I have tried everything possible and 2 hours trying to get this to work.
I'm using the search and replace option in Code (version 1) but just can't get the replace code working.
I want to add a semi colon in between some characters. I search like this:
)}[a-z]
..and that works fine.
and I've tried all of these, but not one keeps the [a-z] part, they all remove/replace it.
)};\1
)};*1
)};[a-z]
)};\1\2
)};$1
)};($1)
)};(\1)
)};[$1]
)};[\1]
I've set it to use the POSIX Basic option, but any option, setting will do. I just need it to simply insert a semicolon
This should work:
s/\)\}([a-z])/)};\1/g
In essence, we literally match a close parenthesis followed by a close brace. We then match a letter between a and z, but the parentheses around the [a-z] make that a capture group that we can reference. We can then (and only then) use \1 in the replacement. Without a capture group, it doesn't know what \1 refers to. With one, it does.
Using it with sed:
% echo 'hello)}world' | sed -Ee 's/\)\}([a-z])/)};\1/g'
hello)};world
Related
I have a code with object.attribute where attribute can be an array
example: object.SIZE_OF_IMAGE[0] or a simple string. I want to search all occurrences "object.attribute" and replace it with self.lowercase(attribute) I want a regular expression on vim to do that.
I can use that :%s/object.*/self./gc and replace it manually but it is very slow.
Here are some examples:
object.SIZE to self.size
object.SIZE_OF_IMAGE[0] to self.size_of_image[0]
You basically just need two things:
Capture groups :help /\( let you store what's matched in between \(...\) and then reference it (via \1, \2, etc.) in the replacement (or even afterwards in the pattern itself).
The :help s/\L special replacement action that makes everything following lowercase.
This gives you the following command:
:%substitute/\<object\.\(\w\+\)/self.\L\1/g
Notes:
I've established a keyword start assertion (\<) at the beginning to avoid matching schlobject as well.
\w\+ matches letters, digits, and underscores (so it fulfills your example); various alternatives are possible here.
sed -E 's/object\.([^ \(]*)(.*)/self.lowercase(\1)\2/g' file_name.txt
above command considers that your attribute is followed by space or "("
you can tweek this command based on your need
Based on your comment above that the attribute part
"finishes by space or [ or (" you could match it with:
/object\.[^ [(]*
So, to replace it with self.attribute use a capturing
group and \L to make everything lowercase:
:%s/\vobject\.([^ [(]*)/self.\L\1/g
In the command mode try this
:1,$ s/object.attribute/self.lowercase(attribute)/g
I am using sublime to search and replace text in a code.
This is what I want to find
pre + <anyvariable> + post
and then replace it like this
sanitize(<anyvariable>)
I don't know how to come up with regexp after finding "pre + "
Came up with something like this
/(\pre \+)(?=.*)/g
Try this:
/^(.*\+)\s*(\S+)\s*(\+.*)$/
This will give you everything in three capture groups, starting from the beginning of the line between the first and second +, and then everything following to the end of the line. If there's other content such as more +'s, then this probably won't work blanketly.
Without a computer, and not sure a out what you want, but if you search for:
pre \+ (.*?) \+ post
And replace with:
sanityze($1) (or \1 I never remember)
It should do what you want :-)
If you want to be able to have linebreaks, replace .* by (?:.|\n)
I've got a file with several (1000+) records like :
lbc3.*'
ssa2.*'
lie1.*'
sld0.*'
ssdasd.*'
I can find them all by :
/s[w|l].*[0-9].*$
What i want to do is to replace the final part of each pattern found with \.*'
I can't do :%s//s[w|l].*[0-9].*$/\\\\\.\*' because it'll replace all the string, and what i need is only replace the end of it from
.'
to
\.'
So the file output is llike :
lbc3\\.*'
ssa2\\.*'
lie1\\.*'
sld0\\.*'
ssdasd\\.*'
Thanks.
