I would like to replace String in [[ Mystring ]] data to MyData.
pattern='[[ Mystring ]]'
replacement='MyData'
sed -i 's/pattern/replacement/g' Myfile.txt
What is wrong with above code? How to search data inside [[]]?
It should be like:
pattern='\[\[ Mystring \]\]'
replacement='MyData'
sed -i "s/$pattern/$replacement/g" Myfile.txt
You have to escape the [
sed -i 's/\[\[\s*Mystring\s*\]\]/foo/g' text.txt
Related
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 am trying to use a here-document with sed with a script.. but I get an weird error.
sed: can't read Some random text.: No such file or directory
FILE=c:/output.file
read -r -d '' VAR <<"EOF"
Some random text.
EOF
./sed -f - "$VAR" > "$FILE" << SED_SCRIPT
s|text|word|g
s|Some|Lots of|g
SED_SCRIPT
You don't use a here document for the sed expression, you just use a parameter in quotes. You normally use the here document for the input.
sed -e 's/text/word/g
s/Some/Lots of|g' <<EOF
Some random text.
EOF
I want to replace a' into à, e' into è, etc. into a file with a script like this:
#!/bin/sh
if [ -e "$1" ]
then
sed 's/a'/\à/g' -i "$1";
sed 's/e'/\è/g' -i "$1";
sed 's/i'/\ì/g' -i "$1";
sed 's/o'/\ò/g' -i "$1";
sed 's/u'/\ù/g' -i "$1";
else
echo "File not found!"
fi
But I get this error:
Syntax error: Unterminated quoted string
I don't know how to wrote '
Yes there is syntax problem, try your sed command as this:
sed -i "s/a'/à/g" "$1"
sed -i "s/e'/è/g" "$1"
sed -i "s/i'/ì/g" "$1"
sed -i "s/o'/ò/g" "$1"
sed -i "s/u'/ù/g" "$1"
Problem was that you were using nested single quote (quote inside code).
Exit the single quotes, then put an escaped quote, then start another single quoted string:
sed 's/a'\''/\à/g' -i "$1"
awk '{ gsub(/e'\''/,"è"); print}' "$1"
Yet another solution available in bash is $'...' quoting, which does allow escaped single quotes:
sed $'s/a\'/\à/g' -i "$1";
I want to use sed to replace all ' with \' and all " with \". Example input:
"a" 'b'
Output:
\"a\" \'b\'
There's no ? character in your post, but I'll assume your question is "How do I do such a replacement?". I just made a quick test file with your input, and this command line seems to work:
sed -e 's#"#\\"#g' -e "s#'#\\\'#g"
Example:
$ cat input
"a" 'b'
$ sed -e 's#"#\\"#g' -e "s#'#\\\'#g" input
\"a\" \'b\'
While using sed is the portable solution, all this can be done using Bash's builtin string manipulation functions as well.
(
#set -xv
#str=$'"a" \'b\''
str='"a" '"'b'" # concatenate 'str1'"str2"
str="${str//\"/\\\"}"
str="${str//\'/\'}"
echo "$str"
)
my question seems to be general, but i can't find any answers.
In sed command, how can you replace the substitution pattern by a value returned by a simple bash function.
For instance, I created the following function :
function parseDates(){
#Some process here with $1 (the pattern found)
return "dateParsed;
}
and the folowing sed command :
myCatFile=`sed -e "s/[0-3][0-9]\/[0-1][0-9]\/[0-9][0-9]/& parseDates &\}/p" myfile`
I found that the caracter '&' represents the current pattern found, i'd like it to be passed to my bash function and the whole pattern to be substituted by the pattern found +dateParsed.
Does anybody have an idea ?
Thanks
you can use the "e" option in sed command like this:
cat t.sh
myecho() {
echo ">>hello,$1<<"
}
export -f myecho
sed -e "s/.*/myecho &/e" <<END
ni
END
you can see the result without "e":
cat t.sh
myecho() {
echo ">>hello,$1<<"
}
export -f myecho
sed -e "s/.*/myecho &/" <<END
ni
END
Agree with Glenn Jackman.
If you want to use bash function in sed, something like this :
sed -rn 's/^([[:digit:].]+)/`date -d #&`/p' file |
while read -r line; do
eval echo "$line"
done
My file here begins with a unix timestamp (e.g. 1362407133.936).
Bash function inside sed (maybe for other purposes):
multi_stdin(){ #Makes function accepet variable or stdin (via pipe)
[[ -n "$1" ]] && echo "$*" || cat -
}
sans_accent(){
multi_stdin "$#" | sed '
y/àáâãäåèéêëìíîïòóôõöùúûü/aaaaaaeeeeiiiiooooouuuu/
y/ÀÁÂÃÄÅÈÉÊËÌÍÎÏÒÓÔÕÖÙÚÛÜ/AAAAAAEEEEIIIIOOOOOUUUU/
y/çÇñÑߢÐð£Øø§µÝý¥¹²³ªº/cCnNBcDdLOoSuYyY123ao/
'
}
eval $(echo "Rogério Madureira" | sed -n 's#.*#echo & | sans_accent#p')
or
eval $(echo "Rogério Madureira" | sed -n 's#.*#sans_accent &#p')
Rogerio
And if you need to keep the output into a variable:
VAR=$( eval $(echo "Rogério Madureira" | sed -n 's#.*#echo & | desacentua#p') )
echo "$VAR"
do it step by step. (also you could use an alternate delimiter , such as "|" instead of "/"
function parseDates(){
#Some process here with $1 (the pattern found)
return "dateParsed;
}
value=$(parseDates)
sed -n "s|[0-3][0-9]/[0-1][0-9]/[0-9][0-9]|& $value &|p" myfile
Note the use of double quotes instead of single quotes, so that $value can be interpolated
I'd like to know if there's a way to do this too. However, for this particular problem you don't need it. If you surround the different components of the date with ()s, you can back reference them with \1 \2 etc and reformat however you want.
For instance, let's reverse 03/04/1973:
echo 03/04/1973 | sed -e 's/\([0-9][0-9]\)\/\([0-9][0-9]\)\/\([0-9][0-9][0-9][0-9]\)/\3\/\2\/\1/g'
sed -e 's#[0-3][0-9]/[0-1][0-9]/[0-9][0-9]#& $(parseDates &)#' myfile |
while read -r line; do
eval echo "$line"
done
You can glue together a sed-command by ending a single-quoted section, and reopening it again.
sed -n 's|[0-3][0-9]/[0-1][0-9]/[0-9][0-9]|& '$(parseDates)' &|p' datefile
However, in contrast to other examples, a function in bash can't return strings, only put them out:
function parseDates(){
# Some process here with $1 (the pattern found)
echo dateParsed
}