How to embed BASH variable in MySQL query in shell script file - mysql

Question

#!/bin/bash
fut=$(date -d "+14 days" +'%Y-%m-%d')
echo $fut
mysql -u user -ppassword base1 << "EOF"
INSERT INTO tableB SELECT * FROM tableA WHERE startdate < $fut;
....many lines of MySQL....
EOF
Mysql engine doesn't understand $fut as a variable. What syntax should be applied to make this work?

Answer 1

I have no problem with using a here-document, but I would like to suggest another possibility which is slightly more verbose, but has advantages when you may have to do more than just substituting variables : piping the output of a function.
#!/bin/bash
sql_statement()
{
local fut=$(date -d "+14 days" +'%Y-%m-%d')
echo "INSERT INTO tableB SELECT * FROM tableA WHERE startdate < $fut;"
echo "....many lines of MySQL...."
}
mysql -u user -ppassword base1 < <(sql_statement)
This is not required in this specific case, but allows the function to have additional logic (e.g. if blocks, loops...) to generate a more complex or variable SQL statement. As a general solution, it has nice advantages.

Answer 2

You need to remove the " from the EOF:
#!/bin/bash
fut=$(date -d "+14 days" +'%Y-%m-%d')
echo $fut
mysql -u user -ppassword base1 << EOF
INSERT INTO tableB SELECT * FROM tableA WHERE startdate < $fut;
....many lines of MySQL....
EOF
From bash docs:
No parameter substitution when the "limit string" is quoted or escaped.