command-line – Fix String Evaluation Not Working in Bash Scripts

bashcommand linescripts

I have the following array in a shell script:

#! usr/bin/bash

schemaPrefix=("aos")
tables_aos=("A" "B" "C")
colnames_aos=("id" "id" "id")

for j in "${!schemaPrefix[@]}"; do

aschema=${schemaPrefix[j]}
schema=$aschema
eval tables=tables_$aschema
echo ${tables[@]}
done

This code should display:

A B C

However, it displays:

tables_aos

What am I doing wrong? Thanks.

Best Answer

In bash version > 4.3, you can declare the tables variable as a nameref:

#! /usr/bin/bash

schemaPrefix=("aos")
tables_aos=("A" "B" "C")

declare -n tables

for j in "${!schemaPrefix[@]}"; do
  aschema=${schemaPrefix[j]}
  tables=tables_$aschema
  echo "${tables[@]}"
done

The loop could be written more simply without indirection as

for aschema in "${schemaPrefix[@]}"; do
  tables=tables_$aschema
  echo "${tables[@]}"
done

Related Solutions

Ubuntu – Execute a command stored into a $variable

Instead of using simple strings, build your command using arrays. Arrays provide a convenient interface: If a is an array, then "${a[@]}" (note the quotes) expands into each element of a, without additional field splitting or globbing (so spaces, and things like wildcards, should remain intact).

Example:

$ a=(printf "|%s|\n" "foo   bar" "*")
$ echo ${a[@]}
printf |%s|\n foo bar bin boot dev etc home lib lib64 lost+found mnt opt proc root run sbin srv sys tmp usr var

Note how the * was expanded, and how the extra spaces between foo and bar were lost. But with the "${a[@]}", these are preserved:

$ echo "${a[@]}"
printf |%s|\n foo   bar *

This is ideal for building commands. Now, you can do:

$ "${a[@]}"
|foo   bar|
|*|

See? The arguments were retained perfectly.

So, do:

COMMAND=(/usr/bin/exiftool "$PATH_NAME")
"${COMMAND[@]}"

How to Echo Differently Based on User Input in Bash Script

The main problem is here:

read $input

In bash, usually, $foo is the value of the variable foo. Here, you don't want the value, but the name of the variable, so it should be just:

read input

Similarly, in the if tests, $yes and $no should be just yes and no, since you just want the strings yes and no there.

You could use a case statement here, which (IMHO) makes it easier to do multiple cases based on input:

case $input in
[Yy]es)    # The first character can be Y or y, so both Yes and yes work
    echo "Hello!"  
    echo "Hello!" | festival --tts
    ;;
[Nn]o)     # no or No
    echo "Are you sure?"
    echo "Are you sure?" | festival --tts
    ;;
*)         # Anything else
    echo "Please answer yes or no."
    echo "Please answer yes or no." | festival --tts
    ;;
esac

You could wrap the two echo statements and the use of festival in a function to avoid repeating yourself:

textAndSpeech ()
{
    echo "$@"
    echo "$@" | festival --tts
}
case $input in
[Yy]es)    # The first character can be Y or y, so both Yes and yes work
    textAndSpeech "Hello!"  
    ;;
[Nn]o)     # no or No
    textAndSpeech "Are you sure?"
    ;;
*)         # Anything else
    textAndSpeech "Please answer yes or no."
    ;;
esac

With $input, bash replaces this with its value, which is nothing initially, so the read command run is:

read

And read by default stores the input in the variable REPLY. So you can, if you want, eliminate the input variable altogether and use $REPLY instead of $input.

Also have a look at the select statement in Bash.

Best Answer

Related Solutions

Ubuntu – Execute a command stored into a $variable

How to Echo Differently Based on User Input in Bash Script

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