通过Shell脚本输出编辑.csv文件

Question

I have a script which takes a .csv file as a an input and runs row by row

kma210,projects.kma210
kma215,projects.kma215_2
KMA3xx,projects.kma3xx
KMI7,projects.kmi7

The above one is a sample for the .csv file. I have two types of process running for each row (example : abc_process and xyz_process).

Now my requirement is once the script runs one row , the output ie, success/fail should print in the same file in third column ( abc_process ) and fourth column ( xyz_process ).

Example please see here:

kma210,projects.kma210,success,success
kma215,projects.kma215_2,fail,success
KMA3xx,projects.kma3xx,success,fail
KMI7,projects.kmi7,fail,fail

Could anyone please suggest?

Answer 1

I think what you are trying to achieve is something of this sort.

From your input file ( input.csv ), use the script as follows:-

#!/bin/bash

while IFS=',' read -r col1 col2
do
    # The below two read commands are to process the success/fail status
    # string as you indicated. If you are receiving a string of the below 
    # form, from an external bash command; you need to use process-substitution 
    # as

    # IFS= read -r _ p1Status <<(p1Command)
    # IFS= read -r _ p1Status <<(p2Command)

    IFS= read -r _ p1Status <<<"abc success"
    IFS= read -r _ p2Status <<<"xyz fail"

    # The below printf just prints the output as you need to stdout
    # to create a .csv out of it, append it to a new file as

    # printf "%s,%s,%s,%s\n" "$col1" "$col2" "$p1Status" "$p2Status" >> output.csv

    printf "%s,%s,%s,%s\n" "$col1" "$col2" "$p1Status" "$p2Status"
done<input.csv

The above command as such writing to stdout , produces an output as

kma210,projects.kma210,success,fail
kma215,projects.kma215_2,success,fail
KMA3xx,projects.kma3xx,success,fail
KMI7,projects.kmi7,success,fail

which you can append to a file using the >> operator with the line I commented out. Or if you want a single file and replace the original file. Add a line

mv -v output.csv input.csv 

as the last line of the script.