bash 查找文件名以指定字符串开头的所有文件？
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Find all files with a filename beginning with a specified string?
提问by RikSaunderson
I have a directory with roughly 100000 files in it, and I want to perform some function on all files beginning with a specified string, which may match tens of thousands of files.
我有一个包含大约 100000 个文件的目录，我想对所有以指定字符串开头的文件执行一些功能，这些文件可能匹配数万个文件。
I have tried
我试过了
ls mystring*
but this returns with the bash error 'Too many arguments'. My next plan was to use
但这会返回 bash 错误“参数过多”。我的下一个计划是使用
find ./mystring* -type f
but this has the same issue.
但这有同样的问题。
The code needs to look something like
代码需要看起来像
for FILE in `find ./mystring* -type f`
do
#Some function on the file
done
回答by Sergio Tulentsev
use
用
find . -name 'mystring*'
回答by jacanterbury
ls | grep "^abc"
will give you all files beginning(which is what the OP specifically required) with the substringabc.
It operates only on the current directory whereas findoperates recursively into sub folders.
将为您提供以 substring开头的所有文件（这是 OP 特别要求的）abc。
它仅在当前目录上find运行，而递归地运行到子文件夹中。
To use findfor only files startingwith your string try
要find仅用于以字符串开头的文件，请尝试
find . -name 'abc'*
找 。-名称'abc'*
回答by matson kepson
If you want to restrict your search only to files you should consider to use -type fin your search
如果您只想将搜索限制为文件，则应考虑-type f在搜索中使用
try to use also -inamefor case-insensitive search
尝试也-iname用于不区分大小写的搜索
Example:
例子：
find /path -iname 'yourstring*' -type f
You could also perform some operations on results without pipe sign or xargs
您还可以对没有管道符号或 xargs 的结果执行一些操作
Example:
例子：
Search for files and show their size in MB
搜索文件并以 MB 为单位显示其大小
find /path -iname 'yourstring*' -type f -exec du -sm {} \;
