I'm trying to find a number of lines that match a regex pattern in grep received as a variable. When I do the grep with the pattern directly in the command substitution, it works. When I use a variable for the pattern, it doesn't.
#!/bin/bash
pattern="'^\\\".*\\\"$'"
echo "pattern : $(echo $pattern)"
NB=$(grep -c -E -v -e ${pattern} abc.txt)
NB2=$(grep -v -c -E -e '^\".*\"$' abc.txt)
echo " -> $NB , $NB2"
Besides what's in the code, I've tried:
NB=$(grep -c -E -v -e $(echo $pattern) abc.txt)
No success.
cmd="grep -c -E -v -e ${pattern} abc.txt"
NB="$($cmd)"
No success.
In the example, abc.txt file contains 3 lines:
"abc"
"abc
abc"
The pattern in the variable seems ok:
pattern : '^\".*\"$'
I'm expecting that the 2 numbers in NB and NB2 are the same. If you look in the code, the actual result is:
pattern : '^\".*\"$'
-> 3 , 2
I expect:
pattern : '^\".*\"$'
-> 2 , 2
NB2=$(grep -v -c -E -e '^\".*\"$' abc.txt)
If that works, then assign that exact regex to $pattern. Don't add more backslashes and quotes.
pattern='^\".*\"$'
It's always a good idea to quote variable expansions to prevent unwanted wildcard expansion and word splitting.
NB=$(grep -c -E -v -e "${pattern}" abc.txt)
# ^ ^
Related
i try to get the char before a substring in bash.
Some examples of the String:
label="LLL:EXT:xxx/Resources/Private/Language/locallang.xlf:flux.content"
or:
label='LLL:EXT:xxx/Resources/Private/Language/locallang.xlf:flux.content'
or:
$this->getLanguageService()->sL('LLL:EXT:xxx/Resources/Private/Language/locallang.xlf:flux.content'
i need to know if there is a single- or doublequote before the LLL. :)
You can get the character using pure Bash code with:
[[ $string =~ (.)LLL ]] && char=${BASH_REMATCH[1]}
See the sections on [[...]] and BASH_REMATCH in the Bash Reference Manual.
You are asking whether or not there is a ' or " before LLL. The shortest answer would be grep -q "['\"]LLL".
If the exit code is 0, then the answer is yes, if the exit code is 1, then the answer is no.
For example:
$ echo 'label="LLL:EXT:xxx/Resources/Private/Language/locallang.xlf:flux.content"' | grep -q "['\"]LLL"
$ echo $?
0
$ echo test | grep -q "['\"]LLL"
$ echo $?
1
You can use grep for this.
grep -E -o '.{0,1}LLL' file
grep -E -o '.{0,1}LLL' file | cut -c-1 # shows only the first character
-E indicates the pattern is an extended regular expression
-o shows only the matching portion
'.{0,1}LLL' specifies the pattern with the first preceding character appended
I am using the sed command on Ubuntu to replace content.
This initial command comes from here.
sed -i '$ s/$/ /replacement/' "$DIR./result/doc.md"
However, as you can see, I have a slash in the replacement. The slash causes the command to throw:
sed: -e expression #1, char 9: unknown option to `s'
Moreover, my replacement is stored in a variable.
So the following will not work because of the slash:
sed -i "$ s/$/ $1/" "$DIR./result/doc.md"
As stated here and in duplicate, I should use another delimiter. If I try with #:
sed -i "$ s#$# $1#" "$DIR./result/doc.md"
It gives the error:
sed: -e expression #1, char 42: unterminated `s' command
My question is:
How can I use a variable in this command as well as other delimiter than / ?
Don't use sed here; perl and awk allow more robust approaches.
sed doesn't allow variables to be passed out-of-band from code, so they always need to be escaped. Use a language without that limitation, and you have code that always works, no matter what characters your data contains.
The Short Answer: Using perl
The below is taken from BashFAQ #21:
inplace_replace() {
local search=$1; shift; local replace=$1; shift
in="$search" out="$replace" perl -pi -e 's/\Q$ENV{"in"}/$ENV{"out"}/g' "$#"
}
inplace_replace '#' "replacement" "$DIR/result/doc.md"
The Longer Answer: Using awk
...or, using awk to do a streaming replacement, and a shell function to make that file replacement instead:
# usage as in: echo "in should instead be out" | gsub_literal "in" "out"
gsub_literal() {
local search=$1 replace=$2
awk -v s="${search//\\/\\\\}" -v r="${rep//\\/\\\\}" 'BEGIN {l=length(s)} {o="";while (i=index($0, s)) {o=o substr($0,1,i-1) r; $0=substr($0,i+l)} print o $0}'
