如果问题显示一些语法错误

Question

#!bin/bash

echo Where?
read Where

echo What?
read What

echo Number?
read Number

for inc in ’seq 1 1 $Number’ do

if [ $Where = Screen ] then
echo $What
fi

if [ $Where = File ] then   
echo $What > test1,txt
fi
done

当我运行它时，输入 Screen，然后输入 Marko，然后输入数字 5 它向我显示此错误 它向我显示意外标记附近的语法错误 if 'if [ $Where = Screen ] then'

Answer 1

你的for陈述缺少一个分号。你要：

for inc in $( seq 1 1 "$Number" ); do

Answer 2

首先运行您的代码https://www.shellcheck.net/