下面是文件的一部分,其中列出了 zimbra 帐户(500+),用空换行符分隔
cn: Jack
displayName: Jack Johnson
givenName: Jack
sn: johnson
zimbraMailDeliveryAddress: Jack@example.com
cn: james ryan
displayName: James Ryan
givenName: James
sn: Ryan
zimbraMailDeliveryAddress: James@example.com
....
我想要包含如下内容的文件,以便我可以使用 zmprove
将它们导入新服务器
cn: Jack displayName: Jack Johnson givenName: Jack sn: johnson zimbraMailDeliveryAddress: Jack@example.com
cn: james ryan displayName: James Ryan givenName: James sn: Ryan zimbraMailDeliveryAddress: James@example.com
我尝试在不删除新行的情况下编写脚本,但到目前为止无法提取
for line in `cat /tmp/account3.txt`;
do
echo $line | grep "zimbraMailDeliveryAddress:" > /dev/null
RESULT=$?
if [ $RESULT -eq 0 ]; then
email=`echo $line | cut -d' ' -f2` > /dev/null
continue
elif echo $line | grep "sn:" > /dev/null
RESULT=$?
if [ $RESULT -eq 0 ]; then
sn=`echo $line | awk '{ print $2; }'` > /dev/null
continue
elif echo $line | grep "givenName:" > /dev/null
RESULT=$?
if [ $RESULT -eq 0 ]; then
givenName=`echo $line | awk '{ print $2; }'` > /dev/null
continue
elif echo $line | grep "displayName:" > /dev/null
RESULT=$?
if [ $RESULT -eq 0 ]; then
displayName=`echo $line | awk '{ print $2; }'` > /dev/null
continue
elif echo $line | grep "cn:" > /dev/null
RESULT=$?
if [ $RESULT -eq 0 ]; then
cn=`echo $line | cut -d' ' -f2` > /dev/null
continue
fi
else
:
fi
echo $email $sn $cn $displayName $givenName
done
# awk '/cn:|displayName:|givenName:|sn:|zimbraMailDeliveryAddress:/{printf "%s ", $0; next} 1' /tmp/account2.txt
最佳答案
$ awk -v RS= '{$1=$1}1' file
cn: Jack displayName: Jack Johnson givenName: Jack sn: johnson zimbraMailDeliveryAddress: Jack@example.com
cn: james ryan displayName: James Ryan givenName: James sn: Ryan zimbraMailDeliveryAddress: James@example.com
关于linux - 删除文件中每个换行符之间的换行符,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38289613/