bash - 查找正好有 n 行的所有文件

Question

我写了一个小脚本 has_n_lines.sh 解决了上述问题，它接收 n 作为参数(在这种情况下它是一组特定的 txt文件):

#!/bin/bash

files=`find . -name "*.txt"`

for file in $files
do
    nlines=`wc -l $file | cut -d " " -f1`
    if [ "$nlines" -eq "$1" ]
    then
        echo $file
    fi
done

有没有更简单的方法来做到这一点？

谢谢

Answer 1

是的，像这样应该可以解决问题

find . -name "*.txt" -exec bash -c 'wc -l < "$1"' -- {} \;

这当然是找到所有 .txt文件。然后它执行 wc -l通过逐个文件发送(< 发送，{} 使用逐个文件)。 \;表示-exec结束。通过 < 发送文件名使 wc -l只输出行数。现在您可以通过以下方式介绍比较:

find . -name "*.txt" -exec bash -c '(($(wc -l < "$1") == $n)) && echo "$1"' -- {} \;

哪里$n是你想要的行数。

Explanation (thanks to John Kugelman): Redirections (<) or if statements cannot be used directly with find ... -exec. This means that you need to explicitly invoke a subshell, inside which you can use them. For extra safety, -- is $0 and {} is $1, the latter is used in subshell. This ensures that filenames containing whitespaces are handled correctly.

Also you need to export $n prior to invoking above command, because a subshell cannot access $n. So if you run bash has_n_lines.sh 123, where 123 is the desired number of lines, prepend this to has_n_lines.sh: n=$1 && export $n;.