}
# usage as in: inplace_replace "in" "out" /path/to/file1 /path/to/file2 ...
inplace_replace() {
local search=$1 replace=$2 retval=0; shift; shift
for file; do
tempfile=$(mktemp "$file.XXXXXX") || { retval |= $?; continue; }
if gsub_literal "$search" "$replace" <"$file" >"$tempfile"; then
mv -- "$tempfile" "$file" || (( retval |= $? ))
else
rm -f -- "$tempfile" || (( retval |= $? ))
fi
done
}
TL;DR:
Try:
sed -i '$ s#$# '"$1"'#' "$DIR./result/doc.md"
Long version:
Let's start with your original code:
sed -i '$ s/$/ /replacement/' "$DIR./result/doc.md"
And let's compare it to the code you referenced:
sed -i '$ s/$/abc/' file.txt
We can see that they don't exactly match up. I see that you've correctly made this substitution:
file.txt --> "$DIR./result/doc.md"
That looks fine (although I do have my doubts about the . after $DIR ). However, the other substitution doesn't look great:
abc --> /replacement
You actually introduced another delimeter /. However, if we replace the delimiters with '#' we get this:
sed -i '$ s#$# /replacement#' "$DIR./result/doc.md"
I think that the above is perfectly valid in sed/bash. The $# will not be replaced by the shell because it is single quoted. The $DIR variable will be interpolated by the shell because it is double quoted.
Looking at one of your attempts:
sed -i "$ s#$# $1#" "$DIR./result/doc.md"
You will have problems due to the shell interpolation of $# in the double quotes. Let's correct that by replacing with single quotes (but leaving $1 unquoted):
sed -i '$ s#$# '"$1"'#' "$DIR./result/doc.md"
Notice the '"$1"'. I had to surround $1 with '' to basically unescape the surrounding single quotes. But then I surrounded the $1 with double quotes so we could protect the string from white spaces.
Use shell parameter expansion to add escapes to the slashes in the variable:
$ cat file
foo
bar
baz
$ set -- ' /repl'
$ sed "s/$/$1/" file
sed: 1: "s/$/ /repl/": bad flag in substitute command: 'r'
$ sed "s/$/${1//\//\\\/}/" file
foo /repl
bar /repl
baz /repl
That is a monstrosity of leaning toothpicks, but it serves to transform this:
sed "s/$/ /repl/"
into
sed "s/$/ \/repl/"
The same technique can be used for whatever you choose as the sed s/// delimiter.
I used many times [``] to capture output of command to a variable. but with following code i am not getting right output.
#!/bin/bash
export XLINE='($ZWP_SCRIP_NAME),$ZWP_LT_RSI_TRIGGER)R),$ZWP_RTIMER'
echo 'Original XLINE'
echo $XLINE
echo '------------------'
echo 'Extract all word with $ZWP'
#works fine
echo $XLINE | sed -e 's/\$/\n/g' | sed -e 's/.*\(ZWP[_A-Z]*\).*/\1/g' | grep ZWP
echo '------------------'
echo 'Assign all word with $ZWP to XVAR'
#XVAR doesn't get all the values
export XVAR=`echo $XLINE | sed -e 's/\$/\n/g' | sed -e 's/.*\(ZWP[_A-Z]*\).*/\1/g' | grep ZWP` #fails
echo "$XVAR"
and i get:
Original XLINE
($ZWP_SCRIP_NAME),$ZWP_LT_RSI_TRIGGER)R),$ZWP_RTIMER
------------------
Extract all word with $ZWP
ZWP_SCRIP_NAME
ZWP_LT_RSI_TRIGGER
ZWP_RTIMER
------------------
Assign all word with $ZWP to XVAR
ZWP_RTIMER
why XVAR doesn't get all the values?
however if i use $() to capture the out instead of ``, it works fine. but why `` is not working?
Having GNU grep you can use this command:
XVAR=$(grep -oP '\$\KZWP[A-Z_]+' <<< "$XLINE")
If you pass -P grep is using Perl compatible regular expressions. The key here is the \K escape sequence. Basically the regex matches $ZWP followed by one or more uppercase characters or underscores. The \K after the $ removes the $ itself from the match, while its presence is still required to match the whole pattern. Call it poor man's lookbehind if you want, I like it! :)
Btw, grep -o outputs every match on a single line instead of just printing the lines which match the pattern.
If you don't have GNU grep or you care about portability you can use awk, like this:
XVAR=$(awk -F'$' '{sub(/[^A-Z_].*/, "", $2); print $2}' RS=',' <<< "$XLINE")
First, the smallest change that makes your code "work":
echo "$XLINE" | tr '$' '\n' | sed -e 's/.*\(ZWP[_A-Z]*\).*/\1/g' | grep ZWP_
The use of tr replaces a sed expression that didn't actually do what you thought it did -- try looking at its output to see.
