how to read file from line x to the end of a file in bash

I would like know how I can read each line of a csv file from the second line to the end of file in a bash script.

I know how to read a file in bash:

while read line
 do   
   echo -e "$line\n"
done < file.csv

But, I want to read the file starting from the second line to the end of the file. How can I achieve this?

tail -n +2 file.csv

From the man page:

-n, --lines=N
     output the last N lines, instead of the last 10
...

If the first character of N (the number of bytes or lines)  is  a  '+',
print  beginning with the Nth item from the start of each file, other-
wise, print the last N items in the file.

In English this means that:

tail -n 100 prints the last 100 lines

tail -n +100 prints all lines starting from line 100

Simple solution with sed:

sed -n '2,$p' <thefile

where 2 is the number of line you wish to read from.

Or else (pure bash)...

{ for ((i=1;i--;));do read;done;while read line;do echo $line;done } < file.csv

Better written:

linesToSkip=1
{
    for ((i=$linesToSkip;i--;)) ;do
        read
        done
    while read line ;do
        echo $line
        done
} < file.csv

This work even if linesToSkip == 0 or linesToSkip > file.csv's number of lines

Edit:

Changed () for {} as gniourf_gniourf enjoin me to consider: First syntax generate a sub-shell, whille {} don't.

of course, for skipping only one line (as original question's title), the loop for (i=1;i--;));do read;done could be simply replaced by read:

{ read;while read line;do echo $line;done } < file.csv

There are many solutions to this. One of my favorite is:

(head -2 > /dev/null; whatever_you_want_to_do) < file.txt

You can also use tail to skip the lines you want:

tail -n +2 file.txt | whatever_you_want_to_do

Depending on what you want to do with your lines: if you want to store each selected line in an array, the best choice is definitely the builtin mapfile:

numberoflinestoskip=1
mapfile -s $numberoflinestoskip -t linesarray < file

will store each line of file file, starting from line 2, in the array linesarray.

help mapfile for more info.

If you don't want to store each line in an array, well, there are other very good answers.

As F. Hauri suggests in a comment, this is only applicable if you need to store the whole file in memory.

Otherwise, you best bet is:

{
    read; # Just a scratch read to get rid (pun!) of the first line
    while read line; do
        echo "$line"
    done
} < file.csv

Notice: there's no subshell involved/needed.

This will work

i=1
while read line
 do   
   test $i -eq 1 && ((i=i+1)) && continue
   echo -e "$line\n"
done < file.csv

I would just get a variable.

#!/bin/bash

i=0
while read line
do
    if [ $i != 0 ]; then
      echo -e $line
    fi
    i=$i+1
done < "file.csv"

UPDATE Above will check for the $i variable on every line of csv. So if you have got very large csv file of millions of line it will eat significant amount of CPU cycles, no good for Mother nature.

Following one liner can be used to delete the very first line of CSV file using sed and then output the remaining file to while loop.

sed 1d file.csv | while read d; do echo $d; done