bash awk 第一列和第三列之后的所有内容
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bash awk first 1st column and 3rd column with everything after
提问by ooXei1sh
I am working on the following bash script:
我正在处理以下 bash 脚本：
# contents of dbfake file
1 100% file 1
2 99% file name 2
3 100% file name 3
#!/bin/bash
# cat out data
cat dbfake |
# select lines containing 100%
grep 100% |
# print the first and third columns
awk '{print , }' |
# echo out id and file name and log
xargs -rI % sh -c '{ echo %; echo "%" >> "fake.log"; }'
exit 0
This script works ok, but how do I print everything in column $3 and then all columns after?
这个脚本可以正常工作，但是如何打印 $3 列中的所有内容，然后打印所有列？
回答by mjuarez
You can use cut instead of awk in this case:
在这种情况下，您可以使用 cut 而不是 awk ：
cut -f1,3- -d ' '
回答by DigitalRoss
awk '{ = ""; print }' # remove col 2
回答by Steve
If you don't mind a little whitespace:
如果你不介意一点空白：
awk '{ ="" }1'
But UUOCand grep:
但是UUOC和grep：
< dbfake awk '/100%/ { ="" }1' | ...
If you'd like to trim that whitespace:
如果你想修剪那个空格：
< dbfake awk '/100%/ { =""; sub(FS "+", FS) }1' | ...
For fun, here's another way using GNU sed:
为了好玩，这是另一种使用方式GNU sed：
< dbfake sed -r '/100%/s/^(\S+)\s+\S+(.*)//' | ...
回答by Steve
Others responded in various ways, but I want to point that using xargs to multiplex output is rather bad idea.
其他人以各种方式回应，但我想指出使用 xargs 来多路输出是相当糟糕的主意。
Instead, why don't you:
相反，你为什么不：
awk '=="100%" { sub("100%[[:space:]]*",""); print; print >>"fake.log"}' dbfake
That's all. You don't need grep, you don't need multiple pipes, and definitely you don't need to fork shell for every line you're outputting.
就这样。您不需要 grep，也不需要多个管道，而且绝对不需要为输出的每一行分叉 shell。
You could do awk ...; print}' | tee fake.log, but there is not much point in forking tee, if awk can handle it as well.
你可以这样做awk ...; print}' | tee fake.log，但是如果 awk 也可以处理它，那么分叉 tee 没有多大意义。
回答by Ed Morton
All you need is:
所有你需要的是：
awk 'sub(/.*100% /,"")' dbfake | tee "fake.log"
