Mac Bash 脚本：每 7 的倍数打开一个终端窗口并输入命令

Question

Essentially all I have is a loop from 0-1000, but I want to make it so that every time a number is a multiple of 7 occurs, open a terminal window and type the following lines:

1. cd desktop/testfolder
2. node testScript.js
3. 7 <---- the multiple of 7 goes here

So for example if the loop gets to 49 the commands would be:

1. cd desktop/testfolder
2. node testScript.js
3. 49

This is what I have so far:

#!/bin/bash

shopt -s nocasematch
read -p " Execute script? (y/n): " response
if [[ $response == y ]]; then
    printf " Loading....\\n"
    for ((x = 0; x<1000; x++)); do

      #add a check if x is a multiple of 7 here...

      printf " Open %s Terminal\\n" $x
      osascript -e 'tell application "Terminal" to do script "cd desktop//testfolder\
node myScript.js\
MULTIPLE OF 7 HERE"' >/dev/null

    done
fi
shopt -u nocasematch

I have got the terminal opening & first two commands sorted but I can't figure out how to add a check for x to make sure its a multiple of 7, and I can't work out how to get the x integer into the terminal "do script" string.

Any help is appreciated!

Answer 1

Instead of

   for ((x = 0; x<1000; x++)); do

      #add a check if x is a multiple of 7 here...

you can do:

for ((x = 0; x<1000; x=x+7)); do

Also, in order to provide this string, if I get correctly what you try to do

tell application "Terminal" to do script "cd desktop//testfolder\
node myScript.js\
7"

you can do this, in the bash shell:

str="tell application \"Terminal\" to do script \"cd desktop//testfolder\\
node myScript.js\\
$x\""
osascript -e "$str"

Answer 2

Got it to work using @thanasisp's for ((x = 0; x<1000; x=x+7)); do for ((x = 0; x<1000; x=x+7)); do

And then to get the $x in the do script string I did the following:

osascript -e 'tell application "Terminal" to do script "cd desktop//testfolder\
node myScript.js\
'$x'"'

Just got a bit confusing with all the single and double quotes I guess...