CODEWITHSUNDEEP
bash
Tom's
Tom's

Reputation: 2506

Initialize an array with a variable contening key

I want to construct an arry with the result of a SQLite query. I do 

result=`sqlite3 "$database" 'SELECT id,value FROM table'`

I obtain

1|value for id 1

2|value for id 2

(By defaut, sqlite use | as separator). What I want is using the id as the key in the bash array, and value as the bash value.

If I type directly 

array=([1]="value for id 1" [2]="value for id 2")

It work great ! So I modify result this way :

result=`sqlite3 "$database" 'SELECT id,value FROM table' | sed 's/^/[' | sed 's/|/]="/' | sed 's/$/"/' | sed ':a;N;$!ba;s/\n/ /g'`

and when i do echo "$result", I succesfully obtain :

[1]="value for id 1" [2]="value for id 2"

But now, when I do

array=($result)
array=("$result")
array=($(echo -n $result))
array=($(echo -n "$result"))
array=(`echo -n $result`)
array=(`echo -n "$result"`)

None of them work. If I do 

for i in "${!array[@]}";do
  echo "key = $i"
  echo "val = ${array[$i]}"
done

I end up with something like

key = 0

val = [1]="value

key = 1

val = for

key = 2

val = id

key = 3

val = 1"

key = 4

val = [2]="value

key = 5

val = for

key = 6

val = id

key = 7

val = 2"

What's wrong ? How can I initalize an array with $result ?

Upvotes: 0

Views: 44

Answers (1)

Charles Duffy
Charles Duffy

Reputation: 295815

Process substitution results aren't evaluated as code. This is, in short, a Very Good Thing -- if you had a value containing $(rm -rf $HOME), you wouldn't want it to be any other way.

Instead, explicitly read your results and assign them to your array:

declare -a array=( )
while IFS='|' read -r key value; do
  array[$key]=$value
done < <(sqlite3 "$database" 'SELECT id,value FROM table')

Upvotes: 1

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