One sane alternative would be to rely on GNU grep's -o option. If you can't do that...
zwpvars=( ) # create a shell array
zwp_assignment_re='[$](ZWP_[[:alnum:]_]+)(.*)' # ...and a regex
content="$XLINE"
while [[ $content =~ $zwp_assignment_re ]]; do
zwpvars+=( "${BASH_REMATCH[1]}" ) # found a reference
content=${BASH_REMATCH[2]} # stuff the remaining content aside
done
printf 'Found variable: %s\n' "${zwpvars[#]}"
I'm trying to write a bash script that iterates over the arguments and builds a string like the following:
Usage:
./myScript a b c d
Expected output:
-e "a" -e "b" -e "c" -e "d"
The script looks like the following:
#!/bin/bash
pattern=""
for arg in "$#" do
pattern=$pattern" -e \"$arg\""
done
echo $pattern
The actual output misses the first -e, i.e., the output is:
"a" -e "b" -e "c" -e "d"
What am I doing wrong? What is the correct way to append -e?
You are doing nothing wrong. It is just that echo takes -e as an argument *.
$ pattern='-e asdf -e ghjk'
$ echo $pattern
asdf -e ghjk
If you quote the variable it works as expected.
$ echo "$pattern"
-e asdf -e ghjk
* man echo
-e enable interpretation of backslash escapes
I'm using
sed -e "s/\*DIVIDER\*/$DIVIDER/g" to replace *DIVIDER* with a user-specified string, which is stored in $DIVIDER. The problem is that I want them to be able to specify escape characters as their divider, like \n or \t. When I try this, I just end up with the letter n or t, or so on.
Does anyone have any ideas on how to do this? It will be greatly appreciated!
EDIT: Here's the meat of the script, I must be missing something.
curl --silent "$URL" > tweets.txt
if [[ `cat tweets.txt` == *\<error\>* ]]; then
grep -E '(error>)' tweets.txt | \
sed -e 's/<error>//' -e 's/<\/error>//' |
sed -e 's/<[^>]*>//g' |
head $headarg | sed G | fmt
else
echo $REPLACE | awk '{gsub(".", "\\\\&");print}'
grep -E '(description>)' tweets.txt | \
sed -n '2,$p' | \
sed -e 's/<description>//' -e 's/<\/description>//' |
sed -e 's/<[^>]*>//g' |
sed -e 's/\&\;/\&/g' |
sed -e 's/\<\;/\</g' |
sed -e 's/\>\;/\>/g' |
sed -e 's/\"\;/\"/g' |
sed -e 's/\&....\;/\?/g' |
sed -e 's/\&.....\;/\?/g' |
sed -e 's/^ *//g' |
sed -e :a -e '$!N;s/\n/\*DIVIDER\*/;ta' | # Replace newlines with *divider*.
sed -e "s/\*DIVIDER\*/${DIVIDER//\\/\\\\}/g" | # Replace *DIVIDER* with the actual divider.
head $headarg | sed G
fi
The long list of sed lines are replacing characters from an XML source, and the last two are the ones that are supposed to replace the newlines with the specified character. I know it seems redundant to replace a newline with another newline, but it was the easiest way I could come up with to let them pick their own divider. The divider replacement works great with normal characters.
You can use bash to escape the backslash like this:
sed -e "s/\*DIVIDER\*/${DIVIDER//\\/\\\\}/g"
The syntax is ${name/pattern/string}. If pattern begins with /, every occurence of pattern in name is replaced by string. Otherwise only the first occurence is replaced.
Maybe:
case "$DIVIDER" in
(*\\*) DIVIDER=$(echo "$DIVIDER" | sed 's/\\/\\\\/g');;
esac
I played with this script:
for DIVIDER in 'xx\n' 'xxx\\ddd' "xxx"
do
echo "In: <<$DIVIDER>>"
case "$DIVIDER" in (*\\*) DIVIDER=$(echo "$DIVIDER" | sed 's/\\/\\\\/g');;
esac
echo "Out: <<$DIVIDER>>"
done
Run with 'ksh' or 'bash' (but not 'sh') on MacOS X:
In: <<xx\n>>
Out: <<xx\\n>>
In: <<xxx\\ddd>>
Out: <<xxx\\\\ddd>>
In: <<xxx>>
Out: <<xxx>>
It seems to be a simple substitution:
$ d='\n'
$ echo "a*DIVIDER*b" | sed "s/\*DIVIDER\*/$d/"
a
b
Maybe I don't understand what you're trying to accomplish.
Then maybe this step could take the place of the last two of yours:
sed -n ":a;$ {s/\n/$DIVIDER/g;p;b};N;ba"
Note the space after the dollar sign. It prevents the shell from interpreting "${s..." as a variable name.
And as ghostdog74 suggested, you have way too many calls to sed. You may be able to change a lot of the pipe characters to backslashes (line continuation) and delete "sed" from all but the first one (leave the "-e" everywhere). (untested)
You just need to escape the escape char.
\n will match \n
\ will match \
\\ will match \
Using FreeBSD sed (e.g. on Mac OS X) you have to preprocess the $DIVIDER user input:
d='\n'
d='\t'
NL=$'\\\n'
TAB=$'\\\t'
d="${d/\\n/${NL}}"
d="${d/\\t/${TAB}}"
echo "a*DIVIDER*b" | sed -E -e "s/\*DIVIDER\*/${d}/"