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I need to create a bash script in order to:
- Empty my root Crontab;
- Insert new Cronjobs via bash script.
For the first point I can use crontab -r
For the second point instead here I found this script:
#!/bin/bash
lines="* * * * * /path/to/command"
(crontab -u root -l; echo "$lines" ) | crontab -u root -
How can I cook this together in a bash script?
Something like this:
#!/bin/bash
crontab -r
line="* * * * * /path/to/command;
* * * * * /path/to/command2;
* * * * * /path/to/command3"
(crontab -u root -l; echo "$line" ) | crontab -u root -
1 Answer
The sample you posted would print current crontab and inject new directives.
If you intend to just inject new directives, wiping the current crontab, instead of your
lines="* * * * * /path/to/command"
( crontab -u root -l; echo "$lines" ) | crontab -u root -
Go with:
lines="* * * * * /path/to/command"
echo "$lines" | crontab -u root -
And, as you pointed it out in the comments, it is wrong, adding multiple crons, to use semicolons as a separator. You can go with:
lines=" line1
line2"
Or:
crontab -u root - <<EOF
line1
line2
EOF
Or:
(
echo line1
echo line2
) | crontab -u root -
