Question

#!/bin/bash

hour="hour"                                 #variable which holds the string for hour
min="min"                                     #variable which holds the string for min
minValue="4"
hourValue="4"

hourEntered="4"
minEntered="4"
###########
# hour    #
###########
if [ ! "$hourEntered" ];  then
  hourValue=$hourValue
  echo "you did not enter an hour"
  #if the user did input the correct values take them over
  #value to be entered 0 or 23....
elif [[ "$hourEntered" =~ ^[0-23]+$ ]]; then
  hourValue=$hourEntered
  echo "you entered for hourEntered:$hourEntered"
  #else provide a message and provide help....
else
  #set the read out values
  hourValue=$hourValue
fi
###########
# min     #
###########
#check if a value was entered if not take the read out values over
if [ ! "$minEntered" ];  then
  minValue=$minValue
  echo "you did not enter an min"
  #if the user did input the correct values take them over
  #value to be entered 0 or 60....
elif [[ "$minEntered" =~ ^[0-59]+$ ]]; then
  minValue=$minEntered
  echo "you entered for minEntered:$minEntered"
  #else provide a message and provide help....
else
  #set the read out values
  minValue=$minValue
fi

所以有一些奇怪的事情，显然是声明

[[ "$hourEntered" =~ ^[0-23]+$ ]];

不起作用，找不到值4。如果你像这样将它改为0-59，那么秒就可以完美地运作。

这是一个错误，或者这里发生了什么？

Answer 1

[0-23]并不意味着从0到23.它意味着从0到2和3.所以它匹配0,1,2和3.没有别的。

比较hourEntered＆gt; = 0和hourEntered＆lt;可能更简单。 24

或使用更复杂的正则表达式

正则表达式没有按预期的数字＆＃34; 4＆＃34;

1 个答案: