I have an executable file called test.script containing this simple bash script:
#!/bin/bash
temp=${1//$'\n'/}
output=${temp//$'\r'/}
printf "$output" > output.txt
When I run
sudo ./test.script "^\r\r\n\n\r\n\r\n\r\n\r\n\n\r\n\rHello World\n\n\r\r\r\n\n\r\r\r$"
in the same directory as test.script, I expect to end up with an output.txt looking like this:
^Hello World$
but when I take a look I instead see this:
^
Hello World
$
Clearly I have a misunderstanding about regex in bash.
Please explain to me what I am missing, then show me how to write the bash so that all newline characters are removed from the string before said string is written to a file. Thanks in advance.
You can "fix" your script like this (although I must say this isn't typical usage of printf):
#!/bin/bash
temp=${1//'\n'/}
output=${temp//'\r'/}
printf "$output"
The argument to your script $1 doesn't contain real newlines or carriage returns, which is what $'\n' and $'\r' are for. Instead, it looks like you just want to remove the literal strings '\n' and '\r'.
To elaborate on my point about printf, normally two (or more) arguments are passed: the format specifier and the variables that are to be inserted. For example, to print a single string you would use something like printf '%s' "$output". In your script, the variable $output is being treated as the format specifer; you're relying on printf to expand your \n and \r into newlines and carriage returns.
You're not actually using regular expressions here by the way; the syntax ${var//match/replace} is a substring replacement, where // means that all occurrences of the substring match in $var are replaced. As you haven't specified anything to replace the substring with, the substring is replaced with nothing (i.e. removed).
Related
sample_file:
this is a test line!
this is a test line!
this is a test line!
this is a test line!
I'm using a Perl one-liner to replace text like this:
perl -pi -e 's{this(.*)\s}{\n\1\n}i && s{.*\n}{}' sample_file
The this(.*)\s part and \1 part are variable and I cannot change that (it comes from user input).
My problem is that I need to adjust the {\n\1\n} part depending on whether the first regex includes the newline character .
For example, if the first regex is {this(.*)\s} I need {\n\1\n}, but if the first is like {(.*)a test} I need {\n\1}.
How can I check whether the newline character is lost and put it back if necessary?
Generally speaking, you want to chomp inputs lines, and add a newline to output lines. -l (in conjunction with -n or -p) will do both.
For example, the following doesn't replace the newline with ! because it was removed by -l (and subsequently re-added by the print).
perl -i -ple's/\s/!/g' file
By the way, \1 ("match what the first capture captured") makes no sense in a substitution. You want $1 (as -w would tell you).
I'm trying to write a bash function that would escape all double quotes within single quotes, eg:
'I need to escape "these" quotes with backslashes'
would become
'I need to escape \"these\" quotes with backslashes'
My take on it was:
Find pairs of single quotes in the input and extract them with grep
Pipe into sed, escape double quotes
Sed again the whole input and replace grep match with sedded match
I managed to get it working to the part of having correctly escaped quotes section, but replacing it in the whole input fails.
The script code copypaste:
# $1 - Full name, $2 - minified name
adjust_quotes ()
{
SINGLE_QUOTES=`grep -Eo "'.*'" $2`
ESCAPED_QUOTES=`echo $SINGLE_QUOTES | sed 's|"|\\\\"|g'`
sed -r "s|'.*'|$ESCAPED_QUOTES|g" "$2" > "$2.escaped"
mv "$2.escaped" $2
echo "Quotes escaped within single quotes on $2"
}
Random additional questions:
In the console, escaping the quote with only two backslashes works, but when code is put in the script - I need four. I'd love to know
Could I modify this code into a loop to escape all pairs of single quotes, one after another until EOF?
Thanks!
P.S. I know this would probably be easier to do in eg. python, but I really need to keep it in bash.
Using BASH string replacement:
s='I need to escape "these" quotes with backslashes'
r="${s//\"/\\\"}"
echo "$r"
I need to escape \"these\" quotes with backslashes
Here's a pure bash solution, which does the transformation on stdin, printing to stdout. It reads the entire input into memory, so it won't work with really enormous files.
escape_enclosed_quotes() (
IFS=\'
read -d '' -r -a fields
for ((i=1; i<${#fields[#]}; i+=2)); do
fields[i]=${fields[i]//\"/\\\"}
done
printf %s "${fields[*]}"
)
I deliberately enclosed the body of the function in parentheses rather than braces, in order to force the body to run in a subshell. That limits the modification of IFS to the body, as well as implicitly making the variables used local.
