Skip blank lines when iterating through file line by line

Question

I am iterating through a file line by line and put each word into a array and that works. But it also picks up blank lines and puts it as an item in the array, how can I skip the blank lines?

example file

      Line 1
line 2

line 3
        line 4 

line 5
   line 6

My code

while read line ; do
            myarray[$index]="$line"
            index=$(($index+1))
    done < $inputfile

Possible psuedo code

while read line ; do
           if (line != space);then
            myarray[$index]="$line"
             fi
            index=$(($index+1))
    done < $inputfile

Solution 1

Remove the blank lines first with sed.

for word in `sed '/^$/d' $inputfile`; do
    myarray[$index]="$word"
    index=$(($index+1))
done

Solution 2

Be more elegant:

echo "\na\nb\n\nc" | grep -v "^$"

cat $file | grep -v "^$" | next transformations...

Solution 3

Implement the same test as in your pseudo-code:

while read line; do
    if [ ! -z "$line" ]; then
        myarray[$index]="$line"
        index=$(($index+1))
    fi
done < $inputfile

The -z test means true if empty. ! negates (i.e. true if not empty).

You can also use expressions like [ "x$line" = x ] or test "x$line" = x to test if the line is empty.

However, any line which contains whitespace will not be considered empty. If that is a problem, you can use sed to remove such lines from the input (including empty lines), before they are passed to the while loop, as in:

sed '/^[ \t]*$/d' $inputfile | while read line; do
    myarray[$index]="$line"
    index=$(($index+1))
done

Solution 4

cat -b -s file |grep -v '^$'

I know it's solved but, I needed to output numbered lines while ignoring empty lines, so I thought of putting it right here in case someone needs it. :)

Solution 5

Use grep to remove the blank lines:

for word in $(cat ${inputfile} | grep -v "^$"); do
   myarray[$index]="${word}"
   index=$(($index+1))
done

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