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如何从 shell 中显示两个文件中的前 n 个不同行？

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如何从 shell 中显示两个文件中的前 n 个不同行？

如何从 shell 中显示两个文件中的前 n 个不同行？我已经尝试过，grep -vf但它并没有真正起作用。

假设 n = 5 此输入：

file1

a
b
c
d
e
f
g
h
i
j
k
l
m
n
o

file2

This line is not the same
b
c
d
This is still not the same
Neither is this
g
h
Nor this
DIFFERENT
k
This is not the same, too
m
another different line
o

将产生输出：

This line is not the same
This is still not the same
Neither is this
Nor this
DIFFERENT

答案1

这是我的建议：

diff -u file1 file2 --unchanged-line-format= --old-line-format= --new-line-format=%L | head -n 5

This line is not the same
This is still not the same
Neither is this
Nor this
DIFFERENT

答案2

假设您的文件不包含制表符（如果包含，请选择另一个明确的分隔符），您可以这样做

$ paste file1 file2 | awk -F'\t' '$2 != $1 {print $2; n++} n==5 {exit}'
This line is not the same
This is still not the same
Neither is this
Nor this
DIFFERENT

答案3

使用 bash 原语，使用固定文件描述符来使事情变得简单。 （未经测试）

# open the two files on fd8 and fd9, should have some error checking
exec 8<file1 9<file2
# start the loop
for(c=0;c<6;)
do
    # read a line from first file, don't worry about EOF 
    IFS="" read -r -u 8 l1
    # read a line from second file, exit the loop if EOF
    read -r -u 9 l2 || break
    # loop if the 2 lines are the same
    [ "$l1" -eq "$l2" ] && continue
    # ok, a different line. Output from file2, bump count and loop
    let c++
    printf '%s\n' "$l2"
done
# If we get here we either have hit EOF on file2 or have printed our 6 lines
# Either way just tidy up
# close the file descriptiors
exec 8<&- 9<&-

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