数字作为命令行参数并打印计数

Question

我想将数字作为命令行参数并打印数量以 0、1、2 等结尾的数字，最多 5 个。

示例：

bash test.sh 12 14 12 15 14

预期输出：

Digit_ends_with   count
0                  0
1                  0
2                  2
3                  0
4                  2
5                  1

Mt 尝试：

read -a integers                                                                                                                                                                                                                                for i in ${integers[@]}                                                                                                 do                                                                                                                              if [[ grep -o '[0-9]' $i ]]                                                                                                     count=$(grep -c $i)                                                                                                     if [ "$count" -ge "0" ]                                                                                                 then 
echo "Digit_ends_with" $i                                                                                                                           echo -e "Count ""$count"                                                                                                fi                                                                                                              fi                                                                                                              done

但这不起作用。我怎样才能达到这个要求？

Answer 1

请您尝试以下操作：

#!/bin/bash

for i in "$@"; do                               # loop over arguments
    (( count[i % 10]++ ))                       # index with the modulo 10
done

printf "%s %s\n" "Digit_ends_with" "count"      # header line
for (( i = 0; i < 6; i++ )); do                 # loop between 0 and 5
    printf "%d\t\t%d\n" "$i" "${count[$i]}"     # print the counts
done

./test.sh 12 14 12 15 14的结果：

Digit_ends_with count
0               0
1               0
2               2
3               0
4               2
5               1

Answer 2

#!/bin/bash
echo "Digit_ends_with count"                                #print table header
for argument in "$@"                                        #loop over all given arguments
do 
    current=$(echo "$argument"  | tail -c 2 )               #get last digit
    if ((  "$current" <= 5 ))                               #check if lower than 6
    then
        echo "$current"                                     #echo if true
    fi
done | sort | uniq -c | sed -E 's/\s+//' | sed -E 's/([0-9]+).?([0-9]+)/\t\t/'  #sort, count, remove leading spaces and switch the fields

示例：

╰─$ ./test.sh 188 182 182 12 13 14 18 15 16 17 18 19 10 0 0 0 0 0 0 0 0 0 0
Digit_ends_with count
0               11
2               3
3               1
4               1
5               1

数字作为命令行参数并打印计数

numbers as command-line arguments and prints the count

bash

shell