In general, the solution is to use a capture. Put \(...\) around the part of the regex that matches what you want to keep, and use \1 to include whatever matched that part of the regex in the replacement string:
s/\(s[w|l].*[0-9].*\)\.\*'$/\1\\.*'/
Since you're really just inserting a backslash between two strings that you aren't changing, you could use a second set of parens and \2 for the second one:
s/\(s[w|l].*[0-9].*\)\(\.\*'\)$/\1\\\2/
Alternatively, you could use \zs and \ze to delimit just the part of the string you want to replace:
s/s[w|l].*p0-9].*\zs\ze\*\'$/\\/
I'm trying to use regex to find single quotes (so I can turn them all into double quotes) anywhere in a line that starts with mySqlQueryToArray (a function that makes a query to a SQL DB). I'm doing the regex in Sublime Text 3 which I'm pretty sure uses Perl Regex. I would like to have my regex match with every single quote in a line so for example I might have the line:
mySqlQueryToArray($con, "SELECT * FROM Template WHERE Name='$name'");
I want the regex to match in that line both of the quotes around $name but no other characters in that line. I've been trying to use (?<=mySqlQueryToArray.*)' but it tells me that the look behind assertion is invalid. I also tried (?<=mySqlQueryToArray)(?<=.*)' but that's also invalid. Can someone guide me to a regex that will accomplish what I need?
To find any number of single quotes in a line starting with your keyword you can use the \G anchor ("end of last match") by replacing:
(^\h*mySqlQueryToArray|(?!^)\G)([^\n\r']*)'
With \1\2<replacement>: see demo here.
Explanation
( ^\h*mySqlQueryToArray # beginning of line: check the keyword is here
| (?!^)\G ) # if not at the BOL, check we did match sth on this line
( [^\n\r']* ) ' # capture everything until the next single quote
The general idea is to match everything until the next single quote with ([^\n\r']*)' in order to replace it with \2<replacement>, but do so only if this everything is:
right after the beginning keyword (^mySqlQueryToArray), or
after the end of the last match ((?!^)\G): in that case we know we have the keyword and are on a relevant line.
\h* accounts for any started indent, as suggested by Xælias (\h being shortcut for any kind of horizontal whitespace).
https://stackoverflow.com/a/25331428/3933728 is a better answer.
I'm not good enough with RegEx nor ST to do this in one step. But I can do it in two:
1/ Search for all mySqlQueryToArray strings
Open the search panel: ⌘F or Find->Find...
Make sure you have the Regex (.* ) button selected (bottom left) and the wrap selector (all other should be off)
Search for: ^\s*mySqlQueryToArray.*$
^ beginning of line
\s* any indentation
mySqlQueryToArray your call
.* whatever is behind
$ end of line
Click on Find All
This will select every occurrence of what you want to modify.
2/ Enter the replace mode
⌥⌘F or Find->Replace...
This time, make sure that wrap, Regex AND In selection are active .
Them search for '([^']*)' and replace with "\1".
' are your single quotes
(...) si the capturing block, referenced by \1 in the replace field
[^']* is for any character that is not a single quote, repeated
Then hit Replace All
I know this is a little more complex that the other answer, but this one tackles cases where your line would contain several single-quoted string. Like this:
mySqlQueryToArray($con, "SELECT * FROM Template WHERE Name='$name' and Value='1234'");
If this is too much, I guess something like find: (?<=mySqlQueryToArray)(.*?)'([^']*)'(.*?) and replace it with \1"\2"\3 will be enough.
You can use a regex like this:
(mySqlQueryToArray.*?)'(.*?)'(.*)
Working demo
Check the substitution section.
You can use \K, see this regex:
mySqlQueryToArray[^']*\K'(.*?)'
Here is a regex demo.
Im trying to use the following regex to search and replace in multiple files in notepad++
([^\n]*)(state="1")([^\n]*)*.
This searches and finds state="1" in the first line of each file and works fine.
However, when I try to replace state="1" using:
Replace with: $1 state="5"
it cuts off the rest of the line.
I thought that it might be possible to get the rest of the line using:
Replace with: $1 state="5" $2
However, $2 doesnt seem to exist as a variable.
Is there some way to attach the rest of the line into variable $2?
Cheers
Heres an image to show how
(?=\A[^\n]*)state="1"
is not working
Ive updated my version of notepad++ and everything
Each capture group, (…), is assigned a number, so $2 represents the second capture group, (state="1"). The remainder of the line is captured in $3.
Either remove the capture group around state="1":
([^\n]*)state="1"([^\n]*)*.
Or use $3:
Replace with: $1 state="5" $3
Also, given the simplicity of the task, I don't see why you couldn't just search for state="1" and replace with state="5". There doesn't seem to be any need for regular expressions here.
Update There's nothing in the pattern listed so far which limits the result to only matching strings on the first line. If you need that I'd recommend using a pattern like this:
(?=\A[^\n]*)state="1"
With these settings:
Update There seems to be some strange behavior with the \A (beginning of text) anchor inside the lookbehind. Removing from the lookbehind seems to work. Try this pattern:
\A([^\n]*)state="1"
And replace with:
$1state="5"
All the other settings should be fine.