BASH script returning command not found

Question

I am very new to bash programming and wanted to create a script that would store each result of find individually into an array. Now I want the command variable to expand on the statement MYRA=($(${Command} $1))

Command = 'find . -iname "*.cpp" -o -iname "*.h"'
declare -a MYRA
    MYRA=($(${Command} $1))
echo ${#MYRA[@]}

However when I try this script I get the result

$ sh script.sh
script.sh: line 1: Command: command not found
0

Any suggestions on how I can fix this ?

First, you may want to include the "shell" you expect this to run as/under, such as #!/bin/bash. I don't know if I'd use "Command" as an identifier, but I do question if you are using a single quote ' versus a single open quote `, as the open quote says to "run this as a command" while a single quote will get you the command you wanted to run as a "string" to put in your array. — Bee Kay
– Bee Kay, Commented Jun 3, 2015 at 17:47
See mywiki.wooledge.org/BashFAQ/050 to discuss safely storing shell commands in variables. — Charles Duffy
– Charles Duffy, Commented Jun 3, 2015 at 17:59
Also, array=( $(find ...) ) fails miserably with filenames that contain whitespace, glob characters, or are in any other way interesting. — Charles Duffy
– Charles Duffy, Commented Jun 3, 2015 at 18:05
BTW, feel free to ask follow-up questions; some of the other tricks I used in my answer (such as reading from a NUL-delimited stream -- since NUL is the only character that can't exist in a UNIX pathname) are documented somewhere other than FAQ 50; that particular one, for instance, is touched in BashFAQ #1 (mywiki.wooledge.org/BashFAQ/001). — Charles Duffy
– Charles Duffy, Commented Jun 3, 2015 at 18:24

larsks · Accepted Answer · 2015-06-03 17:58:27Z

3

Shell assignment statements may not have whitespace around the =. This is valid:

Command='some command'

This is not:

Command = 'some command'

In the second form, bash will interpret Command as a command name.

edited Jun 3, 2015 at 17:58

answered Jun 3, 2015 at 17:44

larsks

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1 Comment

MistyD Over a year ago

Thanks for clearing that up however now I am getting the array size to be zero however if I replace ${Command} with the actual command it works then

Charles Duffy · Accepted Answer · 2015-06-03 18:01:44Z

All of the below requires a #!/bin/bash shebang (which should come as no surprise since you're using arrays, which are a bash-only feature).

Also, see http://mywiki.wooledge.org/BashFAQ/050 for comprehensive discussion.

A best-practices implementation would look something like this:

# commands should be encapsulated in functions where possible
find_sources() { find . '(' -iname '*.cpp' -o -iname '*.h' ')' -print0; }

declare -a source_files
while IFS= read -r -d '' filename; do
  source_files+=( "filename" )
done < <(find_sources)

Now, if you really need to store the command in an array (maybe you're building it up dynamically), doing that would look like this:

# store literal argv for find command in array
# ...if you wanted to build this up dynamically, you could do so.
find_command=( find . '(' -iname '*.cpp' -o -iname '*.h' ')' -print0 )

declare -a source_files
while IFS= read -r -d '' filename; do
  source_files+=( "filename" )
done < <("${find_command[@]}")

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