目录中每个文件的 Linux Shell 脚本获取文件名并执行程序
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Linux Shell Script For Each File in a Directory Grab the filename and execute a program
提问by ThinkCode
Scenario :
设想 ：
A folder in Linux system. I want to loop through every .xls file in a folder.
Linux系统中的一个文件夹。我想遍历文件夹中的每个 .xls 文件。
This folder typically consists of various folders, various filetypes (.sh, .pl,.csv,...).
该文件夹通常由各种文件夹、各种文件类型（.sh、.pl、.csv 等）组成。
All I want to do is loop through all files in the rootand execute a program only on .xls files.
我想要做的就是遍历根目录中的所有文件并仅在 .xls 文件上执行程序。
Edit :
编辑 ：
The problem is the program I have to execute is 'xls2csv' to convert from .xls to .csv format. So, for each .xls file I have to grab the filename and append it to .csv.
问题是我必须执行的程序是“xls2csv”以将 .xls 格式转换为 .csv 格式。因此，对于每个 .xls 文件，我必须获取文件名并将其附加到 .csv。
For instance, I have a test.xls file and the arguments fro xls2csv are : xls2csv test.xls test.csv
例如，我有一个 test.xls 文件，xls2csv 的参数是： xls2csv test.xls test.csv
Did I make sense?
我说的有道理吗？
采纳答案by Ignacio Vazquez-Abrams
bash:
重击：
for f in *.xls ; do xls2csv "$f" "${f%.xls}.csv" ; done
回答by Tom
Look at the findcommand.
查看find命令。
What you are looking for is something like
你正在寻找的是类似的东西
find . -name "*.xls" -type f -exec program
Post edit
发表编辑
find . -name "*.xls" -type f -exec xls2csv '{}' '{}'.csv;
will execute xls2csv file.xls file.xls.csv
将执行 xls2csv file.xls file.xls.csv
Closer to what you want.
更接近你想要的。
回答by ghostdog74
find . -type f -name "*.xls" -printf "xls2csv %p %p.csv\n" | bash
bash 4 (recursive)
bash 4（递归）
shopt -s globstar
for xls in /path/**/*.xls
do
xls2csv "$xls" "${xls%.xls}.csv"
done
回答by AndrewBourgeois
for i in *.xls ; do
[[ -f "$i" ]] || continue
xls2csv "$i" "${i%.xls}.csv"
done
The first line in the dochecks if the "matching" file really exists, because in case nothing matches in your for, the dowill be executed with "*.xls" as $i. This could be horrible for your xls2csv.
在第一行do，如果“匹配”文件确实存在，因为在你的情况没有什么比赛的检查for中，do将用“的* .xls”的执行$i。这对您的xls2csv.
