CODEWITHSUNDEEP
bashshell
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Reputation: 5950

Writing echo to script

In a bash script, I need to create another script file, So I have the following

cat >/etc/script2.sh <<EOL
#!/bin/bash

mysql -e "SHOW TABLES FROM $DATABASE_NAME" > /var/lib/mysql-files/tables.txt

echo "$(tail -n +2 /var/lib/mysql-files/tables.txt)" > /var/lib/mysql-files/tables.txt
EOL

However, the echo at line 5 is executed and the output is added to script2.sh

How can I have script2 to look like

#!/bin/bash

mysql -e "SHOW TABLES FROM $DATABASE_NAME" > /var/lib/mysql-files/tables.txt

echo "$(tail -n +2 /var/lib/mysql-files/tables.txt)" > /var/lib/mysql-files/tables.txt

Without been echo executed?

UPDATE

With my current code when /var/lib/mysql-files/tables.txt does not exist I get tail: cannot open '/var/lib/mysql-files/tables.txt' for reading: No such file or directory

When the file exists I get something like

mysql -e "SHOW TABLES FROM reaction_production" > /var/lib/mysql-files/tables.txt

echo "table1
table2
table3" > /var/lib/mysql-files/tables.txt

Script2 must be evaluated when executed in the future, not when written now.

Thanks

Upvotes: 0

Views: 49

Answers (1)

Cristian Ramon-Cortes
Cristian Ramon-Cortes

Reputation: 1888

You need to scape the $ on the process substitution so that it is treated as a string in the current script and thus, evaluated when executing script2.sh.

#!/bin/bash

  cat > /etc/script2.sh << EOF
#!/bin/bash

mysql -e "SHOW TABLES FROM $DATABASE_NAME" > /var/lib/mysql-files/tables.txt

echo "\$(tail -n +2 /var/lib/mysql-files/tables.txt)" > /var/lib/mysql-files/tables.txt
EOF

Upvotes: 1

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