Linux 如何使用 sed 或 awk 从路径中提取文件名
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How to extract filename from path using sed or awk
提问by Mark McWhirter
I am trying to parse a filename from a modified apache web access log entry that is tab delimited:
我正在尝试从制表符分隔的修改后的 apache 网络访问日志条目中解析文件名：
/common/common/img/pictos/klArrowRight.gif /common/common/img/pictos/klArrowRight.gif 03/Dec/2012:00:00:00 127.0.0.1 03/Dec/2012:00:00:00 us 404
I would like it to come out like this:
我希望它是这样出来的：
klArrowRight.gif /common/common/img/pictos/klArrowRight.gif 03/Dec/2012:00:00:00 127.0.0.1 03/Dec/2012:00:00:00 us 404
I have tried something like this in sed:
我在sed 中尝试过这样的事情：
's:.*/::'
's:.*/::'
However, it is too greedy, and it eats the rest of my line. I have been looking through posts, but so far no luck. Any hints?
然而，它太贪婪了，它吃掉了我的其余部分。我一直在浏览帖子，但到目前为止没有运气。任何提示？
采纳答案by Kent
the input/output in your question is not well formatted. do you need this?
您问题中的输入/输出格式不正确。你需要这个吗？
awk '{gsub(/\/.*\//,"",); print}' file
test
测试
kent$ echo "/common/common/img/pictos/klArrowRight.gif /common/common/img/pictos/klArrowRight.gif 03/Dec/2012:00:00:00 127.0.0.1 03/Dec/2012:00:00:00 us 404"|awk '{gsub(/\/.*\//,"",); print}'
output:
输出：
klArrowRight.gif /common/common/img/pictos/klArrowRight.gif 03/Dec/2012:00:00:00 127.0.0.1 03/Dec/2012:00:00:00 us 404
回答by zb'
using perl regexp and basename (i not think you stuck on sed/awk):
使用 perl regexp 和 basename（我不认为你坚持使用 sed/awk）：
perl -p -e 'use File::Basename;s/([^\s]+\s+)[^\s]+\s+//;print basename()'
example:
例子：
echo "/common/common/img/pictos/klArrowRight.gif /common/common/img/pictos/klArrowRight.gif 03/Dec/2012:00:00:00 127.0.0.1 03/Dec/2012:00:00:00 us 404" |
perl -p -e 'use File::Basename;s/([^\s]+\s+)[^\s]+\s+//;print basename()'
klArrowRight.gif /common/common/img/pictos/klArrowRight.gif 03/Dec/2012:00:00:00 127.0.0.1 03/Dec/2012:00:00:00 us 404
回答by Ed Morton
awk 'BEGIN{FS=OFS="\t"} {sub(/.*\//,"",)} 1' file
回答by Norman Gray
You can do this pretty easily with just sed, as long as you tell it not to be too greedy:
只要你告诉它不要太贪婪，你就可以很容易地用 sed 做到这一点：
% echo '/img/pictos/klArrowRight.gif 03/Dec/2012' | sed 's,^[^ ]*/,,'
klArrowRight.gif 03/Dec/2012
%
(that is, "starting at the beginning of the line, find the longest-possible list of non-space characters, followed by a slash")
（即“从行首开始，找到最长可能的非空格字符列表，后跟斜杠”）
回答by Steve
One way using GNU grep:
一种使用方式GNU grep：
grep -oP "[^/]*\t.*" file
Results:
结果：
klArrowRight.gif /common/common/img/pictos/klArrowRight.gif 03/Dec/2012:00:00:00 127.0.0.1 03/Dec/2012:00:00:00 us 404
回答by user12531
None of the given answers seem to be correct completely when only the extraction of a filename from a given absolute path is desired. Therefore I give here the solution. Let's consider in variable filenamewe have the complete path, e.g., filename=/ABC/DEF/GHIThen,
当只需要从给定的绝对路径中提取文件名时，给出的答案似乎都不是完全正确的。因此我在这里给出了解决方案。让我们考虑在变量文件名中我们有完整的路径，例如，filename=/ABC/DEF/GHI然后，
echo $filename | awk 'BEGIN{FS="/"}{print $NF}'
will result in the filename GHI.
将导致文件名GHI。
