I am trying to make a regular expression that replaces the content of the texts in parentheses.
I have used the following regular expression:
"([A-Za-z ]*)"
But as you can see in the following image does not work:
Thank you and greetings.
Remove the double quotes from your expression and escape the parentheses:
\([A-Za-z ]*\)
Details:
\( - a literal (
[A-Za-z ]* - zero or more ASCII letters or spaces
\) - a literal ).
The unescaped (...) form a capturing group that stores a submatch in the memory buffer that can be used later during matching or replacement via backreferences.
Related
Under Visual Studio 2019 I am trying to replace Parantheses with Square Brackets
For example: fld(126) to fld[126]
Using this regular expression
fld[\(][0-9]*[\)]
matches good what I look for in the code.
But \2 replaces everything between parentheses with '\2' instead of leaving what exists there before.
Any help would be appreciated...
Your example RegEx shows you only want to change the brackets for the function fld.
Use
Find: fld\((\d*)\)
Replace: fld[$1]
If the content between the brackets can be more than numbers use
Find: fld\(([^)]*)\)
Replace: fld[$1]
Try the following find and replace, in regex mode:
Find: ([^(\s]+)\(([^)]+)\)(?!\S)
Replace: $1[$2]
Demo
Here is an explanation of the regex pattern:
([^(\s]+) match AND capture the leading 'fld' term, in $1
this is given by all non '(' and whitespace characters
\( match a literal (
([^)]+) match AND capture the content inside parentheses, in $2
\) match a literal )
(?!\S) assert that what follows the closing ) is a boundary or whitespace
I have the following string:
arg1('value1') arg2('value '')2') arg3('value\'3')
The regex to extract the value looks like:
boost::regex re_arg_values("('[^']*(?:''[^']*)*')");
Now this regex is not able to extract 'value\'3'. How can I modify the regex to consider \' inside the parenthesis as well.
FYI. The value can contain spaces, special characters, and also tabs. The code is in CPP.
Thanks in advance.
boost::regex re_arg_values( "\('([^'\\]|''|\\.)*'\)" );
The \(' and '\) match the bounds.
The (, |, and )* for matching any of the given patterns
The [^'\\] matches normal characters.
The '' matches a pair of single quotes.
The \\. matches any escaped character (including stacked backslashes).
If I use a long regular expression in Notepad++, i.e.:
^([^ ]+) ([^ ]+) ([^ ]+) (\[.*?\]) (".*?") (".*?") (".*?") (".*?") (\d+) (\d+) (\d+)$
(this is for turning an Apache log lines from space-separated to tab-separated)
then I can't successfully use more than nine backreferences for replacing, as \10 yields the content of the first captured group plus a literal "0".
I tried with $10, but that gives the same result.
You can use curly braces for this:
${10}
For reference, Notepad++ uses boost::regex, and you can find its substitution pattern docs here: Boost-Extended Format String Syntax. This replacement mode allows for more complex expressions (like conditionals and common Perl placeholders) in the replacement pattern.
Just use the curly braces:
${10}
This will ensure that the 10th capturing group is being referred, and not the 1st group followed by zero.
I am trying to get hold of regular expressions in Perl. Can anyone please provide any examples of what matches and what doesn't for the below regular expression?
$sentence =~m/.+\/(.+)/s
=~ is the binding operator; it makes the regex match be performed on $sentence instead of the default $_. m is the match operator; it is optional (e.g. $foo =~ /bar/) when the regex is delimited by / characters but required if you want to use a different delimiter.
s is a regex flag that makes . in the regex match any characters; by default . does not match newlines.
The actual regex is .+\/(.+); this will match one or more characters, then a literal / character, then one or more other characters. Because the initial .+ consumes as much as possible while still allowing the regex to succeed, it will match up to the last / in the string that has at least one character after it; then the (.+) will capture the characters that follow that / and make them available as $1.
So it is essentially capturing the final component of a filepath. Of foo/bar it will capture the bar, of foo/bar/ it will capture the bar/. Strings with only one component, like /foo or bar/ or baz will not match.
Any string, including multi-line strings, that contain a slash character somewhere in the middle of the string.
Matches:
foo/bar
asdf\nwrqwer/wrqwerqw # /s modifier allows '.' to match newlines
Doesn't match:
asdfasfdasf # no slash character
/asdfasdf # no characters before the slash
asdfasf/ # no characters after the slash
In addition, the entire substring that follows the last slash in the string will be captured and assigned to the variable $1.
Breakdown:
$sentence =~ — match $sentence with
m/ — the pattern consisting of
. — any character
+ — one or more times
\/ — then a forward-slash
( — and, saving in the $1 capture group,
.+ — any character one or more times
)
/s — allowing . to match newlines
See perldoc perlop for information about operators such as =~ and quote-like operators such as m//, and perldoc perlre about regular expressions and their options such as /s.
I want to replace the below statement
ImageIcon("images/calender.gif");
with
ImageIcon(res.getResource("images/calender.gif"));
Can anyone suggest a regex to do this in eclipse.Instead of "calender.gif" any filename can come.
You can find this pattern (in regex mode):
ImageIcon\(("[^"]+")\)
and replace with:
ImageIcon(res.getResource($1))
The \( and \) in the pattern escapes the braces since they are to match literally. The unescaped braces (…) sets up capturing group 1 which matches the doublequoted string literal, which should not have escaped doublequotes (which I believe is illegal for filenames anyway).
The […] is a character class. Something like [aeiou] matches one of any of the lowercase vowels. [^…] is a negated character class. [^aeiou] matches one of anything but the lowercase vowels.
The + is one-or-more repetition, so [^"]+ matches non-empty sequence of everything except double quotes. We simply surround this pattern with " to match the double-quoted string literal.
So the pattern breaks down like this:
literal( literal)
| |
ImageIcon\(("[^"]+")\)
\_______/
group 1
In replacement strings, $1 substitutes what group 1 matched.
References
regular-expressions.info
Character Class, Repetition, Brackets
Examples/Programming Constructs - Strings - has patterns for strings that may contain escaped doublequotes
Ctrl-F
Find: ImageIcon\("([^\"]*)"\);
Replace with: ImageIcon(res.getResource("\1"));
Check Regular Expressions checkbox.