bash setting variable getting error command not found

Question

I'm writing one small bash script for my Project but i getting some errors

#!/bin/bash
count=$(cat Downloads/datei.csv | wc -l);
for((i=115;i<="122";i++)); do
line$i=$(sed -n $i"p" Downloads/datei.csv);
echo "line$i";
done

i trying to get every line from CSV in some variable

The erroŕ

count.sh: line 4: line115=: command not found

if [[ -z "line$i" ]];
then
      echo "$i is empty";
else
      echo "$i is NOT empty";
fi

the second code gives me the same error

they can be composed together but i get the error command not found. — nullsoft pika, May 14 '22 at 06:50
The problem you ran into is the topic of [BashFAQ #6](https://mywiki.wooledge.org/BashFAQ/006). — Charles Duffy, May 14 '22 at 13:04
Oneliner: `mapfile -s 114 -n 8 -t linearray < Downloads/datei.csv` this puts the lines 115 to 122 in an ARRAY and you can access e.g. line 119 it by: `echo ${linearray[119]}` — Marco, May 15 '22 at 06:10

Victor Lee · Answer 1 · 2022-05-14T07:03:06.480

Suggest compose the code together, just update line$i to a temporary normal variable without other var like line_get.

One more thing, you need to update -z "line$i" to -z "$line_get". As in bash, you should use $ and the var name to get the value. Looks like:

#!/bin/bash
count=$(cat Downloads/datei.csv | wc -l)
for((i=115;i<=122;i++)); do
    line_get=$(sed -n $i"p" Downloads/datei.csv)
    if [[ -z "$line_get" ]];
    then
        echo "$i is empty"
    else
        echo "$i is NOT empty"
    fi
done

If you need to dynamic generate the variable name. Try this:

declare "line$i=$(sed -n $i"p" Downloads/datei.csv)" 
var="line$i"
echo "${!var}"

And echo "line$i"; just print the text like line115 but not the value of the variable line$1.

I need this var for the rest of my script – nullsoft pika May 14 '22 at 06:43 — nullsoft pika, May 14 '22 at 06:43

bash setting variable getting error command not found

1 Answers1