bash while loop not printing what is expected

Question

I am trying this simple while loop in bash.

My text file

# cat test.txt
line1:21
line2:25
line5:27
These are all on new line

My Script

# cat test1.sh
while read line
do
 awk -F":" '{print 2ドル}'
done < test.txt

Output

# ./test1.sh
25
27

The output isn't printing the first line 2ドル value. Could anyone please help me understand this case?

• Stéphane Chazelas: See also Why is using a shell loop to process text considered bad practice?

Answer 1

You don't need that loop:

$ awk -F ':' '{ print 2ドル }' test.txt
21
25
27

awk will process the input line by line.

---

With your loop, the read will get the first line of the file, which is lost since it's not used/outputted. The awk will then take over the standard input of the loop and read the other two lines in the file (so the loop will only ever do one single iteration).

Your loop, annotated:

while read line # first line read ($line never used)
do
 awk -F ':' '{ print 2ドル }' # reads from standard input, which will
 # contain the rest of the test.txt file
done <test.txt

Answer 2

I able to fix your code by adding echo. Reason for this have described there, question why it's printing other two values.

while read line;
do
 echo "$line" | awk -F":" '{print 2ドル}'
done < test.txt

Answer 3

while IFS=":" read z x; do 
 echo $x; 
done<test.txt

or

sed "s/^.*://g" test.txt