[SOLVED] how can I skip a specific number of lines of a script

Issue

I have created a bash script named call.sh

#!/bin/bash
termux-tts-speak whom doyou want to call
var="$(termux-speech-to-text)"
if [ "$var" = "Pappu" ]
then
termux-tts-speak calling to pappu
termux-telephony-call xxxxxxxxxx

elif [ "$var" = "call me" ]
then
 termux-tts-speak calling to you
termux-telephony-call xxxxxxxxxx

else
termux-tts-speak sorry I can not understand say it again
fi

I runs the script using the command

$bash call.sh 

But if I want to execute the script skipping first 3 line without modify the bash file. How can I do that help me please?

Solution

Use the tail command and pipe it to bash:

tail -n +4 call.sh | bash

But a better idea would be to use command-line arguments and test in the script.

#!/bin/bash

if [ "$1" != "--noask" ]
then
    termux-tts-speak whom do you want to call
fi

var="$(termux-speech-to-text)"
if [ "$var" = "Pappu" ]
then
    termux-tts-speak calling to pappu
    termux-telephony-call xxxxxxxxxx
elif [ "$var" = "call me" ]
then
    termux-tts-speak calling to you
    termux-telephony-call xxxxxxxxxx
else
    termux-tts-speak sorry I cannot understand say it again
fi

Then you can run the script like this to skip the first command:

./call.sh --noask

Answered By - Barmar

Answer Checked By - Katrina (WPSolving Volunteer)