The function uses the read builtin to read the entire input (since the line delimiter is set to NUL with -d '') into an array (-a) using a single quote as the field separator (IFS=\'). The result is that the parts of the input surrounded with single quotes are in the odd positions of the array, so the function loops over the odd indices to do the substitution only for those fields. I use bash's find-and-replace syntax instead of deferring to an external utility like sed.
This being bash, there are a couple of gotchas:
If the file contains a NUL, the rest of the file will be ignored.
If the last line of the file does not end with a newline, and the last character of that line is a single quote, it will not be output.
Both of the above conditions are impossible in a portable text file, so it's probably OK. All the same, worth taking note.
The supplementary question: why are the extra backslashes needed in
ESCAPED_QUOTES=`echo $SINGLE_QUOTES | sed 's|"|\\\\"|g'`
Answer: It has nothing to do with that line being in a script. It has to do with your use of backticks (...) for command substitution, and the idiosyncratic and often unpredictable handling of backslashes inside backticks. This syntax is deprecated. Do not use it. (Not even if you see someone else using it in some random example on the internet.) If you had used the recommended $(...) syntax for command substitution, it would have worked as expected:
ESCAPED_QUOTES=$(echo $SINGLE_QUOTES | sed 's|"|\\"|g')
(More information is in the Bash FAQ linked above.)
I'll keep this explanation of why I need help to a mimimum. One of my file directories got hacked through XSS and placed a long string at the beginning of all php files. I've tried to use sed to replace the string with nothing but it won't work because the pattern to match includes many many characters that would need to be escaped.
I found out that I can use fgrep to match a fixed string saved in a pattern file, but I'd like to replace the matched string (NOT THE LINE) in each file, but grep's -v inverts the result on the line, rather than the end of the matched string.
This is the command I'm using on an example file that contains the hacked
fgrep -v -f ~/hacked-string.txt example.php
I need the output to contain the <?php that's at the end of the line (sometimes it's a <style> tag), but the -v option inverts at the end of that line, so the output doesn't contain the <?php at the beginning.
NOTE
I've tried to use the -o or --only-matching which outputs nothing instead:
fgrep -f ~/hacked-string.txt example.php --only-matching -v
Is there another option in grep that I can use to invert on the end of the matched pattern, rather than the line where the pattern was matched? Or alternatively, is there an easier option to replace the hacked string in all .php files?
Here is a small snippet of what's in hacked-string.txt (line breaks added for readability):
]55Ld]55#*<%x5c%x7825bG9}:}.}-}!#*<%x55c%x7825)
dfyfR%x5c%x7827tfs%x5c%x7c%x785c%x5c%x7825j:^<!
%x5c%x7825w%x5c%x7860%x5c%x785c^>Ew:25tww**WYsb
oepn)%x5c%x7825bss-%x5c%x7825r%x5c%x7878B%x5c%x
7825h>#]y3860msvd},;uqpuft%x5c%x7860msvd}+;!>!}
%x5c%x7827;!%x5c%x7825V%x5c%x7827{ftmfV%x5e56+9
9386c6f+9f5d816:+946:ce44#)zbssb!>!ssbnpe_GMFT%
x5c5c%x782f#00#W~!%x5c%x7825t2w)##Qtjw)#]82#-#!
#-%x5c%x7825tmw)%x5c%x78w6*%x5c%x787f_*#fubfsdX
k5%x5c%xf2!>!bssbz)%x5c%x7824]25%x5c%x7824-8257
-K)fujs%x5c%x7878X6<#o]o]Y%x5c%x78257;utpI#7>-1
-bubE{h%x5c%x7825)sutcvt)!gj!|!*bubEpqsut>j%x5c
%x7825!*72!%x5c%x7827!hmg%x5c%x78225>2q%x5c%x7
Thanks in advance!
I think what you are asking is this:
"Is it possible to use the grep utility to remove all instances of a fixed string (which might contain lots of regex metacharacters) from a file?"
In that case, the answer is "No".
What I think you wanted to ask was:
"What is the easiest way to remove all instances of a fixed string (which might contain lots of regex metacharacters) from a file?"
