列出目录中每个文件的前几行

I'm trying to create a really simple bash script, which will list the first few lines of every file in a specific directory. The directory should be specified by the argument. I think that the Grep command should be used, but I have really no idea how.

我正在尝试创建一个非常简单的bash脚本,它将列出特定目录中每个文件的前几行。该目录应由参数指定。我认为应该使用Grep命令,但我真的不知道如何。

My existing script does not seem to work at all, so it's no use putting it in here.

我现有的脚本似乎根本不起作用,所以把它放在这里是没用的。

5 个解决方案

#1

Use head command:

使用head命令:

head -3 /path/to/dir/*

#2

For any answer using head and *, redirect stderr to /dev/null unless you want to see errors like:

对于使用head和*的任何答案,将stderr重定向到/ dev / null除非你想看到如下错误:

head: error reading ‘tmp’: Is a directory

#3

for file in dir/*; do
    echo "-- $file --"
    head "$file"
    echo
done

#4

If you want the first few lines of all files ending in .txt, try

如果你想要以.txt结尾的所有文件的前几行,请尝试

head *.txt

head --lines=3 *.txt

#5

Because bash does filename expansion (globbing) by default, you can just let your shell expand input and let head do the rest:

因为bash默认情况下会进行文件名扩展(globbing),所以你可以让你的shell扩展输入,让head完成剩下的工作:

head *

The * wildcard expands to all the filenames in the working directory. On zsh you can see this nicely, when it autocompletes your commandline when you press tab.

*通配符扩展为工作目录中的所有文件名。在zsh上,当你按Tab键自动填充命令行时,你可以很好地看到它。

You can change the amount of lines with the -n argument to head.

您可以使用-n参数更改行数。

If you want to do this recursively:

如果你想以递归方式执行此操作:

find . \! -type d -exec head '{}' +

#1