bash,replace,count , shell script for counting replacements


shell script for counting replacements

Question:

Tag: bash,replace,count

Ten files located in a directory. Each file content has "apple" word. write a bash shell script to replace "apple" with "banana" in all ten files and Print the number of replacements for each file. Have tried in this way but dont know how to get number of replacement. can anyone help in this ? and sed is only modifying in display actual file not getting modified

#!/bin/bash 
for f in *.txt 
do sed 's/apple/banana/g' $f 
done

Answer:

Assuming you want to replace the word 'apple' with 'banana' (exact match) in the contents of the files and not on the names of the files (see my comment above) and that you are using the bash shell:

#!/bin/bash

COUNTER=0

    for file in *.txt ; do
        COUNTER=$(grep -o "\<apple\>" $file | wc -l)
        sed -i 's/\<apple\>/banana/g' $file
        echo "RESULT: $COUNTER replacements in file $file"
        let COUNTER=0
    done

exit 0;

Explanation:


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