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up vote 1 down vote favorite

I have the snippet below in a larger script to basically take my balances out of ledger into two arrays so I can print them the way I want to view them. 

#!/bin/bash
assets=("assets:checking" "assets:google wallet" "assets:savings" "assets:cash")
assets-bal=()

num=${#assets[@]}
for $i in {0..${num}}
do
  read -a tmp <<< `ledger -f finances balance "${assets[${i}]}"`
  assets-bal[${i}]=tmp[0]
  echo "${assets[${i}]}   ${assets-bal[${i}]}"
done

Every time I try runnning the script I get the error: 

syntax error near unexpected token `num=${#assets[@]}'
`num=${#assets[@]}'

From my searching, there should be nothing wrong with that line and I just keep coming up empty trying to find a reason why it won't work.

Can anybody point out where I'm wrong?

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1  
Take a look at shellcheck.net –  Cyrus Dec 9 '14 at 20:54

4 Answers 4

up vote 4 down vote accepted

The problem is probably

assets-bal=()

Variable names must not contain a dash. You are limited to underscores.

I don't know how bash interprets assets-bal=() but it considers that an incomplete command which has to be finished on another line. Just run that line in a shell to see what I mean.

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1  
This line is interpreted as the start of a function declaration. After assets-bal=(), try { echo wibble; } then assets-bal=. Bash allows function names with nonstandard characters like - and =, but it doesn't allow function bodies to be a simple command. –  Gilles Dec 9 '14 at 23:53
up vote 2 down vote

Brace expansion does not support variables since it is done before variables are expanded. Use seq instead, if you absolutely must:

for i in $(seq 0 ${num})

Or (much more preferable) get the array keys directly:

for i in "${!assets[@]}"

(Also, for i in .., not for $i in ...)

Hauke Laging got the first error. Once you fix that, this will probably be the next error.

And instead of 

read -a tmp <<< `ledger -f finances balance "${assets[${i}]}"`

Consider using:

tmp=($(ledger -f finances balance "${assets[${i}]}"))

The () outside the command substitution converts it into an array.

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@Hauke Very sorry for the typo! –  muru Dec 9 '14 at 21:05
    
Thanks, I'll definitely use both of those. I probably would have caught the $i error, I just hadn't made it that far with the script failing earlier. –  user66330 Dec 9 '14 at 21:06
up vote 1 down vote

You have a syntax ERROR : you can't use a - in the name of a variable.

assets-bal

should be replaced by

 assets_bal

For the loop, you can use a for loop in the C style :

for ((i=0; i<${num}; i++)); do
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You have a typo in variable name –  sputnick Dec 9 '14 at 21:01
up vote 0 down vote

I think maybe you're relying a little too heavily on array types there - and variables in general. If I understand what you're doing up there correctly, I suspect this should do much of the same:

for a in checking google\ wallet savings cash
do  a=assets:$a
    printf "$a\t%d\n" "$(ledger -f finances balance "$a")"
done

That could probably be improved upon a great deal by somehow getting a reading of all values for $a above at once in a stream and then arranging it with a scripted editor like sed or similar. In general a shell variable is a pretty awful place to keep information of any value - or of any significant size for that matter. A shell variable should store only enough information to make retrieving actual information from a file more manageable.

Written as a shell function, the above might look like:

assets() if   [ "$#" -gt 0 ]
         then while [ "$#" -gt 0 ]
              do    printf "assets:%s\t%d\n" "$1" "$(
                    ledger -f finances balance "assets:$1")"
              shift;done
         else assets checking google\ wallet savings cash
         fi

...and could be called like:

assets

...for your default list, or like...

assets cash checking

...for a different list without every setting a single shell variable value at all.

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