Here's one reasonably simple solution:
delete_string() {
awk -v s="$the_string" '{while(i=index($0,s))$0=substr($0,1,i-1)substr($0,i+length(s))}1'
}
delete_string 'some_hideous_string_with*!"_inside' < original_file > new_file
The shell syntax is slightly fragile; it will break if the string contains an apostrophe ('). However, you can read a raw string from stdin into a variable with:
$ IFS= read -r the_string
absolutely anything here
which will work with any string which doesn't contain a newline or a NUL character. Once you have the string in a variable, you can use the above function:
delete_string "$the_string" < original_file > new_file
Here's another possible one liner, using python:
delete_string() {
python -c 'import sys;[sys.stdout.write(l.replace(r"""'"$1"'""","")) for l in sys.stdin]'
}
This won't handle strings which have three consecutive quotes (""").
Is the hacked string the same in every file?
If the length of hacked string in chars was 1234 then you can use
tail -c +1235 file.php > fixed-file.php
for each infected file.
Note that tail c +1235 tells to start output at 1235th character of the input file.
With perl:
perl -i.hacked -pe "s/\Q$(<hacked-string.txt)\E//g" example.php
Notes:
The $(<file) bit is a bash shortcut to read the contents of a file.
The \Q and \E bits are from perl, they treat the stuff in between as plain characters, ignoring regex metachars.
The -i.hacked option will edit the file in-place, creating a backup "example.php.hacked"
I have a file that looks like this:
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
I want it to look like this
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
I thought I could use sed to do this but I can't figure out how to store something in a buffer and then modify it.
Am I even using the right tool?
Thanks
You don't have to get tricky with regular expressions and replacement strings: use sed's p command to print the line intact, then modify the line and let it print implicitly
sed 'p; s/\.png//'
Glenn jackman's response is OK, but it also doubles the rows which do not match the expression.
This one, instead, doubles only the rows which matched the expression:
sed -n 'p; s/\.png//p'
Here, -n stands for "print nothing unless explicitely printed", and the p in s/\.png//p forces the print if substitution was done, but does not force it otherwise
That is pretty easy to do with sed and you not even need to use the hold space (the sed auxiliary buffer). Given the input file below:
$ cat input
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
you should use this command:
sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
The result:
$ sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
This commands is just a replacement command (s///). It matches anything starting with #" followed by non-period chars ([^.]*) and then by .png",. Also, it matches all non-period chars before .png", using the group brackets \( and \), so we can get what was matched by this group. So, this is the to-be-replaced regular expression:
#"\([^.]*\)\.png",
So follows the replacement part of the command. The & command just inserts everything that was matched by #"\([^.]*\)\.png", in the changed content. If it was the only element of the replacement part, nothing would be changed in the output. However, following the & there is a newline character - represented by the backslash \ followed by an actual newline - and in the new line we add the #" string followed by the content of the first group (\1) and then the string ",.
This is just a brief explanation of the command. Hope this helps. Also, note that you can use the \n string to represent newlines in some versions of sed (such as GNU sed). It would render a more concise and readable command:
sed 's/#"\([^.]*\)\.png",/&\n#"\1",/' input
I prefer this over Carles Sala and Glenn Jackman's:
sed '/.png/p;s/.png//'
Could just say it's personal preference.
or one can combine both versions and apply the duplication only on lines matching the required pattern
sed -e '/^#".*\.png",/{p;s/\.png//;}' input
The regex:
^ *x *\=.*$
means "match a literal x preceded by an arbitrary count of spaces, followed by an arbitrary count of spaces, then an equal sign and then anything up to the end of line." Sed was invoked as:
sed -r -e 's|^ *x *\=.*$||g' file
However it doesn't find a single match, although it should. What's wrong with the regex?
To all: thanks for the answers and effort! It seems that the problem was in tabs present in input file, which are NOT matched by the space specifier ' '. However the solution with \s works regardless of present tabs!
^\s*x\s*=.*$
Maybe you must escape some chars, figure it out one by one.
BTW: Regex tags should really have three requirements:
what is the input string, what is the output string and your platform/language.
sed processes the file line-by-line, and executes the given program for each. The simplest program that does what you want is
sed -re '/^ *x *=.*$/!d' file
"/^ *x *=.*$/" selects each line that matches the pattern.
"!" negates the result.
"d" deletes the line.
sed will by default print all lines unless told otherwise. This effectively prints lines that matches the pattern.
One alternative way of writing it is:
sed -rne '/^ *x *=.*$/p' file
"/^ *x *=.*$/" selects each line that matches the pattern.
"p" prints the line.
The difference here is that I used the "-n" switch to suppress the automatic printing of lines, and instead print only the lines I have selected.
You can also use "grep" for this task:
grep -E '^ *x *=.*$' file
or maybe '^[ ]*x[ ]*='. It's a bit more compatible, but will not match tabs or the like. And, if you don't need groups, why bother about the rest of the line?