dump all mysql tables into separate files automagically?

Question

I'd like to get dumps of each mysql table into separate files. The manual indicates that the syntax for this is

mysqldump [options] db_name [tbl_name ...] 

Which indicates that you know the table names before hand. I could set up the script that knows each table name now, but say I add a new table down the road and forget to update the dump script. Then I'm missing dumps for one or more table.

Is there a way to automagically dump each existing table into a separate file? Or am I going to have to do some script-fu; query the database, get all the table names, and dump them by name.

If I go the script-fu route, what scripting langauges can access a mysql database?

Answer 1

Here's a script that dumps table data as SQL commands into separate, compressed files. It does not require being on the MySQL server host, doesn't hard-code the password in the script, and is just for a specific db, not all db's on the server:

#!/bin/bash  # dump-tables-mysql.sh # Descr: Dump MySQL table data into separate SQL files for a specified database. # Usage: Run without args for usage info. # Author: @Trutane # Ref: http://stackoverflow.com/q/3669121/138325 # Notes: #  * Script will prompt for password for db access. #  * Output files are compressed and saved in the current working dir, unless DIR is #    specified on command-line.  [ $# -lt 3 ] && echo "Usage: $(basename $0) <DB_HOST> <DB_USER> <DB_NAME> [<DIR>]" && exit 1  DB_host=$1 DB_user=$2 DB=$3 DIR=$4  [ -n "$DIR" ] || DIR=. test -d $DIR || mkdir -p $DIR  echo -n "DB password: " read -s DB_pass echo echo "Dumping tables into separate SQL command files for database '$DB' into dir=$DIR"  tbl_count=0  for t in $(mysql -NBA -h $DB_host -u $DB_user -p$DB_pass -D $DB -e 'show tables')  do      echo "DUMPING TABLE: $DB.$t"     mysqldump -h $DB_host -u $DB_user -p$DB_pass $DB $t | gzip > $DIR/$DB.$t.sql.gz     tbl_count=$(( tbl_count + 1 )) done  echo "$tbl_count tables dumped from database '$DB' into dir=$DIR"