bash Unix：合并多个文件，同时删除所有文件的第一行
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Unix: merge many files, while deleting first line of all files
提问by Abdel
I have >100 files that I need to merge, but for each file the first line has to be removed. What is the most efficient way to do this under Unix? I suspect it's probably a command using catand sed '1d'. All files have the same extension and are in the same folder, so we probably could use *.extension to point to the files. Many thanks!
我有超过 100 个文件需要合并，但对于每个文件，必须删除第一行。在 Unix 下执行此操作的最有效方法是什么？我怀疑这可能是一个使用cat和sed '1d'的命令。所有文件都有相同的扩展名并且都在同一个文件夹中，所以我们可能可以使用 *.extension 来指向这些文件。非常感谢！
回答by xpapad
Assuming your filenames are sorted in the order you want your files appended, you can use:
假设您的文件名按您希望文件附加的顺序排序，您可以使用：
ls *.extension | xargs -n 1 tail -n +2
EDIT: After Sorin and Gilles comments about the possible dangers of piping lsoutput, you could use:
编辑：在 Sorin 和 Gilles 评论管道ls输出的可能危险之后，您可以使用：
find . -name "*.extension" | xargs -n 1 tail -n +2
回答by sorpigal
Everyone has to be complicated. This is really easy:
每个人都必须是复杂的。这真的很容易：
tail -q -n +2 file1 file2 file3
And so on. If you have a large number of files you can load them in to an array first:
等等。如果您有大量文件，您可以先将它们加载到数组中：
list=(file1 file2 file3)
tail -q -n +2 "${list[@]}"
All the files with a given extension in the current directory?
当前目录中具有给定扩展名的所有文件？
list=(*.extension)
tail -q -n +2 "${list[@]}"
Or just
要不就
tail -q -n +2 *.extension
回答by Douglas Leeder
Just append each file after removing the first line.
删除第一行后只需附加每个文件。
#!/bin/bash
DEST=/tmp/out
FILES=space separated list of files
echo "" >$DEST
for FILE in $FILES
do
sed -e'1d' $FILE >>$DEST
done
回答by Gilles 'SO- stop being evil'
tailoutputs the last lines of a file. You can tell it how many lines to print, or how many lines to omit at the beginning (-n +Nwhere N is the number of the first line to print, counting from 1 — so +2omits one line). With GNU utilities (i.e. under Linux or Cygwin), FreeBSD or other systems that have the -qoption:
tail输出文件的最后几行。您可以告诉它要打印多少行，或者在开头省略多少行（-n +N其中 N 是要打印的第一行的编号，从 1 开始计数——因此+2省略了一行）。使用 GNU 实用程序（即在 Linux 或 Cygwin 下）、FreeBSD 或其他具有以下-q选项的系统：
tail -q -n +2 *.extension
tailprints a header before each file, and -qis not standard. If your implementation doesn't have it, or to be portable, you need to iterate over the files.
tail在每个文件之前打印一个标题，-q这不是标准的。如果您的实现没有它，或者是可移植的，您需要遍历文件。
for x in *.extension; do tail -n +2 <"$x"; done
Alternatively, you can call Awk, which has a way to identify the first line of each file. This is likely to be faster if you have a lot of small files and slower if you have many large files.
或者，您可以调用 Awk，它有一种方法可以识别每个文件的第一行。如果您有很多小文件，这可能会更快，如果您有很多大文件，这可能会更慢。
awk 'FNR != 1' *.extension
回答by Vijay
ls -1 file*.txt | xargs nawk 'FNR!=1'
