Problem with putting value in array in bash

Question

I would like to make array which put users in a time using for loop. For example:

y[1]="user1"

y[2]="user2"

...

y[n]="usern"

I tried to do it like this

#!/bin/bash

x=$(who | cut -d " " -f1 | sort | uniq | wc -l)

for (( i=1; i<=$x; i++ )); do
        y[$i]=$(who | cut -d " " -f1 | sort | uniq | sed -n '$ip')
        p[$i]=$(lsof -u ${y[$i]} | wc -l)

        echo "Users:"
        echo ${y[$i]}
        echo -e "Number of launched files:\n" ${p[$i]}
done

Most likely I'm using command "sed" wrong. Can you help me?

It seems that you're appending the number to end of the user output. Try this: y[$i]=$(who | cut -d " " -f1 | sort | uniq | sed "s/$/${ip}". — sungtm
– sungtm, Commented Nov 27, 2019 at 4:11

Bayou · Accepted Answer · 2019-11-27 07:43:59Z

2

Indeed your sed command seems to be a bit off. I can't really guess what you're trying to do there. Besides that, I'm wondering why you're executing who twice. You can make use of the data first obtained in the following manner.

#!/bin/bash

# define two arrays
y=()
p=()
#x=0

while read -r username; do
        y+=("$username")
        p+=($(lsof -u $(id -u "$username") | wc -l))

        echo -e "User:\n${y[-1]}"
        echo -e "Open files:\n${p[-1]}"

        # The -1 index is the last index in the array, but you
        # could uncomment the x=0 variable and the line below:
        #((x++))

done <<< $(who | cut -d " " -f1 | sort | uniq)

echo "Amount of users: $x"
exit 0

edited Nov 27, 2019 at 7:43

answered Nov 27, 2019 at 7:10

Bayou

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1 Comment

Gordon Davisson Over a year ago

I'd also recommend double-quoting references to $username, just in case of weird characters.

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