Best way to group filename based on filename in Bash? - bash

Question

I have a folder with the following files:
DA-001-car.jpg
DA-001-dog.jpg
DA-001-coffee.jpg
DA-002-house.jpg
DA-003-coffee.jpg
DA-003-cat.jpg
...
I want to generate this (CSV) output:
SKU, IMAGE
DA-001, "DA-001-car.jpg, DA-001-dog.jpg, DA-001-coffee.jpg"
DA-002, "DA-001-house.jpg"
DA-003, "DA-001-coffee.jpg, DA-001-cat.jpg"
I tried to program this in Bash:
#!/bin/bash
echo "SKU, FILE" >> tmp.csv
for file in /home/calvin/test/*.jpg
do
SKU_NAME="${file##*/}"
echo ${SKU_NAME:0:6}, \"inner for-loop?, ?, ?\" >> tmp.csv
done
uniq tmp.csv output.csv
As you can see I'm a noob as for programming :)
Please help me out, thanks in advance!

Answer 1

This will do the trick. This requires GNU awk to output in ascending order. If you don't care about the order, you can use any old awk and remove the PROCINFO line
#!/bin/bash
awk -F- '
BEGIN{
print "SKU, IMAGE"
}
{
sep=!a[$2]?"":", "
a[$2]=a[$2] sep $0
}
END{
PROCINFO["sorted_in"] = "#ind_str_asc" # GNU only feature
for(i in a){print "DA-" i ", " "\"" a[i] "\""}
}' <(find /home/calvin/test -type f -name "*.jpg" -printf "%f\n") > ./tmp.csv
Example Output
$ cat ./tmp.csv
SKU, IMAGE
DA-001, "DA-001-coffee.jpg, DA-001-car.jpg, DA-001-dog.jpg"
DA-002, "DA-002-house.jpg"
DA-003, "DA-003-coffee.jpg, DA-003-cat.jpg"

Answer 2

If the filenames don't contain spaces, you can use sed instead of an inner loop:
printf '%s\n' *.jpg \
| cut -f1,2 -d- \
| sort -u \
| while IFS= read -r sku ; do
echo "$sku",\"$(echo "$sku"* | sed 's/ /, /')\"
done
With the inner loop, you can switch to printf from echo. Sed is used to remove the trailing comma.
printf '%s\n' *.jpg \
| cut -f1,2 -d- \
| sort -u \
| while IFS= read -r sku ; do
printf %s "$sku, \""
for f in "$sku"* ; do
printf '%s, ' "$f"
done | sed 's/, $//'
printf '"\n'
done
If you don't want to parse the output of ls and run sort, you can store the prefixes in an associative array:
#!/bin/bash
declare -A prefix
for jpg in *.jpg ; do
p1=${jpg%%-*}
jpg=${jpg#*-}
p2=${jpg%%-*}
prefix[$p1-$p2]=1
done
for sku in "${!prefix[#]}" ; do
printf '%s, "' "$sku"
for f in "$sku"* ; do
printf '%s, ' "$f"
done | sed 's/, $//'
printf '"\n'
done

Answer 3

awk '
BEGIN {
OFS = ", "
print "SKU", "IMAGE"
for (i=1; i<ARGC; i++) {
curr = fname = ARGV[i]
sub(/-[^-]+$/,"",curr)
if ( curr != prev ) {
if ( i > 1 ) {
print prev, "\"" fnames "\""
}
prev = curr
fnames = ""
}
fnames = (fnames == "" ? "" : fnames OFS) fname
}
print prev, "\"" fnames "\""
exit
}
' /home/calvin/test/*.jpg
SKU, IMAGE
DA-001, "DA-001-car.jpg, DA-001-coffee.jpg, DA-001-dog.jpg"
DA-002, "DA-002-house.jpg"
DA-003, "DA-003-cat.jpg, DA-003-coffee.jpg"

Answer 4

As result of all the replies and advises I'm using this code to achieve the desired output:
#!/bin/bash
echo "SKU, IMAGES" >> output.csv
ls *.jpg | cut -f1,2 -d- | sort -u | while read SKU
do
echo $SKU, \"$(echo "$SKU"* | sed 's/ /, /g')\" >> output.csv
done
Thanks all!