Awk：在一行中删除最后一个记录分隔符

Question

I have a simple command (part of a bash script) that I'm piping through awk but can't seem to suppress the final record separator without then piping to sed. (Yes, I have many choices and mine is sed.) Is there a simpler way without needing the last pipe?

dolls = $(egrep -o 'alpha|echo|november|sierra|victor|whiskey' /etc/passwd \
| uniq | awk '{IRS="\n"; ORS=","; print}'| sed s/,$//);

Without the sed, this produces output like echo,sierra,victor, and I'm just trying to drop the last comma.

Answer 1

You don't need awk, try:

egrep -o ....uniq|paste -d, -s

Here is another example:

kent$  echo "a
b
c"|paste -d, -s
a,b,c

Also I think your chained command could be simplified. awk could do all things in an one-liner.

Answer 2

除了egrep，uniq，awk，sed等之外，所有这些都可以在一个awk命令中完成：

awk -F":" '!($1 in a){l=l $1 ","; a[$1]} END{sub(/,$/, "", l); print l}' /etc/password

Answer 3

Here is a small and quite straightforward one-liner in awk that suppresses the final record separator:

echo -e "alpha\necho\nnovember" | awk 'y {print s} {s=$0;y=1} END {ORS=""; print s}' ORS=","

Gives:

alpha,echo,november

So, your example becomes:

dolls = $(egrep -o 'alpha|echo|november|sierra|victor|whiskey' /etc/passwd | uniq | awk 'y {print s} {s=$0;y=1} END {ORS=""; print s}' ORS=",");

The benefit of using awk over paste or tr is that this also works with a multi-character ORS.

Answer 4

Since you tagged it bash here is one way of doing it:

#!/bin/bash

# Read the /etc/passwd file in to an array called names
while IFS=':' read -r name _; do 
  names+=("$name"); 
done < /etc/passwd

# Assign the content of the array to a variable
dolls=$( IFS=, ; echo "${names[*]}")

# Display the value of the variable
echo "$dolls"

Answer 5

echo "a
b
c" |

 mawk 'NF-= _==$NF' FS='\n' OFS=, RS=

a,b,c