Bash脚本，用于对目录中的所有文件执行命令

Could somebody please provide the code to do the following: Assume there is a directory of files, all of which need to be run through a program. The program outputs the results to standard out. I need a script that will go into a directory, execute the command on each file, and concat the output into one big output file.

谁能提供以下代码:假设有一个文件目录，所有文件都需要通过程序运行。程序将结果输出到标准输出。我需要一个脚本，它将进入一个目录，在每个文件上执行命令，并将输出转换为一个大的输出文件。

For instance, to run the command on 1 file:

例如，在一个文件上运行命令:

$ cmd [option] [filename] > results.out

6 个解决方案

#1

258

The following bash code will pass $file to command where $file will represent every file in /dir

下面的bash代码将把$file传递给命令，其中$file表示/dir中的每个文件

for file in /dir/*
do
  cmd [option] "$file" >> results.out
done

Example

例子

el@defiant ~/foo $ touch foo.txt bar.txt baz.txt
el@defiant ~/foo $ for i in *.txt; do echo "hello $i"; done
hello bar.txt
hello baz.txt
hello foo.txt

#2

123

How about this:

这个怎么样:

find /some/directory -maxdepth 1 -type f -exec cmd option {} \; > results.out

-maxdepth 1 argument prevents find from recursively descending into any subdirectories. (If you want such nested directories to get processed, you can omit this.)
-maxdepth 1参数阻止find递归地下降到任何子目录中。(如果您希望处理这样的嵌套目录，可以忽略它。)
-type -f specifies that only plain files will be processed.
-type -f指定只处理纯文件。
-exec cmd option {} tells it to run cmd with the specified option for each file found, with the filename substituted for {}
-exec cmd选项{}告诉它运行cmd，每个文件都有指定的选项，文件名被替换为{}
\; denotes the end of the command.
\;表示命令的结束。
Finally, the output from all the individual cmd executions is redirected to results.out
最后，所有cmd执行的输出都被重定向到result .out

However, if you care about the order in which the files are processed, you might be better off writing a loop. I think find processes the files in inode order (though I could be wrong about that), which may not be what you want.

但是，如果您关心文件处理的顺序，您最好编写一个循环。我认为按inode顺序查找文件(尽管我可能会错)，这可能不是您想要的。

#3

I'm doing this on my raspberry pi from the command line by running:

我在命令行中的树莓派上运行:

for i in *;do omxplayer "$i";done

#4

Based on @Jim Lewis's approach:

基于@Jim Lewis的方法:

Here is a quick solution using find and also sorting files by their modification date:

这里有一个快速解决方案，使用查找和排序文件的修改日期:

$ find  directory/ -maxdepth 1 -type f -print0 | \
  xargs -r0 stat -c "%y %n" | \
  sort | cut -d' ' -f4- | \
  xargs -d "\n" -I{} cmd -op1 {}

For sorting see:

排序:

http://www.commandlinefu.com/commands/view/5720/find-files-and-list-them-sorted-by-modification-time

#5

I needed to copy all .md files from one directory into another, so here is what I did.

我需要将所有的。md文件从一个目录复制到另一个目录中，所以我这样做了。

for i in **/*.md;do mkdir -p ../docs/"$i" && rm -r ../docs/"$i" && cp "$i" "../docs/$i" && echo "$i -> ../docs/$i"; done

我在**/*.md，做mkdir -p ../docs/ $i & rm -r/文档/“$i”& cp“$i”/docs/$i" & echo "$i -> .. ./docs/$i";完成

Which is pretty hard to read, so lets break it down.

这很难读，我们把它分解一下。

first cd into the directory with your files,

第一张cd进入你的文件目录，

for i in **/*.md; for each file in your pattern

因为我在* * / *。海事;对于模式中的每个文件

mkdir -p ../docs/"$i"make that directory in a docs folder outside of folder containing your files. Which creates an extra folder with the same name as that file.

mkdir - p . ./docs/“$i”将目录放在包含文件的文件夹之外的docs文件夹中。创建一个与该文件同名的额外文件夹。

rm -r ../docs/"$i" remove the extra folder that is created as a result of mkdir -p

rm - r . ./docs/“$i”删除mkdir -p结果创建的额外文件夹

cp "$i" "../docs/$i" Copy the actual file

cp " $我" . ./docs/$i"复制实际文件。

echo "$i -> ../docs/$i" Echo what you did

回声“$ i - > . ./docs/$i“和你做的一样

; done Live happily ever after

;从此过上了幸福的生活。

#6

One quick and dirty way which gets the job done sometimes is:

一种快速而肮脏的方式，有时完成工作是:

find directory/ | xargs  Command

For example to find number of lines in all files in the current directory, you can do:

例如，要在当前目录中的所有文件中查找行数，可以这样做:

find . | xargs wc -l

#1

258