如何检查输入是否为空格并为空或不是？

Question

这是一个示例bash作为你在两个方面看：

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，我该如何检查输入参数是否为white space或null？

我想在bash中检查white space或null。

感谢

Answer 1

你可以尝试这样的事情：

#!/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 

Answer 2

这部分检查输入字符串为NULL

if [ "##"${input}"##" = "####" ] 
then 
echo "You have input a NULL string" 
fi 

，这部分检查一个或多个空格caharacters被输入

newinput=$(echo ${input} | tr -s " ") # There was a typo on thjis line. should be fixed now 
if [ "##"${newinput}"##" = "## ##" ] 
then 
echo "You have input one (or more) whitespace character(s)" 
fi 

在aorder你把他们在完成所有比较后，在if - fi块中查看拟合和设置标志以进行评估。