根据 Bash 中的文件名对文件名进行分组的最佳方法？

Question

我有一个包含以下文件的文件夹：

DA-001-car.jpg
DA-001-dog.jpg
DA-001-coffee.jpg
DA-002-house.jpg
DA-003-coffee.jpg
DA-003-cat.jpg
...

我想生成这个 (CSV) 输出：

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"

我尝试在 Bash 中对此进行编程：

#!/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

如您所见，我是编程菜鸟 :) 请帮帮我，提前谢谢！

Answer 1

这样就可以了。这需要 GNU awk 以升序输出。如果您不关心订单，可以使用任何旧的awk 并删除PROCINFO 行

#!/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

示例输出

$ 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

如果文件名不包含空格，您可以使用 sed 代替内部循环：

printf '%s\n' *.jpg \
| cut -f1,2 -d- \
| sort -u \
| while IFS= read -r sku ; do
    echo "$sku",\"$(echo "$sku"* | sed 's/ /, /')\"
done

使用内部循环，您可以从echo 切换到printf。 sed 用于删除结尾的逗号。

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

如果您不想解析ls 的输出并运行sort，您可以将前缀存储在关联数组中：

#!/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

作为所有回复和建议的结果，我正在使用此代码来实现所需的输出：

#!/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

谢谢大家！