CODEWITHSUNDEEP
bashscripting
user1777907
user1777907

Reputation: 1375

Bash: while terminate early when reading from file

I am trying to run a file that contains a sequence of commands/scripts to run with arguments, like:

ls /etc/
cat /etc/hosts/
script.sh some parameters 
...

This seems to work fine but in some cases the while loop will end prematurely. This seems to be the case only when the scripts it is executing contains SSH/SCP at the end. The code to read the file:

while IFS= read -r line
do
    # Cut command and parameters
    IFS=', ' read -a parameters <<< "$line"
    cmd="${parameters[0]}"
    unset parameters[0]
    runScriptAndCheckError "$cmd" "${parameters[@]}"
done < "$SCRIPT_FILENAME"

When using set -x:

+ checkError 0 'ERROR: script.sh failed'
+ '[' 0 -ne 0 ']'
+ IFS=
+ read -r line

It looks like there is no more input although there is still lines in the file. If I comment out runScriptAndCheckError "$cmd" "${parameters[@]}" then it does print a lot more lines.

I am not sure what is wrong with this code. I'd be really helpful if someone could please help.

Upvotes: 1

Views: 970

Answers (2)

automorphic
automorphic

Reputation: 800

Stealing Input

The key is as @chepner states, that the statements inside:

read line; do
   <command>
done

Will interfere with the loop if they attempt to read from stdin.

If you don't want your script reading from stdin, prevent it from doing so as follows:

read line; do
   cmd < /dev/null
done

Now your read loop will not be missing any input.

Upvotes: 1

chepner
chepner

Reputation: 531165

If runScriptAndCheckError also reads from standard input, it will read lines from $SCRIPT_FILENAME. Have the read command in the while loop use a different file descriptor:

while IFS= read -r line <&3; do
  ...
done 3< "$SCRIPT_FILENAME"

Upvotes: 4

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