linux - 如何在 bash 脚本中传递两个参数或参数

Question

<分区>

Answer 1

作为开始，您可以这样做:

#!/bin/bash
dayoffset=$1
processMode=$2
echo "Do something with $dayoffset and $processMode."

用法:

./start.sh 1 true

另一个:

#!/bin/bash

while [[ $# -gt 0 ]]; do
    case "$1" in
    -dayoffset)
        day_offset=$2
        shift
        ;;
    -processMode)
        if [[ $2 != true && $2 != false ]]; then
            echo "Option argument to '-processMode' can only be 'true' or 'false'."
            exit 1
        fi
        process_mode=$2
        shift
        ;;
    *)
        echo "Invalid argument: $1"
        exit 1
    esac
    shift
done

echo "Do something with $day_offset and $process_mode."

用法:

./start.sh -dayoffset 1 -processMode true

使用日期偏移量解析的示例参数:

#!/bin/bash 
dayoffset=$1
date -d "now + $dayoffset days"

测试:

$ bash script.sh 0
Fri Aug 15 09:44:42 UTC 2014
$ bash script.sh 5
Wed Aug 20 09:44:43 UTC 2014