'read' command for unix not working without while loop even for single line input

Asked May 18 '20 at 07:34

Active May 18 '20 at 07:34

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I am not able to understand why the read command is not working without the while loop as shown in the code below. I referred this Using “while read…” in a linux script question where the answer also mentioned that since the input is a single line

This is to say that having a while loop here is not really accomplishing anything.

 55 echo "Some sample text" | read var1 var2
 56
 57 echo "Var1: ${var1}"
 58 echo "Var2: ${var2}"
 59
 60 echo "----------------------------"
 61
 62
 63 echo "Some sample text" | while read v1 v2
 64 do
 65        echo "Var1: ${v1}"
 66        echo "Var2: ${v2}"
 67 done
 68
 69 echo "----------------------------"

Output:

Var1:
Var2:
----------------------------
Var1: Some
Var2: sample text
----------------------------

asked May 18 '20 at 07:34

Manav Saxena

you send three arguments "Some", "sample", "text" but read in two variables. – kofemann May 18 '20 at 07:40
The commands in a pipeline are executed in a subshell, variable changes don't affect the original shell. – Barmar May 18 '20 at 07:45
@kofemann The last variable gets all the remaining words. – Barmar May 18 '20 at 07:45
`shopt -s lastpipe` for that to work or just use a herestring `read -r var1 var2 <<< 'some sample text'` – Jetchisel May 18 '20 at 07:47
@kofemann, the number of given variables is not the problem, it will work even with just one variable, subshell is the culprit. – Jetchisel May 18 '20 at 07:51
Well, even without a variable the builtin variable `$REPLY` will/should catch all that is given to `read` – Jetchisel May 18 '20 at 07:57
@Jetchisel I tried the $REPLY as will and it doesn't catch the input aswell. – Manav Saxena May 18 '20 at 09:12
`subshell` is the problem you need to solve, and everything will work out.... – Jetchisel May 18 '20 at 09:14

'read' command for unix not working without while loop even for single line input

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