How to loop over a file with space/tab

Question

I have a file "sample.txt" looks like:

apple 1
banana 10

and I'm using the following shell code to loop over lines like:

for line in $(cat sample.txt)
do
   echo $(echo $line| cut -f1)
done

My expected output is

apple
banana

But I got:

apple
1
banana
10

I can guess that shell takes each line as a list. Is it possible to remedy this?

Answer 1

Try the following code:

while read line; do
  echo "$line" | cut -d " " -f1
  #                  ??????
  #                  |
  #                  ? Split at empty space
done <sample.txt

Answer 2

You can eliminate the use of the cut utility by using the shell builtin read command as shown below:

#!/bin/bash

while read first rest
do
    echo $first
done < sample.txt

which outputs:

apple
banana

The key is in how the read command is used. From the bash manpage:

read [-ers] [-a aname] [-d delim] [-i text] [-n nchars] [-N nchars] [-p prompt] [-t timeout] [-u fd] [name ...]
    One line is read from the standard input, or from the file descriptor fd supplied as an argument to the -u option, split into
    words as described above under Word Splitting, and the first word is assigned to the first name, the second word to the second
    name, and so on. If there are more words than names, the remaining words and their intervening delimiters are assigned to the
    last name. If there are fewer words read from the input stream than names, the remaining names are assigned empty values.

In our case, we are interested in the first word, which is assigned by read to the shell variable first, while the rest of the words in the line are assigned to the shell variable rest. We then just output the contents of the shell variable first to get the desired output.