Bash array- substitue 0 for NULL

Question

I'm filling up an array from bash like this:

array[0]=$(awk '/Item/' logfile.log | awk '{print $21}' | awk -F'"' '{ print $2 }')

array[1]=$(awk '/Item/' logfile.log | awk '{print $26}' | awk -F'"' '{ print $2 }')

array[2]=$(awk '/Item/' logfile.log | awk '{print $31}' | awk -F'"' '{ print $2 }')

For some I get a value as a digit for others, there is no output so I believe this populates the array elements with a NULL.
I want to find the array elements that contain NULL and fill them with "0"

I have tried a few different things, but I can't seem to find the correct method here.

One thing I tried: if [[ ${array[$i]} ]]; then array[$i]=0;fi

I think I'm missing qoutes or brackets but just can't seem to hit on the correct syntax.

Thanks

What do you mean by NULL? Bash doesn't have pointers. Do you mean an empty string? — Keith Thompson
– Keith Thompson, Commented Jul 12, 2013 at 17:21

that other guy · Accepted Answer · 2013-07-12 17:22:38Z

1

You just have the condition the wrong way. [[ something ]] checks that something is non-empty. Use [[ -z thing ]] to check if the string is empty instead:

array[0]="foo"
array[1]=""
array[2]="bar"
for i in {0..2}
do
    if [[ -z ${array[$i]} ]]; then array[$i]=0;fi
done
echo "${array[@]}"

This prints foo 0 bar

answered Jul 12, 2013 at 17:22

that other guy

125k12 gold badges187 silver badges214 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

geezer Over a year ago

yeah it was the -z that did the trick. BRain farts on Friday I guess. Thanks Much!

Collectives™ on Stack Overflow

Bash array- substitue 0 for NULL

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related