bash awk第一列和第三列包含所有内容

Question

I am working on the following bash script:

# 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 $1, $3}' |

# 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?

Answer 1

在这种情况下，您可以使用cut而不是awk：

  cut -f1,3- -d ' '

Answer 2

awk '{ $2 = ""; print }' # remove col 2

Answer 3

If you don't mind a little whitespace:

awk '{ $2="" }1'

But UUOC and grep :

< dbfake awk '/100%/ { $2="" }1' | ...

If you'd like to trim that whitespace:

< dbfake awk '/100%/ { $2=""; sub(FS "+", FS) }1' | ...

---

---

For fun, here's another way using GNU sed :

< dbfake sed -r '/100%/s/^(\S+)\s+\S+(.*)/\1\2/' | ...

Answer 4

Others responded in various ways, but I want to point that using xargs to multiplex output is rather bad idea.

Instead, why don't you:

awk '$2=="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.

You could do awk ...; print}' | tee fake.log awk ...; print}' | tee fake.log awk ...; print}' | tee fake.log , but there is not much point in forking tee, if awk can handle it as well.

Answer 5

所有你需要的是：

awk 'sub(/.*100% /,"")' dbfake | tee "fake.log"