如何将第N个文件写入新文件夹

Question

我有这段代码可以扫描文件夹，并将每个文件夹中的所有文件移动到一个新文件夹。

我如何做到只有第N个文件被移动？

#!/bin/bash

# Save this file in the directory containing the folders (bb in this case)
# Then to run it, type:
# ./rencp.sh

# The first output frame number
let "frame=1"

# this is where files will go. A new directory will be created if it doesn't exist
outFolder="collected"

# print info every so many files.
feedbackFreq=250

# prefix for new files
namePrefix="ben_timelapse"

#new extension (uppercase is so ugly)
ext="jpg"

# this will make sure we only get files from camera directories
srcPattern="ND850"

mkdir -p $outFolder
for f in *${srcPattern}/*
do
mv $f `printf "$outFolder/$namePrefix.%05d.$ext" $frame`
if ! ((frame % $feedbackFreq)); then
    echo "moved and renamed $frame files to $outFolder"
fi
let "frame++"
done

相当确定我需要for f in *${srcPattern}/*编辑for f in *${srcPattern}/*的行for f in *${srcPattern}/*但是不确定正确的语法

Answer 1

在do之后尝试使用此命令而不是mv命令：

if ! ((frame % 5)); then
  a=$((frame / 5));
  mv $f `printf "$outFolder/$namePrefix.%05d.$ext" $a`
fi

它将frame = 5,10，依此类推，移至$outFolder/$namePrefix.00001.$ext ， $outFolder/$namePrefix.00002.$ext等。

Answer 2

如果列出后ND850文件夹中的文件是连续的（即，填充的帧号），并且文件夹本身是按顺序排列的，则以下代码应该起作用。

#!/bin/bash

# Maintain a counter, and the output frame number
let "frame=1"
let "outframe=1"

outFolder="collected"

# frequency
gap=5

namePrefix="ben_timelapse"

#new extension (uppercase is so ugly)
ext="jpg"

srcPattern="ND850"

echo "Copying and renaming 1 in every $gap files"

mkdir -p "$outFolder"
for f in *${srcPattern}/*
do
if ! ((frame % $gap)); then
    outfile=`printf "$outFolder/$namePrefix.%05d.$ext" $outframe`
    cp $f "$outfile"
    echo "copied $f to $outfile"
    let "outframe++"
fi
let "frame++"
done