CODEWITHSUNDEEP
shell
Tofuwarrior
Tofuwarrior

Reputation: 669

shell script variable being truncated - why

I have been trying to customise this very useful (in principle) backup to s3 script. I really am not a shell scripter to any real level and I can't work out why this line is truncating the variable.

so e.g. 

DB=abcdefg
abcdefg_USER=testuser

USER=$(eval echo \$${DB}_USER)

The eval statement is returning bcdefg_USER so is truncating the variable and echoing out bcdefg_USER not abcdefg_USER and so isn't evaluating the variable abcdefg_USER

Running on an amazon linux ec2 instance.

Anyone explain to me what I am missing, I've tried playing around with the escaping and braces etc and echoing out each stage in the process but can't get a handle on what is going on.

Thanks

full script below:

    ## Specify data base schemas to backup and credentials
DATABASES="wp myotherdb"

## Syntax databasename as per above _USER and _PW
wp_USER=username
wp_PW=password
myotherdb_USER=username
myotherdb_PW=password

## Specify directories to backup (it's clever to use relaive paths)
DIRECTORIES="/var/www root etc/cron.daily etc/cron.monthly etc/apache2 etc/mysql etc/php5" 

## Initialize some variables
DATE=$(date +%d)
BACKUP_DIRECTORY=/tmp/backups
S3_CMD="s3cmd"

## Specify where the backups should be placed
S3_BUCKET_URL=s3://mybackupbucket/$DATE/

## The script
cd /
mkdir -p $BACKUP_DIRECTORY
rm -rf $BACKUP_DIRECTORY/*

## Backup MySQL:s
for DB in $DATABASES
do
BACKUP_FILE=$BACKUP_DIRECTORY/${DB}.sql
USER=$(eval echo \$${DB}_USER)
PASSWORD=$(eval echo \$${DB}_PW)
/usr/bin/mysqldump -v -u $USER --password=$PASSWORD -h localhost -r $BACKUP_FILE $DB 2>&1
gzip $BACKUP_FILE 2>&1
$S3_CMD put ${BACKUP_FILE}.gz $S3_BUCKET_URL 2>&1
done

## Backup of config directories
for DIR in $DIRECTORIES
do
BACKUP_FILE=$BACKUP_DIRECTORY/$(echo $DIR | sed 's/\//-/g').tgz
tar zcvf ${BACKUP_FILE} $DIR 2>&1
$S3_CMD put ${BACKUP_FILE} $S3_BUCKET_URL 2>&1
done

Upvotes: 0

Views: 2947

Answers (2)

Kaz
Kaz

Reputation: 58598

You're running into some weird bug involving command substitution and echo.

When using eval to access a computed variable name, it is not necessary to complicate things by involving echo wrapped in a process substitution. Try this pattern, which should work pretty much in any POSIX-like shell.

eval FINAL_VALUE=\$${COMPUTED_VAR_PREFIX}_FIXED_SUFFIX

That is to say, just generate the source code of the desired variable assignment, and eval that code.

Upvotes: 0

glenn jackman
glenn jackman

Reputation: 246847

Assuming that you are using bash, this is how to avoid eval:

$ DB=abcdefg
$ abcdefg_USER=testuser
$ tmpvar=${DB}_USER
$ USER=${!tmpvar}
$ echo $USER
testuser

If you have bash version 4, consider using associative arrays:

$ declare -A users
$ users[abcdefg]=testuser
$ echo "${users[$DB]}"
testuser

Upvotes: 3

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