bash - 解析bash函数参数

Question

我有一个 bash 功能，比如说foo ()

我在字符串中传递了一些参数，例如user=user1 pass=pwd address=addr1 other= 参数可能会丢失或以随机顺序传递我需要在foo

USER=...
PASSWORD=...
ADDRESS=...

我该怎么做？

我可以grep多次使用，但这种方式不好，例如

foo ()
{
  #for USER
  for par in $* ; do 
    USER=`echo $par | grep '^user='` 
    USER=${USER#*=} 
  done
  #for PASSWORD
  for ...
}

Answer 1

你是这个意思吗？

#!/bin/sh
# usage: sh tes.sh username password addr

# Define foo
# [0]foo [1]user [2]passwd [3]addr
foo () {
    user=$1
    passwd=$2
    addr=$3
    echo ${user} ${passwd} ${addr}
}

# Call foo
foo $1 $2 $3

结果：

$ sh test.sh username password address not a data
username password address

您的问题也已经在这里得到解答：

将参数传递给 Bash 函数

---

显然上面的答案与问题无关，那么这个怎么样？

#!/bin/sh

IN="user=user pass=passwd other= another=what?"

arr=$(echo $IN | tr " " "\n")

for x in $arr
do
    IFS==
    set $x
    [ $1 = "another" ] && echo "another = ${2}"
    [ $1 = "pass" ] && echo "password = ${2}"
    [ $1 = "other" ] && echo "other = ${2}"
    [ $1 = "user" ] && echo "username = ${2}"
done

结果：

$ sh test.sh
username = user
password = passwd
other =
another = what?