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Top Forums UNIX for Advanced & Expert Users [SOLVED] Code does not run when assigned to a variable Post 302738185 by in2nix4life on Friday 30th of November 2012 01:34:53 PM
Old 11-30-2012
Not sure what OS or shell you're using but the output might be a bit much for a normal variable. Instead, you could try storing the returned output in an array:

Code:
declare -a array=( $(su nbadmin -c "ssh -t servery /usr/openv/netbackup/bin/bplist -C servery -t 19 -l -R -s 11/01/2012 -e 11/01/2012 /vol/root/") )

Then iterate the array to display the results:

for (( i = 0; i < ${#array[@]}; i++ ))
do
    echo ${array[$i]}
done

 

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RESET(3)								 1								  RESET(3)

reset - Set the internal pointer of an array to its first element

SYNOPSIS
mixed reset (array &$array) DESCRIPTION
reset(3) rewinds $array's internal pointer to the first element and returns the value of the first array element. PARAMETERS
o $array - The input array. RETURN VALUES
Returns the value of the first array element, or FALSE if the array is empty. EXAMPLES
Example #1 reset(3) example <?php $array = array('step one', 'step two', 'step three', 'step four'); // by default, the pointer is on the first element echo current($array) . "<br /> "; // "step one" // skip two steps next($array); next($array); echo current($array) . "<br /> "; // "step three" // reset pointer, start again on step one reset($array); echo current($array) . "<br /> "; // "step one" ?> SEE ALSO
current(3), each(3), end(3), next(3), prev(3). PHP Documentation Group RESET(3)
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