I cant figure out how to change this:
\usepackage{scrpage2}
\usepackage{pgf} \usepackage[latin1]{inputenc}\usepackage{times}\usepackage[T1]{fontenc}
\usepackage[colorlinks,citecolor=black,filecolor=black,linkcolor=black,urlcolor=black]{hyperref}
to this using sed only
REPLACED
REPLACED REPLACEDREPLACEDREPLACED
REPLACED
Im trying stuff like sed 's!\\.*\([.*]\)\?{.\+}!REPLACED!g' FILE
but that gives me
REPLACED
REPLACED
REPLACED
I think .* gets used and everything else in my pattern is just ignored, but I can't figure out how to go about this.
After I learned how to format a regex like that, my next step would be to change it to this:
\usepackage{scrpage2}
\usepackage{pgf}
\usepackage[latin1]{inputenc}
\usepackage{times}
\usepackage[T1]{fontenc}
\usepackage[colorlinks,citecolor=black,filecolor=black,linkcolor=black,urlcolor=black]{hyperref}
So I would appreciate any pointers in that direction too.
Here's some code that happens to work for the example you gave:
sed 's/\\[^\\[:space:]]\+/REPLACED/g'
I.e. match a backslash followed by one or more characters that are not whitespace or another backslash.
To make things more specific, you can use
sed 's/\\[[:alnum:]]\+\(\[[^][]*\]\)\?{[^{}]*}/REPLACED/g'
I.e. match a backslash followed by one or more alphanumeric characters, followed by an optional [ ] group, followed by a { } group.
The [ ] group matches [, followed by zero or more non-bracket characters, followed by ].
The { } group matches {, followed by zero or more non-brace characters, followed by }.
Perl to the rescue! It features the "frugal quantifiers":
perl -pe 's!\\.*?\.?{.+?}!REPLACED!g' FILE
Note that I removed the capturing group as you didn't use it anywhere. Also, [.*] matches either a dot or an asterisk, but you probably wanted to match a literal dot instead.
Related
I want to replace
$this->input->post("product_name");
with
$post_data["product_name"];
I want to use notepad++ regex, but I couldn't find proper solution
In find --> $this->input->post("[\*w\]");
In replace --> $post_data["$1"];
but its not working
The $this->input->post("[\*w\]"); pattern does not work because:
$ is a special char matching the end of a line, you need to use \$ to match it as a literal char
[\*w'\] is a malformed pattern as there is no matching unescaped ] for the [ that opens a character class. Also, w just matches w, not any letter, digit or underscore, \w does that.
You may use
Find What: \$this->input->post\("(\w*)"\);
Replace With: $post_data["$1"];
If there can be any char inside double quotes use .*? instead of \w*:
Find What: \$this->input->post\("(.*?)"\);
Regulex graph:
NPP test:
Use this pattern to match desired text \$this->input->post\(("[^"]+")\);
And replace it with pattern \$post_data\[\1\]
Explanation:
\$this->input->post - matach $this->input->post literally
\(("[^"]+")\); - match (literally, then match double quates and everything between them with "[^"]+" and store inside first capturing group, then match ); literally
To replace
$this->input->post("product_name");
by
$post_data["product_name"];
do replace, with regex activated
this->input->post\("(.*)"\);
by
post_data\["\1"\];
The \x with x a number, corresponds to the x-th match catched with the parenthesis. Here we catch any character inside this->input->post(XXXX);
Don't forget to escape special character with \.
Your special characters were []()
I have a function, translate(), takes multiple parameters. The first param is the only required and is a string, that I always wrap in single quotes, like this:
translate('hello world');
The other params are optional, but could be included like this:
translate('hello world', true, 1, 'foobar', 'etc');
And the string itself could contain escaped single quotes, like this:
translate('hello\'s world');
To the point, I now want to search through all code files for all instances of this function call, and extract just the string. To do so I've come up with the following grep, which returns everything between translate(' and either ') or ',. Almost perfect:
grep -RoPh "(?<=translate\(').*?(?='\)|'\,)" .
The problem with this though, is that if the call is something like this:
translate('hello \'world\', you\'re great!');
My grep would only return this:
hello \'world\
So I'm looking to modify this so that the part that currently looks for ') or ', instead looks for the first occurrence of ' that hasn't been escaped, i.e. doesn't immediately follow a \
Hopefully I'm making sense. Any suggestions please?
You can use this grep with PCRE regex:
grep -RoPh "\btranslate\(\s*\K'(?:[^'\\\\]*)(?:\\\\.[^'\\\\]*)*'" .
Here is a regex demo
RegEx Breakup:
\b # word boundary
translate # match literal translate
\( # match a (
\s* # match 0 or more whitespace
\K # reset the matched information
' # match starting single quote
(?: # start non-capturing group
[^'\\\\]* # match 0 or more chars that are not a backslash or single quote
) # end non-capturing group
(?: # start non-capturing group
\\\\. # match a backslash followed by char that is "escaped"
[^'\\\\]* # match 0 or more chars that are not a backslash or single quote
)* # end non-capturing group
' # match ending single quote
Here is a version without \K using look-arounds:
grep -oPhR "(?<=\btranslate\(')(?:[^'\\\\]*)(?:\\\\.[^'\\\\]*)*(?=')" .
RegEx Demo 2
I think the problem is the .*? part: the ? makes it a non-greedy pattern, meaning it'll take the shortest string that matches the pattern. In effect, you're saying, "give me the shortest string that's followed by quote+close-paren or quote+comma". In your example, "world\" is followed by a single quote and a comma, so it matches your pattern.
