查找目录和子目录中的所有零字节文件

如何找到目录及其子目录中的所有零字节文件？

我这样做了:

#!/bin/bash
lns=`vdir -R *.* $dir| awk '{print $8"\t"$5}'`
temp=""
for file in $lns; do
if test $file = "0"; then
printf $temp"\t"$file"\n"
fi
temp=$file
done

但是，我只能在工作目录上得到结果，而不是 subdirs, 如果任何文件名包含一个空格，那么我只得到第一个单词后面跟着制表符

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最佳答案

To print the names of all files in and below $dir of size 0:

find "$dir" -size 0

Note that not all implementations of find will produce output by default, so you may need to do:

find "$dir" -size 0 -print

Two comments on the final loop in the question:

Rather than iterating over every other word in a string and seeing if the alternate values are zero, you can partially eliminate the issue you're having with whitespace by iterating over lines. eg:

printf '1 f1\n0 f 2\n10 f3\n' | while read size path; do
test "$size" -eq 0 && echo "$path"; done

Note that this will fail in your case if any of the paths output by ls contain newlines, and this reinforces 2 points: don't parse ls, and have a sane naming policy that doesn't allow whitespace in paths.

Secondly, to output the data from the loop, there is no need to store the output in a variable just to echo it. If you simply let the loop write its output to stdout, you accomplish the same thing but avoid storing it.

小开

No, you don't have to bother grep.

find $dir -size 0 ! -name "*.xml"
小开

As addition to the answers above:

If you would like to delete those files

find $dir -size 0 -type f -delete
小开

Bash 4+ tested - This is the correct way to search for size 0:

find /path/to/dir -size 0 -type f -name "*.xml"

Search for multiple file extensions of size 0:

find /path/to/dir -size 0 -type f \( -iname \*.css -o -iname \*.js \)

Note: If you removed the \( ... \) the results would be all of the files that meet this requirement hence ignoring the size 0.


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