How to copy one row from a text file at a time in a bash for loop?

In the following bash loop:

for i in {0..40}; do
    *random code* X -o file_{i}.dat

for each file, I am trying to replace the X with a row from another file, resulting in the following output:

*random code* ROW1 -o file_0.dat
*random code* ROW2 -o file_1.dat
*random code* ROW3 -o file_2.dat
...et cetera

Do you know any way to do it?

There are many possibilities. Here are a couple of the more plausible ones:

Option 1

Read the wanted rows of the other file into an array in advance, then access the elements by index in the loop. For example:

declare -a my_array
mapfile -t -n 41 my_array < other_file
for i in {0..40}; do
    do_something "${my_array[$i]}" -o file_"$i".dat
done

Option 2

Read lines from the other file on-demand as you progress through the loop. For example:

for i in {0..40}; do
    read -r line
    do_something "$line" -o file_"$i".dat
done < other_file

You probably want to beef up either one of those with appropriate behavior for the case where the other file doesn't contain enough lines.

It could be done like this

#!/bin/bash
#read file line by line to a max of 41 and store each line in an array
lc=0
while IFS='' read -r line || [[ -n "$line" ]]  && [ $lc -le 40 ]; do
    array+=("$line")
    lc=$((lc+1))
done < "input.txt"

#if array is greater than 40
if [ ${#array[@]} -gt 40 ]; then
    for i in {0..40}; do        
        printf "*random code* %s -o file_$i.dat\n" "${array[$i]}"    
    done            
fi

To copy one row from a text file at a time in a bash for loop, you can use the read command to read each line of the file and then use a loop to process each line.

#!/bin/bash

# Open the file for reading
exec 3< file.txt

# Loop through each line in the file
while read -u 3 line; do
    # Do something with the line, for example, copy it to another file
    echo "$line" >> new_file.txt
done

# Close the file
exec 3<&-