String not getting replaced with sed command

Question

In my script, I am using below sed command

while read line; do
  echo "$line"
  sed -r "s/where /where $line ;/" Query.cql > Query-2.cql
done < $1

the value of $line is

where Column11 in ('Value1','Value2','Vlaue3') and Column12 in ('Value11','Value22','Vlaue32')

File Content ::

capture 'data.csv'

select * from test where

capture off;

After Executing

sed -r "s/where /where $line ;/" Query.cql > Query-2.cql

OUTPUT is ::

capture 'data.csv'

select * from test where

capture off;

Here, the string is not getting replaced. What went wrong?

Put the sed statement in quotes when you are expanding variables. — Raman Sailopal, Mar 29 '21 at 09:21
No need to embed `sed` inside a `while read` loop, imagine if that file contains 1k lines, then `sed` will have to process each line, so `sed` will run 1k times too.... — Jetchisel, Mar 29 '21 at 09:26
@apapillon... its working out side the script but in script it isn't — R_S_C, Mar 29 '21 at 09:26
@Jetchisel.. I need to prepare where query for each where clause and there will be at most 100 clause... that will be 100 lines... so is there any other way to do ? — R_S_C, Mar 29 '21 at 09:28
Generating a single SQL file with all your commands would seem a lot simpler as well as more robust and efficient. If the input is complex, maybe switch from `sed` to Awk or a more modern scripting language. — tripleee, Mar 29 '21 at 09:50

KamilCuk · Accepted Answer · 2021-03-31T09:15:37.623

sed: -e expression #1, char 19: unterminated `s' command

Sure the s command is s/<something>/<else>/ - there is a trailing /. Do:

sed -r 's/where /$line ;/'
                        ^ - trailing /

The -r option seems unused - where has no extended regular expressions. If so, remove it.

Your command uses ' quotes, so $line is not expanded. Research the difference between single and double quotes in shell, most probably you meant to use " here.

Note that each loop > qWithWhere.cql is recreating result and overwriting the result from previous loop. You might just run the loop on the last line them.

Read how to read a file line by line in bash and how to Escape a string for a sed replace pattern .

The following code with a space after where in input:

cat <<EOF >input
capture 'data.csv'
select * from test where 
capture off;
EOF
line="where Column11 in ('Value1','Value2','Vlaue3') and Column12 in ('Value11','Value22','Vlaue32')"
sed -r "s/where /where $line ;/" input

outputs:

capture 'data.csv'
select * from test where where Column11 in ('Value1','Value2','Vlaue3') and Column12 in ('Value11','Value22','Vlaue32') ;
capture off;

its not working .. i'll update the question to show you output — R_S_C, Mar 31 '21 at 09:01
`What went wrong?` Well, is there a `where` __and a space after it__ in your source file? You are matching `/where /` — KamilCuk, Mar 31 '21 at 09:14

String not getting replaced with sed command

1 Answers1