Why does this sed command do nothing? [closed] - bash

Question

Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question appears to be off-topic because it lacks sufficient information to diagnose the problem. Describe your problem in more detail or include a minimal example in the question itself.
Closed 9 years ago.
Improve this question
I am trying to cat a file to create a copy of itself, but at the same time replace some values
my command is:
cat ${FILE} | sed "s|${key}|${value}|g" > ${TEMP_FILE}
However, when I open the temp file, none of the keys have been replaced- just a straight swap. I have echoed the values of key and value and they are correct - they come from an array element.
Yet if I use a plain string not a variable, it works fine for one type of key - i.e:
cat ${FILE} | sed "s|example_key|${value}|g" > ${TEMP_FILE}
The example_key instances within the file are replaced which is what I want.
However, when I try to use my array $key parameter, it does nothing. No idea why :-(
Command usage:
declare -a props
...
....
for x in "${props[#]}"
do
key=`echo "${x}" | cut -d '=' -f 1`
value=`echo "${x}" | cut -d '=' -f 2`
# global replace on the $FILE
cat ${FILE} | sed "s|${key}|${value}|g" > ${TEMP_FILE}
#cat ${FILE} | sed "s|example_key|${value}|g" > ${TEMP_FILE}
done
array elements are stored in the following format: $key=$value

Answer 1

key='echo "${x}" | cut -d '=' -f 1
value='echo "${x}" | cut -d '=' -f 2
Use back-ticks, not single-quotes, if you want to do command substitution.
key=`echo "${x}" | cut -d '=' -f 1`
value=`echo "${x}" | cut -d '=' -f 2`
Also note that as you loop over the series of key=value pairs, you're overwriting your temp file each time, using only one substitution applied to the original file.. So after the loop is finished, the best you can hope for is that only the last substitution will be applied.
I'd also suggest not doing this in multiple passes -- do it by passing multiple expressions to sed:
for x in "${props[#]}" ; do
subst="$subst -e 's=$x=g'"
done
sed $subst "${FILE}" > "${TEMP_FILE}"
I'm using a trick, by using = as the delimiter for the sed substitution expression, we don't have to separate the key from the value. The command simply becomes:
sed -e 's=foo=1=g' -e 's=bar=2=g' "${FILE}" > "${TEMP_FILE}"

Answer 2

Thanks to #BillKarwin for spotting the crux of the problem: each iteration of the loop wipes out the previous iterations' replacements, because the result of a single key-value pair replacement replaces the entire output file every time.
Try the following:
declare -a props
# ...
cp "$FILE" "$TEMP_FILE"
for x in "${props[#]}"; do
IFS='=' read -r key value <<<"$x"
sed -i '' "s|${key}|${value}|g" "${TEMP_FILE}"
done
Copies the input file to the output file first, then replaces the output file in-place (using sed's -i option) in every iteration of the loop.
I also streamlined the code to parse each line into key and value, using read.
Also note that I consistently double-quoted all variable references.
#anubhava makes a good general point: depending on the variable values, a different regex delimiter may be needed (in your case: if the keys or values contained '|', you couldn't use '|' to delimit the regexes).
Update: #BillKarwin makes a good point: performing the replacements one by one, in a loop, is inefficient.
Here's a one-liner that avoids loops altogether:
sed -f <(awk -F'=' '{ if ($0) print "s/" $1 "/" substr($0, 1+length($1)) "/g" }' \
"$FILE") "$FILE" > "$TEMP_FILE"
Uses awk to build up the entire set of substitution commands for sed (one per line).
Then feeds the result via process substitution as a command file to sed with -f.
Handles the case where values have embedded = chars. correctly.