目前,我使用exec 22>& 2 21>& 1 1> $log 2>& 1将我的bash脚本写入stderr和stdout到日志文件.我想为每个条目添加一个时间戳,但我还没有找到一个简单的方法来做到这一点.理想情况下,对当前命令进行简单的更改,然后在其余部分之前将时间写在同一行上.
@H_403_2@这是我正在使用命令的脚本：
#!/bin/bash
#This script takes the server to rysnc as an argument. You can also tell
#the script to check the server_status.txt file.
#
#Example: /path/to/script/sync.sh grail true
#
#The arguments are order senstive. The server name must come before the status
#check value.
#Logfile
LOG=/var/log/sync.log
DIRECTORYS="auth/ keys/ log/mailwhen/ intranet/ www/calmaa/data/ www/admatch/data/ www/sfhsa/data/ www/hfa3_org www/padmatch/ www/serverdown/"
if [ "x$2" == "xfalse" ]; then
return 0
elif [ "x$2" == "xtrue" ]; then
if [ `cat /srv/www/wan*/server_status.txt` == "primary" ]; then
exit 0
fi
else
echo "Please use \"true\" or \"false\" for the second value."
exit 1
fi
# Copy stdout and stderr,and then open the logfile
exec 22>&2 21>&1 1>$log 2>&1
# Here is how to restore stdout and stderr:
# exec 2>&22 1>&21
for DIRECTORY in $DIRECTORYS; do
rsync -azu --delete --bwlimit=500 $1:/srv/$DIRECTORY /srv/$DIRECTORY
done
解决方法
如果没有看到更多的脚本,我无法告诉您满足您特定需求的最佳方式.然而,这是一种可以根据您的需求进行调整的一般方法.
exec > >(while read -r line; do printf '%s %s\n' "$(date --rfc-3339=seconds)" "$line"; done)@H_403_2@输出的每行文本都将具有前置时间的时间戳.输出看起来像这样：
2013-09-04 21:32:14-05:00 An event occurred and this is the message 2013-09-04 21:32:37-05:00 Some time passed,another event produced a message
