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I'm trying to get a software version by using the bash script. The Command is simple: 

/usr/local/sbin/keepalived -v

This is the output:

Keepalived v2.0.13 (02/18,2019), git commit v2.0.12-53-ga9ed1993+

Copyright(C) 2001-2019 Alexandre Cassen,

Built with kernel headers for Linux 3.10.0 Running on Linux 3.10.0-957.5.1.el7.x86_64 #1 SMP Fri Feb 1 14:54:57 UTC 2019

Config options: NFTABLES LVS VRRP VRRP_AUTH OLD_CHKSUM_COMPAT FIB_ROUTING SNMP_V3_FOR_V2 SNMP_VRRP SNMP_CHECKER SNMP_RFCV2 SNMP_RFCV3 FILE_LOGGING LOG_FILE_APPEND EINTR_CHECK

System options: PIPE2 SIGNALFD INOTIFY_INIT1 VSYSLOG EPOLL_CREATE1 IPV6_ADVANCED_API LIBNL3 RTA_ENCAP RTA_EXPIRES RTA_PREF FRA_TUN_ID RTAX_CC_ALGO RTAX_QUICKACK FRA_OIFNAME IFA_FLAGS IP_MULTICAST_ALL NET_LINUX_IF_H_COLLISION LIBIPTC_LINUX_NET_IF_H_COLLISION LIBIPVS_NETLINK VRRP_VMAC CN_PROC SOCK_NONBLOCK SOCK_CLOEXEC O_PATH GLOB_BRACE INET6_ADDR_GEN_MODE SO_MARK SCHED_RT SCHED_RESET_ON_FORK

From this output i need only the first row:

Keepalived v2.0.13 (02/18,2019), git commit v2.0.12-53-ga9ed1993+

I tried the following methods, but none of them works for me:

$ /usr/local/sbin/keepalived -v | head -n 1
$ echo "$(/usr/local/sbin/keepalived -v)" | head -n 1

In both cases, I'm receiving the full output instead of the first line only.

I tried to send stdout into file:

$ /usr/local/sbin/keepalived -v > /tmp/keepalived.txt
$ echo $(/usr/local/sbin/keepalived -v) > /tmp/keepalived.txt
$ /usr/local/sbin/keepalived -v | tee /tmp/keepalived.txt

But, I'm getting an empty file.

Can someone explain why is that happening and how can I get the first line only?

Ilie Soltanici
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1 Answers1

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It's happening because the output of /usr/local/sbin/keepalived -v is going to stderr instead of stdout.

You can redirect the output to stdout before extracting the data you want by doing keepalived -v 2>&1 | head -n 1

Daniel Simões
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