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Top Forums Shell Programming and Scripting Trouble Assigning Variable with Function Post 302919334 by Corona688 on Tuesday 30th of September 2014 02:17:31 PM
Old 09-30-2014
Use "${array[@]}" not "@{!array[@]}

Also, for does not work that way. i becomes a string, not a number index.

Also, you can put a ! in the expression instead of an empty 'if'.

Code:
for i in "${array[@]}"
do
        if [[ ! -e "$i" ]]
        then
                echo "$i not found">&2
        fi
done

This User Gave Thanks to Corona688 For This Post:
 

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IMPLODE(3)								 1								IMPLODE(3)

implode - Join array elements with a string

SYNOPSIS
string implode (string $glue, array $pieces) DESCRIPTION
string implode (array $pieces) Join array elements with a $glue string. Note implode(3) can, for historical reasons, accept its parameters in either order. For consistency with explode(3), however, it may be less confusing to use the documented order of arguments. PARAMETERS
o $glue - Defaults to an empty string. o $pieces - The array of strings to implode. RETURN VALUES
Returns a string containing a string representation of all the array elements in the same order, with the glue string between each ele- ment. EXAMPLES
Example #1 implode(3) example <?php $array = array('lastname', 'email', 'phone'); $comma_separated = implode(",", $array); echo $comma_separated; // lastname,email,phone // Empty string when using an empty array: var_dump(implode('hello', array())); // string(0) "" ?> NOTES
Note This function is binary-safe. SEE ALSO
explode(3), preg_split(3), http_build_query(3). PHP Documentation Group IMPLODE(3)
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