bash的总平均...更多

Question

我试图让这个

Enter Incidents 
Mon 14 
Tue 5 
Wed 12 

Incidents 

Mon 14 
Tue 5 
Wed 12 

Total 38 

Average 7.6 
------------------------ 

Mon incidents is the highest 11 
Tue incidents is the lowest 2 
------------------------ 

Enter a Day that you'd like to see how many incidents occurred? Wed 
12 Incidents 
------------------------ 

到目前为止，我做了总平均，最大值和最小值。

的问题是“周一如何输出最高11，我能找到的最小和最大，但不能打印出当天最大和最小的值匹配....

#!/bin/bash 

echo " ----Enter Incidents-----" 



echo -n "Mon= " 
read days[0] 


echo -n "Tue= " 
read days[1] 


echo -n "Wed= " 
read days[2] 






echo " ------Incidents--------" 


total=`expr ${days[0]} + ${days[1]} + ${days[2]}` 


echo "Toatal is= " $total 


ave=`expr $total \/ 3` 


echo "Average= " $ave 



max=${days[0]} 

min=${days[0]} 



for i in "${days[@]}" 

do 

if [[ "$i" -gt "$max" ]]; then 

    max="$i" 
fi 


if [[ "$i" -lt "$min" ]]; then 

    min="$i" 

fi 

done 


echo "Max is:" $max 

echo "Min is:" $min 


this is killing me lol 
Mon incidents is the highest 11 
Tue incidents is the lowest 2 
------------------------ 

Enter a Day that you'd like to see how many incidents occurred? Wed 
12 Incidents 
------------------------------- 

Answer 1

取代

echo -n "Mon= " read days[0] 

通过

read -p "Mon= " days[0] 

也更换

expr something 

通过

$((somthing)) # space matters 

，并请格式化你的代码，因此它在这个问题可读。

Answer 2

假设你的天，你写（您可以使用索引，但是你可以用“星期二”“星期三”等，为您阵列）：

for i in ${!days[@]}; do 
    if [[ ${days[$i]} -gt "$max" ]]; then 
     maxday="$i" 
    fi 
    if [[ ${days[$i]} -lt "$min" ]]; then 
     minday="$i" 
    fi 
done 

现在你可以打印minday，和天[$ minday ]。如果您将您的密钥作为索引，则需要将索引转换为日期。如果您希望您的阵列很好地格式化：

declare -A days  
echo Tue= 
read days['Tue'] 

等，如果你定义了天。因此，用上述循环，可以打印结果为：

echo $maxday yields the maximum ${days[$maxday]}