Bash script - "tar czf ..." command ignore parameters

Question

I have a simple bash script called script.sh

!#/bin/bash
tar czf /var/log/apache2/Backup_$(date "+%d-%m-%Y").tar.gz /var/log/apache2/ --exclude='backup*' --exclude='Backup*'

When I rund this line in SSH-Console, all works fine but when I run the .sh-script at linux console with bash ./script.sh

then the tar czf ... command works but ignore the --exclude parameters

Whats wrong? I don't know... :/

what ignores the exclude parameter? the "bash script" or "tar". — Max Gunter, Mar 03 '14 at 12:21
You have a typo in your shebang, it should read `#!/bin/bash` instead of `!#/bin/bash`. — Adrian Frühwirth, Mar 03 '14 at 13:42

score 0 · Answer 1 · edited May 23 '17 at 12:25

0

try this one:

UPDATE

#!/bin/bash
tar  -cZf /var/log/apache2/Backup_$(date +%Y%m%d).tgz.gz /var/log/apache2/ --exclude='backup*' --exclude='Backup*'

exclude is a pattern... see these helps: help1 help2 help3

edited May 23 '17 at 12:25

Community

answered Mar 03 '14 at 12:12

MLSC

No, that does not change anything, only the Filename but the parameter "exclude=" will be ignored, too. – Max Gunter Mar 03 '14 at 12:20

score 0 · Answer 2 · answered Mar 03 '14 at 21:01

0

you have to use :

!#/bin/bash
tar -czf /var/log/apache2/Backup_`date "+%d-%m-%Y"`.tar.gz /var/log/apache2/ --exclude='backup*' --exclude='Backup*'

answered Mar 03 '14 at 21:01

PersianGulf

2 Answers2