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}'
done < test.txt
Output
# ./test1.sh
25
27
The output isn't printing the first line $2 value. Could anyone please help me understand this case?
You don't need that loop:
$ awk -F ':' '{ print $2 }' test.txt
21
25
27
awk will process the input line by line.
With your loop, the read will get the first line of the file, which is lost since it's not used/outputted. The awk will then take over the standard input of the loop and read the other two lines in the file (so the loop will only ever do one single iteration).
Your loop, annotated:
while read line # first line read ($line never used)
do
awk -F ':' '{ print $2 }' # reads from standard input, which will
# contain the rest of the test.txt file
done <test.txt
I able to fix your code by adding echo. Reason for this have described there, question why it's printing other two values.
while read line;
do
echo "$line" | awk -F":" '{print $2}'
done < test.txt
while IFS=":" read z x; do
echo $x;
done<test.txt
or
sed "s/^.*://g" test.txt
