这是一个示例bash,您可以从两个方面看到:
1)首先
#!/bin/bash
number=0
echo "Your number is: $number"
IFS=' ' read -t 2 -p "Press space to add one to your number: " input
if [ "$input" -eq $IFS ]; then #OR ==> if [ "$input" -eq ' ' ]; then
let number=number+1
echo $number
else
echo wrong
fi
2)第二:
#!/bin/bash
number=0
echo "Your number is: $number"
read -t 2 -p "Press space to add one to your number: " input
case "$input" in
*\ * )
let number=$((number+1))
echo $number
;;
*)
echo "no match"
;;
esac
现在的问题是:
有了这两种方法,如何检查输入参数是
white space
还是null
?我想在bash中同时检查
white space
和null
。谢谢
最佳答案
你可以试试这样的方法:
#!/bin/bash
number=0
echo "Your number is: $number"
IFS= read -t 2 -p "Press space to add one to your number: " input
# Check for Space
if [[ $input =~ \ + ]]; then
echo "space found"
let number=number+1
echo "$number"
# Check if input is NULL
elif [[ -z "$input" ]]; then
echo "input is NULL"
fi
关于linux - 如何检查输入是否为空格并且为空?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/21950315/