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
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 have been working in bash, and need to create a string argument. bash is a newish for me, to the point that I dont know how to build a string in bash from a list.
// foo.txt is a list of abs file names.
/foo/bar/a.txt
/foo/bar/b.txt
/delta/test/b.txt
should turn into: a.txt,b.txt,b.txt
OR: /foo/bar/a.txt,/foo/bar/b.txt,/delta/test/b.txt
code
s = ""
for file in $(cat foo.txt);
do
#what goes here? s += $file ?
done
myShellScript --script $s
I figure there was an easy way to do this.
with for loop:
for file in $(cat foo.txt);do echo -n "$file",;done|sed 's/,$/\n/g'
with tr:
cat foo.txt|tr '\n' ','|sed 's/,$/\n/g'
only sed:
sed ':a;N;$!ba;s/\n/,/g' foo.txt
This seems to work:
#!/bin/bash
input="foo.txt"
while IFS= read -r var
do
basename $var >> tmp
done < "$input"
paste -d, -s tmp > result.txt
output: a.txt,b.txt,b.txt
basename gets you the file names you need and paste will put them in the order you seem to need.
The input field separator can be used with set to create split/join functionality:
# split the lines of foo.txt into positional parameters
IFS=$'\n'
set $(< foo.txt)
# join with commas
IFS=,
echo "$*"
For just the file names, add some sed:
IFS=$'\n'; set $(sed 's|.*/||' foo.txt); IFS=,; echo "$*"
I need to substitute a unique string in a json file: {FILES} by a bash variable that contains thousands of paths: ${FILES}
sed -i "s|{FILES}|$FILES|" ./myFile.json
What would be the most elegant way to achieve that ? The content of ${FILES} is a result of an "aws s3" command. The content would look like :
FILES="/file1.ipk, /file2.ipk, /subfolder1/file3.ipk, /subfolder2/file4.ipk, ..."
I can't think of a solution where xargs would help me.
The safest way is probably to let Bash itself expand the variable. You can create a Bash script containing a here document with the full contents of myFile.json, with the placeholder {FILES} replaced by a reference to the variable $FILES (not the contents itself). Execution of this script would generate the output you seek.
For example, if myFile.json would contain:
{foo: 1, bar: "{FILES}"}
then the script should be:
#!/bin/bash
cat << EOF
{foo: 1, bar: "$FILES"}
EOF
You can generate the script with a single sed command:
sed -e '1i#!/bin/bash\ncat << EOF' -e 's/\$/\\$/g;s/{FILES}/$FILES/' -e '$aEOF' myFile.json
Notice sed is doing two replacements; the first one (s/\$/\\$/g) to escape any dollar signs that might occur within the JSON data (replace every $ by \$). The second replaces {FILES} by $FILES; the literal text $FILES, not the contents of the variable.
Now we can combine everything into a single Bash one-liner that generates the script and immediately executes it by piping it to Bash:
sed -e '1i#!/bin/bash\ncat << EOF' -e 's/\$/\\$/g;s/{FILES}/$FILES/' -e '$aEOF' myFile.json | /bin/bash
Or even better, execute the script without spawning a subshell (useful if $FILES is set without export):
sed -e '1i#!/bin/bash\ncat << EOF' -e 's/\$/\\$/g;s/{FILES}/$FILES/' -e '$aEOF' myFile.json | source /dev/stdin
Output:
{foo: 1, bar: "/file1.ipk, /file2.ipk, /subfolder1/file3.ipk, /subfolder2/file4.ipk, ..."}
Maybe perl would have fewer limitations?
perl -pi -e "s#{FILES}#${FILES}#" ./myFile.json
It's a little gross, but you can do it all within shell...
while read l
do
if ! echo "$l" | grep -q '{DATA}'
then
echo "$l"
else
echo "$l" | sed 's/{DATA}.*$//'
echo "$FILES"
echo "$l" | sed 's/^.*{DATA}//'
fi
done <./myfile.json >newfile.json
#mv newfile.json myfile.json
Obviously I'd leave the final line commented until you were confident it worked...
Maybe just don't do it? Can you just :
echo "var f = " > myFile2.json
echo $FILES >> myFile2.json
And reference myFile2.json from within your other json file? (You should put the global f variable into a namespace if this works for you.)
Instead of putting all those variables in an environment variable, put them in a file. Then read that file in perl:
foo.pl:
open X, "$ARGV[0]" or die "couldn't open";
shift;
$foo = <X>;
while (<>) {
s/world/$foo/;
print;
}
Command to run:
aws s3 ... >/tmp/myfile.$$
perl foo.pl /tmp/myfile.$$ <myFile.json >newFile.json
Hopefully that will bypass the limitations of the environment variable space and the argument length by pulling all the processing within perl itself.
I am trying to remove newlines from a file. My file is like this (it contains backward slashes):
line1\|
line2\|
I am using the following script to remove newlines:
#!/bin/bash
INPUT="file1"
while read line
do
: echo -n $line
done < $INPUT
I get the following output:
line1|line2|
It removes the backslashes. How can I retain those backslashes?
The -r option to read prevents backslash processing of the input.
while read -r line
do
echo -n "$line"
done < $INPUT
But if you just want to remove all newlines from the input, the tr command would be better:
tr -d '\n' < $INPUT
Try sed 's/\n//' /path/to/file
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
}