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Pass array to shell and print

 
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# 1  
Old 03-18-2014
Pass array to shell and print
How do i pass an array from test4.sh to a function in another shell script test5.sh, basically i am sourcing the test5.sh in test4.sh and printing the contents, but not working below are my trial scripts, please help, thank you.

Code:
#!/bin/bash
# /usr/local/dw/archive/test5.sh

print_array()
{
echo $1
i=1

while [ $i -le $# ];
do
  arr[$i]="$2"
  shift
  i=$(( $i + 1 ))
done

printf "%s\n" "$arr[@]"
}

Code:
#!/bin/bash
# test4.sh
. /usr/local/dw/archive/test5.sh
array[0]="Line 1"
array[1]="Line 2"
array[2]="Line 3"

print_array "john cena" "${array[@]}"

Output:
Code:
john cena
[@]

# 2  
Old 03-18-2014
This is wrong: "$arr[@]" It won't assume the [@] is part of the array without { }.

It needs to be: "${arr[@]}"
# 3  
Old 03-18-2014
Is there a way to pass just the name but not need eval or a child ksh to expand it? Perhaps a (local?) alias not a function?
# 4  
Old 03-19-2014
Quote:
Originally Posted by DGPickett
Is there a way to pass just the name but not need eval or a child ksh to expand it? Perhaps a (local?) alias not a function?
Probably not safely. The same kind of doublethink needed to reprocess variable names also processes things like backticks. Read can be used to write to arbitrary variable names, but not read from them, and isn't so useful for arrays either.
# 5  
Old 03-19-2014
If I understand what you're trying to do, try replacing test5.sh with:
Code:
#!/bin/bash
# /usr/local/dw/archive/test5.sh

print_array()
{
	printf "%s\n" "$1"
	shift

	while [ $# -gt 1 ]
	do	printf "%s " "$1"
		shift
	done
	printf "%s\n" "$1"
}

If you then run test4.sh, you get:
Code:
john cena
Line 1 Line 2 Line 3

# 6  
Old 03-19-2014
At least fix this little error:
Code:
printf "%s\n" "${arr[@]}"

 
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