bash 在 Linux 上将一组文件重命名为 001、002、...
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Renaming a set of files to 001, 002, ... on Linux
提问by DisgruntledGoat
I originally had a set of images of the form image_001.jpg, image_002.jpg, ...
我最初有一组形式为 image_001.jpg、image_002.jpg、...
I went through them and removed several. Now I'd like to rename the leftover files back to image_001.jpg, image_002.jpg, ...
我浏览了它们并删除了几个。现在我想将剩余的文件重命名回 image_001.jpg、image_002.jpg、...
Is there a Linux command that will do this neatly? I'm familiar with rename but can't see anything to order file names like this. I'm thinking that since ls *.jpglists the files in order (with gaps), the solution would be to pass the output of that into a bash loop or something?
是否有一个 Linux 命令可以巧妙地做到这一点？我熟悉重命名，但看不到任何可以对这样的文件名进行排序的内容。我在想，由于ls *.jpg按顺序列出文件（有间隙），解决方案是将其输出传递到 bash 循环或其他什么？
采纳答案by Matthew Flaschen
If I understand right, you have e.g. image_001.jpg, image_003.jpg, image_005.jpg, and you want to rename to image_001.jpg, image_002.jpg, image_003.jpg.
如果我理解正确，您有例如 image_001.jpg、image_003.jpg、image_005.jpg，并且您想重命名为 image_001.jpg、image_002.jpg、image_003.jpg。
EDIT: This is modified to put the temp file in the current directory. As Stephan202 noted, this can make a significant difference if temp is on a different filesystem. To avoid hitting the temp file in the loop, it now goes through image*
编辑：修改后将临时文件放在当前目录中。正如 Stephan202 所指出的，如果 temp 位于不同的文件系统上，这可能会产生显着差异。为了避免循环中的临时文件，它现在通过图像*
i=1; temp=$(mktemp -p .); for file in image*
do
mv "$file" $temp;
mv $temp $(printf "image_%0.3d.jpg" $i)
i=$((i + 1))
done
回答by Stephan202
A simple loop (test with echo, execute with mv):
一个简单的循环（用 测试，用echo执行mv）：
I=1
for F in *; do
echo "$F" `printf image_%03d.jpg $I`
#mv "$F" `printf image_%03d.jpg $I` 2>/dev/null || true
I=$((I + 1))
done
(I added 2>/dev/null || trueto suppress warnings about identical source and target files. If this is not to your liking, go with Matthew Flaschen's answer.)
（我添加2>/dev/null || true了抑制有关相同源文件和目标文件的警告。如果这不符合您的喜好，请使用Matthew Flaschen的答案。）
回答by Paul Weibert
Try the following script:
尝试以下脚本：
This code snipped should do the job:
剪下的这段代码应该可以完成这项工作：
./numerate.sh -d <your image folder> -b <start number> -L 3 -p image_ -s .jpg -o numerically -r
回答by lhunath
Some good answers here already; but some rely on hiding errors which is not a good idea (that assumes mvwill only error because of a condition that is expected - what about all the other reaons mvmight error?).
这里已经有一些很好的答案；但有些人依赖于隐藏错误，这不是一个好主意（假设mv只会因为预期的条件而出错 - 所有其他原因mv可能出错呢？）。
Moreover, it can be done a little shorter and should be better quoted:
此外，它可以做得更短一点，应该更好地引用：
for file in *; do
printf -vsequenceImage 'image_%03d.jpg' "$((++i))"
[[ -e $sequenceImage ]] || \
mv "$file" "$sequenceImage"
done
Also note that you shouldn't capitalize your variablesin bash scripts.
回答by ennuikiller
This does the reverse of what you are asking (taking files of the form *.jpg.001 and converting them to *.001.jpg), but can easily be modified for your purpose:
这与您所要求的相反（获取 *.jpg.001 格式的文件并将它们转换为 *.001.jpg），但可以根据您的目的轻松修改：
for file in *
do
if [[ "$file" =~ "(.*)\.([[:alpha:]]+)\.([[:digit:]]{3,})$" ]]
then
mv "${BASH_REMATCH[0]}" "${BASH_REMATCH[1]}.${BASH_REMATCH[3]}.${BASH_REMATCH[2]}"
fi
done
回答by jess
I was going to suggest something like the above using a for loop, an iterator, cut -f1 -d "_", then mv i i.iterator. It looks like it's already covered other ways, though.
我打算使用 for 循环，迭代器，cut -f1 -d "_"，然后 mv i i.iterator 来建议类似上面的内容。不过，它似乎已经涵盖了其他方式。
