我有一个大文本文件，如下例所示。它有更多的行组由空行分隔。
Aggr2_N1_SATA
Normal
192.168.1.2:/floluesxprd5_ds_vol1
Unknown
522.50 GB
478.69 GB
NFS
10/14/2020 3:21:52 PM
Enabled
Disabled
Not supported
boot_lun_svr1
Normal
NETAPP Fibre Channel Disk (naa.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx):3
Non-SSD
5.00 GB
4.29 GB
VMFS5
2/10/2020 4:26:37 PM
Enabled
Disabled
Supported
我想将其转换为如下所示：
Aggr2_N1_SATA,Normal,192.168.1.2:/floluesxprd5_ds_vol1,Unknown,522.50 GB,478.69 GB,NFS,10/14/2020,3:21:52 PM,Enabled,Disabled,Not supported
boot_lun_svr1,Normal,NETAPP Fibre Channel Disk (naa.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx):3,Non-SSD,5.00 GB,4.29 GB,VMFS5,2/10/2020 4:26:37 PM,Enabled,Disabled,Supported
感谢任何想法！
答案 0 :(得分：8)
perl的：
perl -00 -lpe 's/\n/,/g' file
答案 1 :(得分：6)
awk -v RS= -v OFS=, '{$1 = $1} 1' file
输出：
Aggr2_N1_SATA,Normal,192.168.1.2:/floluesxprd5_ds_vol1,Unknown,522.50 GB,478.69 GB,NFS,10/14/2020 3:21:52 PM,Enabled,Disabled,Not supported
boot_lun_svr1,Normal,NETAPP Fibre Channel Disk (naa.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx):3,Non-SSD,5.00 GB,4.29 GB,VMFS5,2/10/2020 4:26:37 PM,Enabled,Disabled,Supported
之间有空格：
awk -v RS= -v OFS=, '{$1 = $1} NR > 1 { print "" } 1' file
输出：
Aggr2_N1_SATA,Normal,192.168.1.2:/floluesxprd5_ds_vol1,Unknown,522.50 GB,478.69 GB,NFS,10/14/2020 3:21:52 PM,Enabled,Disabled,Not supported
boot_lun_svr1,Normal,NETAPP Fibre Channel Disk (naa.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx):3,Non-SSD,5.00 GB,4.29 GB,VMFS5,2/10/2020 4:26:37 PM,Enabled,Disabled,Supported
每行后面都有空格：
awk -v RS= -v OFS=, -v ORS='\n\n' '{$1 = $1} 1' file
答案 2 :(得分：2)
这是使用GNU sed完成路由的方法。请注意，这仅供参考，您最好使用glenn jackman's perl解决方案，因为它易于理解且易于理解。
简单的方法是在一个由,分隔的长字符串中读取整个文件，然后在看到两个连续的,时添加换行符。
sed '
:a; # Create a label a for a loop
$!N; # Append the next line to pattern space if it is not the last line
s/\n/,/; # Substitute the newline with ,
ta; # If the substitution modified pattern space repeat the loop
s/,,/\n\n/g # At the end, substitute two , with two newlines
' file
sed ':a;$!N;s/\n/,/;ta;s/,,/\n\n/g' file
但是，在一个长字符串中读取整个文件并不是一种有效的解决方案。因此sed有一个段落模式，允许您一次处理一个段落。
sed '
/./ { # If it is not a blank line
H; # Append the pattern space to hold space
$!d # Delete it if it is not the last line
}
x # When we encounter a blank line, we swap the pattern and hold space
s/\n// # We remove the first empty newline
s/\n/,/g # Replace all newlines with ,
$!G # If it not the last line swap hold and pattern space for blank lines
' file
sed '/./{H;$!d};x;s/\n//;s/\n/,/g;$!G' file
Aggr2_N1_SATA,Normal,192.168.1.2:/floluesxprd5_ds_vol1,Unknown,522.50 GB,478.69 GB,NFS,10/14/2020 3:21:52 PM,Enabled,Disabled,Not supported
boot_lun_svr1,Normal,NETAPP Fibre Channel Disk (naa.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx):3,Non-SSD,5.00 GB,4.29 GB,VMFS5,2/10/2020 4:26:37 PM,Enabled,Disabled,Supported
答案 3 :(得分：0)
Perl解决方案：
perl -pe '$comma and !/^$/ and print "," ;$comma = !/^$/ and chomp or print "\n"' input
答案 4 :(得分：0)
也许这可能有效
perl -00 -F"\n" -lanE 'say join(",", @F)'
打印
Aggr2_N1_SATA,Normal,192.168.1.2:/floluesxprd5_ds_vol1,Unknown,522.50 GB,478.69 GB,NFS,10/14/2020 3:21:52 PM,Enabled,Disabled,Not supported
boot_lun_svr1,Normal,NETAPP Fibre Channel Disk (naa.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx):3,Non-SSD,5.00 GB,4.29 GB,VMFS5,2/10/2020 4:26:37 PM,Enabled,Disabled,Supported
和
perl -00 -F"\n" -lanE 'say join(",", @F),"\n"'
打印
Aggr2_N1_SATA,Normal,192.168.1.2:/floluesxprd5_ds_vol1,Unknown,522.50 GB,478.69 GB,NFS,10/14/2020 3:21:52 PM,Enabled,Disabled,Not supported
boot_lun_svr1,Normal,NETAPP Fibre Channel Disk (naa.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx):3,Non-SSD,5.00 GB,4.29 GB,VMFS5,2/10/2020 4:26:37 PM,Enabled,Disabled,Supported
答案 5 :(得分：0)
根据这个答案（Format output in columns [bash, grep, sed, awk]），如果每十一行都被转置：
printf "%s,%s,%s,%s,%s,%s,%s,%s,%s,%s,%s\n" $(sed 's/ /_/g' file_name) | sed 's/_/ /g'