In these cases, I like to use something like the following reasoning:
A string is a quote, zero or more characters, and a quote: '.*'
A character is anything that isn't a quote (because a quote terminates the string): '[^']*'
Except that you can put a quote in a string by escaping it with a backslash, so a character is either "backslash followed by a quote" or, failing that, "not a quote": '(\\'|[^'])*'
Put it all together and you get
grep -RoPh "(?<=translate\(')(\\'|[^'])*(?='\)|'\,)" .
I am using Notepad++ to remove some unwanted strings from the end of a pattern and this for the life of me has got me.
I have the following sets of strings:
myApp.ComboPlaceHolderLabel,
myApp.GridTitleLabel);
myApp.SummaryLabel + '</b></div>');
myApp.NoneLabel + ')') + '</label></div>';
I would like to leave just myApp.[variable] and get rid of, e.g. ,, );, + '...', etc.
Using Notepad++, I can match the strings themselves using ^myApp.[a-zA-Z0-9].*?\b (it's a bit messy, but it works for what I need).
But in reality, I need negate that regex, to match everything at the end, so I can replace it with a blank.
You don't need to go for negation. Just put your regex within capturing groups and add an extra .*$ at the last. $ matches the end of a line. All the matched characters(whole line) are replaced by the characters which are present inside the first captured group. .
matches any character, so you need to escape the dot to match a literal dot.
^(myApp\.[a-zA-Z0-9].*?\b).*$
Replacement string:
\1
DEMO
OR
Match only the following characters and then replace it with an empty string.
\b[,); +]+.*$
DEMO
I think this works equally as well:
^(myApp.\w+).*$
Replacement string:
\1
From difference between \w and \b regular expression meta characters:
\w stands for "word character", usually [A-Za-z0-9_]. Notice the inclusion of the underscore and digits.
(^.*?\.[a-zA-Z]+)(.*)$
Use this.Replace by
$1
See demo.
http://regex101.com/r/lU7jH1/5
I have a file which have the data something like this
34sdf, 434ssdf, 43fef,
34sdf, 434ssdf, 43fef, sdfsfs,
I have to identify the sdfsfs, and replace it and/or print the line.
The exact condition is the tokens are comma separated. target expression starts with a non numeric character, and till a comma is met.
Now i start with [^0-9] for starting with a non numeric character, but the next character is really unknown to me, it can be a number, a special char, an alphabet or even a space. So I wanted a (anything)*. But the previous [] comes into play and spoils it. [^0-9]* or [^0-9].*, or [^0-9]\+.*, or [^0-9]{1}*, or [^0-9][^,]* or [^0-9]{1}[^\,]*, nothing worked till now. So my question is how to write a regex for this (starting character a non numeric, then any character except a comma or any number of character till comma) I am using grep and sed (gnu). Another question is for posix or non-posix, any difference comes there?
Something like that maybe?
(?:(?:^(\D.*?))|(?:,\s(\D.*?))),
This captures the string that starts with a non-numeric character. Tested here.
I'm not sure if sed supports \D, but you can easily replace it with [^0-9] if not, which you already know.
EDIT: Can be trimmed to:
(?:\s|^)(\D.*?),
With sed, and slight modifications to your last regex:
sed -n 's/.*,[ ]*\([^ 0-9][^\,]*\),/\1/p' input
I think pattern (\s|^)(\D[^,]+), will catch it.
It matches white-space or start of string and group of a non-digit followed by anything but comma, which is followed by comma.
You can use [^0-9] if \D is not supported.
This might work for you (GNU sed):
sed '/\b[^0-9,][^,]*/!d' file # only print lines that match
or:
sed -n 's/\b[^0-9,][^,]*/XXX/gp' file # substitute `XXX` for match
I want to do this:
%s/shop_(*)/shop_\1 wp_\1/
Why doesn't shop_(*) match anything?
There's several issues here.
parens in vim regexen are not for capturing -- you need to use \( \) for captures.
* doesn't mean what you think. It means "0 or more of the previous", so your regex means "a string that contains shop_ followed by 0+ ( and then a literal ). You're looking for ., which in regex means "any character". Put together with a star as .* it means "0 or more of any character". You probably want at least one character, so use .\+ (+ means "1 or more of the previous")
Use this: %s/shop_\(.\+\)/shop_\1 wp_\1/.
Optionally end it with g after the final slash to replace for all instances on one line rather than just the first.
If I understand correctly, you want %s/shop_\(.*\)/shop_\1 wp_\1/
Escape the capturing parenthesis and use .* to match any number of any character.
(Your search is searching for "shop_" followed by any number of opening parentheses followed by a closing parenthesis)
If you would like to avoid having to escape the capture parentheses and make the regex pattern syntax closer to other implementations (e.g. PCRE), add \v (very magic!) at the start of your pattern (see :help \magic for more info):
:%s/\vshop_(*)/shop_\1 wp_\1/
#Luc if you look here: regex-info, you'll see that vim is behaving correctly. Here's a parallel from sed:
echo "123abc456" | sed 's#^([0-9]*)([abc]*)([456]*)#\3\2\1#'
sed: -e expression #1, char 35: invalid reference \3 on 's' command's RHS
whereas with the "escaped" parentheses, it works:
echo "123abc456" | sed 's#^\([0-9]*\)\([abc]*\)\([456]*\)#\3\2\1#'
456abc123
I hate to see vim maligned - especially when it's behaving correctly.
PS I tried to add this as a comment, but just couldn't get the formatting right.