使用echo / printf在awk / sed脚本中打印出shell变量和“”

Question

A printout question: We need to print out the following awk into a .sh file. (Remove all the row that has "0" in the 12th column)

awk -F, '$12 != "0"' output.csv >> output2.csv

Here is our script (The case loop is necessary for another purpose. This is a small step of a bigger script. Assume $TargetIDs=123):

case $TargetIDs in ($P1) echo "awk -F, '$12 != "0"' ${TargetIDs}_output.csv >> ${TargetIDs}_output2.csv" >> output.sh;

(*)  ;;

esac

This will print out the following in the "output.sh". The "$1" in "$12" disappears : (

awk -F, '2 != "0"' 123_output.csv >> 123_output2.csv

We try an array:

 V="12"

 b=($V)

case $TargetIDs in ($P1) echo "awk -F, '${b[1]} != "0"' ${TargetIDs}_output.csv >> ${TargetIDs}_output2.csv" >> output.sh;

(*)  ;;

esac

This will print out the following in the "output.sh". $12 disappear:

    awk -F, ' != 0' output.csv >> output2.csv

The second question is similar: we use a sed way:

We want to print out this sed:

    sed -i.temp '/"0"/d' 123_output.csv.temp
    mv 123_output.csv.temp 123_output.csv

Here is the script

    case $TargetIDs in ($P1) printf "sed -i.temp '/"0"/d' ${TargetIDs}_output.csv.temp\n mv ${TargetIDs}_output.csv.temp ${TargetIDs}_output.csv\n" >> output.sh ;;

    (*)  ;;

    esac

This will print out the following in the "output.sh":

sed -i.temp '/0/d' 123_output.csv.temp
mv 123_output.csv.temp 123_output.csv

The sed command becomes this in the "output.sh"

sed -i.temp '/0/d'

instead of

sed -i.temp '/"0"/d'

"" is gone in the "output.sh" and hence all the rows that have zero will be removed by

sed -i.temp '/0/d'

Wonder if gurus might have some solutions for this? Thanks!

Answer 1

You need to watch your quoting. Inside double quotes, single quotes aren't special, but dollar signs are. The nice bit is that you can concatenate multiple quoted segments, so you can change between different quoting styles in the same parameter.

Try this:

echo "awk -F, '"'$12 != "0"'"' ${TargetIDs}_output.csv >> ${TargetIDs}_output2.csv"

If TargetIDs might include whitespace or shell metacharacters, you'd also want to insert some more quotes into the generated text:

echo "awk -F, '"'$12 != "0"'"' \"${TargetIDs}_output.csv\" >> \"${TargetIDs}_output2.csv\""

As per the discussion in the comments, there is a difference between:

awk -F, '$12 != "0"' output.csv >> output2.csv

and

awk -F, '$12 != 0' output.csv >> output2.csv

in the case that field 12 is 00 . However, neither are correct if you want to remove lines where the 12th field is "0" , which seems to be the point of your sed line in your second question. In order to catch that, you'd need:

awk -F, '$12 != "\"0\""' output.csv >> output2.csv

Make sure that you are using the correct version.

Answer 2

Try this:

echo "awk -F, '\$12 != 0' ${TargetIDs}_output.csv >> ${TargetIDs}_output2.csv"

This of course assumes that you are treating column 12 as a numeric value.