variable interpolation in shell script

Question

I wrote the following shell script for a pg backup:

#!/bin/bash

PG_USER=donato
DATABASE=mydb
SERVER=216.58.219.174
DIR="$HOME/pg_bak"
DATE=$(date +"%m_%d_%y")
FILE="$DATABASE_$DATE"
ERROR_FILE="$HOME/pg_bak/error_bak/$FILE_error.txt"
# pass @ .pgpass

PG_BAK_NOW () {
  pg_dump -h $SERVER -U $PG_USER $DATABASE >$FILE 2>$ERROR_FILE
  code=$?
  if [ $code -ne 0 ]; then
    echo 1>&2 "The backup failed (exit code $code), check for errors in $ERROR_FILE"
  fi
}

echo "Ready to dump to $FILE" >> "$HOME/pg_status" 

cd $DIR
if [ -f "$FILE" ];
then
  rm $FILE
  PG_BAK_NOW 
else
  PG_BAK_NOW
fi

When I execute it, I know it executes for a bit of time:

$ pgrep -fl pg_bak.sh
4603 pg_bak.sh

But then it does crash:

$ ./pg_bak.sh
The backup failed (exit code 1), check for errors in /home/viggy/pg_bak/error_bak/.txt

Notice the .txt part. The name of the error file was supposed to be mydb_6_11_2016_error.txt, not .txt. Why did the bash script not interpolate the variable $FILE and the hardcoded string "_error"? It did interpolate $FILE in the dump file correctly, but not the error file. Why?

Answer 1

A very common mistake:

ERROR_FILE="$HOME/pg_bak/error_bak/${FILE}_error.txt"

Answer 2

As jlliagre pointed out you need to use braces around the part of the string you want bash to treat as a variable name. Otherwise it will only stop treating the string as the name when it gets to a character that can't be part of the name.

In this case FILE_error is being treated as the name of the variable because _ is allowed in variable names. You actually only want the FILE part, so you have to tell bash that specifically by putting braces around that part as ${FILE}_error.

In your example FILE_error was never set, so it expands to an empty string, so you get just .txt as the filename.