How to substitute `$variable` within bash cat `<< 'EOF' >`? [duplicate]

Question

I write bash cat << 'EOF' > script to create files.script as below:

#!/bin/bash for sitename in (site1,site2,site3) do cat << 'EOF' >/home/$sitename/conf DEPLOY_DIR="/var/www/$sitename" git --work-tree="DEPLOY_DIR" EOF done

The result after run this script should like:

[root@localhost]cat home/site1/conf DEPLOY_DIR="/var/www/site1" git --work-tree="$DEPLOY_DIR"

The key is I need to substitute $sitename in DEPLOY_DIR="/var/www/$sitename" and keep git --work-tree="DEPLOY_DIR" as same.
How to do it?

I added a new answer to the duplicate explaining how to avoid expansion of some strings. Also the for loop looks wrong here -- you want for sitename in site1 site2 site3; do — tripleee
– tripleee, Commented Jan 30, 2019 at 7:07

NiteRain · Accepted Answer · 2019-01-30 07:07:24Z

5

#!/bin/bash for sitename in site1 site2 site3;do cat << EOF > /home/$sitename/conf DEPLOY_DIR="/var/www/$sitename" git --work-tree="\$DEPLOY_DIR" EOF done

answered Jan 30, 2019 at 7:07

NiteRain

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2 Comments

kittygirl Over a year ago

That's workable, but if you have lots of $variables , hard work.

NiteRain Over a year ago

This is the only way you will be able to extrapolate some variables while not extrapolating others. The other way would require more typing.

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