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I need to create a bash script in order to:

  1. Empty my root Crontab;
  2. 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 -
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NineCattoRules
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1 Answers1

1

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 -
SYN
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