从 bash 运行 mysql 查询； 其中查询存储在一个单独的文件中，并在 bash 中设置了变量

Question

I have a bash script that is responsible for querying a database and piping the results to a file:

date=`date +"%Y/%m/%d"`

/usr/bin/mysql -u $db_user -p$db_pass -h $db_name $db_schema << MYSQLEOF > $output_file

  select *
  from table
  where date = $date

MYSQLEOF

Now expanding this project into other areas, for the sake of re-usability/orthogonality I wanted to house the query in it's own file and call with something like:

cmd=`echo $sql_file`

/usr/bin/mysql -u $db_user -p$db_pass -h $db_name $db_schema -e "$cmd" > $output_file

sql file:

  select *
  from table
  where date = $date

I am having trouble finding out how I can (in bash) do this, while still being able to have and adjust variables ($date) inside the sql file. Echoing the .sql file into a bash variable, I haven't been able to get "$date" to not be taken as a literal string.

Is there a solution in bash, or should I look into something like perl to handle this?

Answer 1

Use the mySQL CLI's source command.

queryfile=my_query.sql
/usr/bin/mysql -u $db_user -p$db_pass -h $db_name $db_schema -e "source $queryfile" > $output_file

Answer 2

I found a solution using command-line Perl to search/replace on the fly.

Bash script:

date=`date +"%Y/%m/%d"`
export date

cmd=`perl -lpe 's/DATE_VAR/$ENV{date}/g' "$sql_file"`

/usr/bin/mysql -u $db_user -p$db_pass -h $db_name $db_schema -e "$cmd" > $output_file

and in sql_file:

select *
from table
where date = 'DATE_VAR'