shell script variable being truncated - why

Question

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

When I run your example, USER is set to testuser as expected. Does this not happen for you? Can you make a test case, a shorter version of your script that shows the problem? (see the bash tag wiki for tips on what you can leave out) — that other guy
– that other guy, Commented Apr 17, 2014 at 20:15
Note that the eval command does not echo anything; the echo command does. The gluing together of ${DB} and _testuser is done as part of the expansion prior to the evaluation of the echo. If a character is being chopped at that point, it's quite a mystery; it generates the wrong variable name and so then the echo doesn't echo anything (or even raises an error if we trap uses of undefined variables with set -u). — Kaz
– Kaz, Commented Apr 17, 2014 at 20:28
"amazon linux ec2" isn't very helfpul: it is proving difficult to obtain any detailed specifications about what is the Amazon distribution of Linux. Are you using #!/bin/sh or #!/bin/bash in the script? What shell exactly is it using, and what version. — Kaz
– Kaz, Commented Apr 17, 2014 at 20:40
thank you for your comments and pointers, very much appreciated. I will know how to specify question better by detailing which shell etc next time. — Tofuwarrior
– Tofuwarrior, Commented Apr 18, 2014 at 14:57

glenn jackman · Accepted Answer · 2014-04-17 20:15:23Z

3

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

answered Apr 17, 2014 at 20:15

glenn jackman

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3 Comments

glenn jackman Over a year ago

Reference: look for "variable indirection" at gnu.org/software/bash/manual/…

glenn jackman Over a year ago

Also, get out of the habit of using ALL_CAPS_VARNAMES -- someday you'll accidentally use PATH=... and break your script.

Tofuwarrior Over a year ago

used arrays hence why accepted this answer. Thanks very much.

Kaz · Accepted Answer · 2014-04-17 20:38:13Z

0

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.

answered Apr 17, 2014 at 20:38

Kaz

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