egrep.
Try: .*\n*.*<FooBar> assuming you are also allowing blank newlines. As you are allowing any character including nothing before <FooBar>.
I had the same problem and solved it in probably not the best way but it works. I replaced all line breaks before I did my real match:
mystring = Regex.Replace(mystring, "\r\n", "")
I am manipulating HTML so line breaks don't really matter to me in this case.
I tried all of the suggestions above with no luck. I am using .NET 3.5 FYI.
I wanted to match a particular if block in Java:
...
...
if(isTrue){
doAction();
}
...
...
}
If I use the regExp
if \(isTrue(.|\n)*}
it included the closing brace for the method block, so I used
if \(!isTrue([^}.]|\n)*}
to exclude the closing brace from the wildcard match.
Often we have to modify a substring with a few keywords spread across lines preceding the substring. Consider an XML element:
<TASK>
<UID>21</UID>
<Name>Architectural design</Name>
<PercentComplete>81</PercentComplete>
</TASK>
Suppose we want to modify the 81, to some other value, say 40. First identify .UID.21..UID., then skip all characters including \n till .PercentCompleted.. The regular expression pattern and the replace specification are:
String hw = new String("<TASK>\n <UID>21</UID>\n <Name>Architectural design</Name>\n <PercentComplete>81</PercentComplete>\n</TASK>");
String pattern = new String ("(<UID>21</UID>)((.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
String replaceSpec = new String ("$1$2$440$6");
// Note that the group (<PercentComplete>) is $4 and the group ((.|\n)*?) is $2.
String iw = hw.replaceFirst(pattern, replaceSpec);
System.out.println(iw);
<TASK>
<UID>21</UID>
<Name>Architectural design</Name>
<PercentComplete>40</PercentComplete>
</TASK>
The subgroup (.|\n) is probably the missing group $3. If we make it non-capturing by (?:.|\n) then the $3 is (<PercentComplete>). So the pattern and replaceSpec can also be:
pattern = new String("(<UID>21</UID>)((?:.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
replaceSpec = new String("$1$2$340$5")
and the replacement works correctly as before.
Typically searching for three consecutive lines in PowerShell, it would look like:
$file = Get-Content file.txt -raw
$pattern = 'lineone\r\nlinetwo\r\nlinethree\r\n' # "Windows" text
$pattern = 'lineone\nlinetwo\nlinethree\n' # "Unix" text
$pattern = 'lineone\r?\nlinetwo\r?\nlinethree\r?\n' # Both
$file -match $pattern
# output
True
Bizarrely, this would be Unix text at the prompt, but Windows text in a file:
$pattern = 'lineone
linetwo
linethree
'
Here's a way to print out the line endings:
'lineone
linetwo
linethree
' -replace "`r",'\r' -replace "`n",'\n'
# Output
lineone\nlinetwo\nlinethree\n
Option 1
One way would be to use the s flag (just like the accepted answer):
/(.*)<FooBar>/s
Demo 1
Option 2
A second way would be to use the m (multiline) flag and any of the following patterns:
/([\s\S]*)<FooBar>/m
or
/([\d\D]*)<FooBar>/m
or
/([\w\W]*)<FooBar>/m
Demo 2
RegEx Circuit
jex.im visualizes regular expressions:
My regex pattern looks something like
<xxxx location="file path/level1/level2" xxxx some="xxx">
I am only interested in the part in quotes assigned to location. Shouldn't it be as easy as below without the greedy switch?
/.*location="(.*)".*/
Does not seem to work.
You need to make your regular expression lazy/non-greedy, because by default, "(.*)" will match all of "file path/level1/level2" xxx some="xxx".
Instead you can make your dot-star non-greedy, which will make it match as few characters as possible:
/location="(.*?)"/
Adding a ? on a quantifier (?, * or +) makes it non-greedy.
Note: this is only available in regex engines which implement the Perl 5 extensions (Java, Ruby, Python, etc) but not in "traditional" regex engines (including Awk, sed, grep without -P, etc.).
location="(.*)" will match from the " after location= until the " after some="xxx unless you make it non-greedy.
So you either need .*? (i.e. make it non-greedy by adding ?) or better replace .* with [^"]*.
[^"] Matches any character except for a " <quotation-mark>
More generic: [^abc] - Matches any character except for an a, b or c
How about
.*location="([^"]*)".*
This avoids the unlimited search with .* and will match exactly to the first quote.
Use non-greedy matching, if your engine supports it. Add the ? inside the capture.
/location="(.*?)"/
Use of Lazy quantifiers ? with no global flag is the answer.
Eg,
If you had global flag /g then, it would have matched all the lowest length matches as below.
Here's another way.
Here's the one you want. This is lazy [\s\S]*?
The first item:
[\s\S]*?(?:location="[^"]*")[\s\S]* Replace with: $1
Explaination: https://regex101.com/r/ZcqcUm/2
For completeness, this gets the last one. This is greedy [\s\S]*
The last item:[\s\S]*(?:location="([^"]*)")[\s\S]*
Replace with: $1
Explaination: https://regex101.com/r/LXSPDp/3
There's only 1 difference between these two regular expressions and that is the ?
The other answers here fail to spell out a full solution for regex versions which don't support non-greedy matching. The greedy quantifiers (.*?, .+? etc) are a Perl 5 extension which isn't supported in traditional regular expressions.
If your stopping condition is a single character, the solution is easy; instead of
a(.*?)b
you can match
a[^ab]*b
i.e specify a character class which excludes the starting and ending delimiiters.
In the more general case, you can painstakingly construct an expression like
start(|[^e]|e(|[^n]|n(|[^d])))end
to capture a match between start and the first occurrence of end. Notice how the subexpression with nested parentheses spells out a number of alternatives which between them allow e only if it isn't followed by nd and so forth, and also take care to cover the empty string as one alternative which doesn't match whatever is disallowed at that particular point.
Of course, the correct approach in most cases is to use a proper parser for the format you are trying to parse, but sometimes, maybe one isn't available, or maybe the specialized tool you are using is insisting on a regular expression and nothing else.
Because you are using quantified subpattern and as descried in Perl Doc,
By default, a quantified subpattern is "greedy", that is, it will
match as many times as possible (given a particular starting location)
while still allowing the rest of the pattern to match. If you want it
to match the minimum number of times possible, follow the quantifier
with a "?" . Note that the meanings don't change, just the
"greediness":
*? //Match 0 or more times, not greedily (minimum matches)
+? //Match 1 or more times, not greedily
Thus, to allow your quantified pattern to make minimum match, follow it by ? :
/location="(.*?)"/
import regex
text = 'ask her to call Mary back when she comes back'
p = r'(?i)(?s)call(.*?)back'
for match in regex.finditer(p, str(text)):
print (match.group(1))
Output:
Mary