Im tring to assign echo value which to a variable but im getting error
Var='(echo $2 | sed -e 's/,/: chararray /g'| sed -e 's/$/: chararray/')'
echo $var
Input : sh load.sh file 1,2,3,4
Error load.sh: line 1: chararray: command not found
Var=$(echo "$2" | sed -e 's/,/: chararray /g' | sed -e 's/$/: chararray/')
echo "$Var"
OR
Var=`echo "$2" | sed -e 's/,/: chararray /g' | sed -e 's/$/: chararray/'`
echo "$Var"
Use either $(…) or perhaps `…` backtick notation. However, the backtick notation is deprecated and should be avoided. Also, check the comments by mmgross, Etan Reisner and svlasov to your question. They are all correct.
Related
I'm trying to execute the following bash script but it gives me invalid arithmetic operator error in line 8.
#!/bin/bash
criteria=$1
re_match=$2
replace=$3
for i in $( ls *$criteria* );
do
src=$i
tgt=$[echo $i | sed -e "s/$re_match/$replace/"]
mv $src $tgt
done
You need to do:
$(echo $i | sed -e "s/$re_match/$replace/")
instead of
$[echo $i | sed -e "s/$re_match/$replace/"]
$() is used for variable expansion. [] is used for doing arithmetic.
I have a black list to save tag id list, e.g. 1-3,7-9, actually it represents 1,2,3,7,8,9. And could expand it by below shell
for i in {1..3,7..9}; do for j in {$i}; do echo -n "$j,"; done; done
1,2,3,7,8,9
but first I should convert - to ..
echo -n "1-3,7-9" | sed 's/-/../g'
1..3,7..9
then put it into for expression as a parameter
echo -n "1-3,7-9" | sed 's/-/../g' | xargs -I # for i in {#}; do for j in {$i}; do echo -n "$j,"; done; done
zsh: parse error near `do'
echo -n "1-3,7-9" | sed 's/-/../g' | xargs -I # echo #
1..3,7..9
but for expression cannot parse it correctly, why is so?
Because you didn't do anything to stop the outermost shell from picking up the special keywords and characters ( do, for, $, etc ) that you mean to be run by xargs.
xargs isn't a shell built-in; it gets the command line you want it to run for each element on stdin, from its arguments. just like any other program, if you want ; or any other sequence special to be bash in an argument, you need to somehow escape it.
It seems like what you really want here, in my mind, is to invoke in a subshell a command ( your nested for loops ) for each input element.
I've come up with this; it seems to to the job:
echo -n "1-3,7-9" \
| sed 's/-/../g' \
| xargs -I # \
bash -c "for i in {#}; do for j in {\$i}; do echo -n \"\$j,\"; done; done;"
which gives:
{1..3},{7..9},
Could use below shell to achieve this
# Mac newline need special treatment
echo "1-3,7-9" | sed -e 's/-/../g' -e $'s/,/\\\n/g' | xargs -I# echo 'for i in {#}; do echo -n "$i,"; done' | bash
1,2,3,7,8,9,%
#Linux
echo "1-3,7-9" | sed -e 's/-/../g' -e 's/,/\n/g' | xargs -I# echo 'for i in {#}; do echo -n "$i,"; done' | bash
1,2,3,7,8,9,
but use this way is a little complicated maybe awk is more intuitive
# awk
echo "1-3,7-9,11,13-17" | awk '{n=split($0,a,","); for(i=1;i<=n;i++){m=split(a[i],a2,"-");for(j=a2[1];j<=a2[m];j++){print j}}}' | tr '\n' ','
1,2,3,7,8,9,11,13,14,15,16,17,%
echo -n "1-3,7-9" | perl -ne 's/-/../g;$,=",";print eval $_'
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
}
I have a fairly simple bash shell scripting problem.
I want to sed a piece of text and then assign the result of the sed to a variable.
#!/bin/bash
MOD_DATE=echo $(date) | sed 's/\ /_/g'
echo $MOD_DATE // should show date with spaces replaced with underscores.
I have tried the above and it doesn't work. Can anyone point out what I'm doing wrong?
To collect the output in stdout into a variable, use a command substitution:
MOD_DATE=`echo $(date) | sed 's/\ /_/g'`
# ^ ^
or
MOD_DATE=$(echo $(date) | sed 's/\ /_/g')
# ^^ ^
Maybe this can help:
mod_date = "$(date +"%d_%m_%Y")"
echo "$mod_date"
I'm using
sed -e "s/\*DIVIDER\*/$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}/"