我正在尝试搜索如何在Bash函数中传递参数，但是出现的问题始终是如何从命令行传递参数。
我想在我的脚本中传递参数。我试过了：
myBackupFunction("..", "...", "xx")
function myBackupFunction($directory, $options, $rootPassword) {
...
}
但语法不正确，如何将参数传递给我的函数？
答案 0 :(得分：1390)
声明函数有两种典型方法。我更喜欢第二种方法。
function function_name {
command...
}
或
function_name () {
command...
}
使用参数调用函数：
function_name "$arg1" "$arg2"
该函数引用的是它们的位置（而不是名称）传递的参数，即$ 1，$ 2等等。 $ 0 是脚本本身的名称。
示例：
function_name () {
echo "Parameter #1 is $1"
}
此外，您需要在声明之后调用您的函数。
#!/usr/bin/env sh
foo 1 # this will fail because foo has not been declared yet.
foo() {
echo "Parameter #1 is $1"
}
foo 2 # this will work.
<强>输出：
./myScript.sh: line 2: foo: command not found
Parameter #1 is 2
答案 1 :(得分：54)
高级编程语言（C / C ++ / Java / PHP / Python / Perl ...）的知识会向外行人建议bash函数应该像在其他语言中那样工作。 而不是，bash函数就像shell命令一样工作，并希望将参数传递给它们的方式与将命令传递给shell命令（ls -l）的方式相同。实际上，bash中的函数参数被视为位置参数（$1, $2..$9, ${10}, ${11}，依此类推）。考虑到getopts的工作原理，这并不奇怪。在bash中调用函数不需要括号。
（注意：我目前正在开发Open Solaris。）
# bash style declaration for all you PHP/JavaScript junkies. :-)
# $1 is the directory to archive
# $2 is the name of the tar and zipped file when all is done.
function backupWebRoot ()
{
tar -cvf - $1 | zip -n .jpg:.gif:.png $2 - 2>> $errorlog &&
echo -e "\nTarball created!\n"
}
# sh style declaration for the purist in you. ;-)
# $1 is the directory to archive
# $2 is the name of the tar and zipped file when all is done.
backupWebRoot ()
{
tar -cvf - $1 | zip -n .jpg:.gif:.png $2 - 2>> $errorlog &&
echo -e "\nTarball created!\n"
}
#In the actual shell script
#$0 $1 $2
backupWebRoot ~/public/www/ webSite.tar.zip
答案 2 :(得分：37)
如果您更喜欢命名参数，可以（通过一些技巧）将命名参数实际传递给函数（也可以传递数组和引用）。
我开发的方法允许您定义传递给函数的命名参数：
function example { args : string firstName , string lastName , integer age } {
echo "My name is ${firstName} ${lastName} and I am ${age} years old."
}
您还可以将参数注释为@required或@readonly，创建... rest参数，从顺序参数创建数组（使用例如string[4]）并可选择在多行中列出参数：
function example {
args
: @required string firstName
: string lastName
: integer age
: string[] ...favoriteHobbies
echo "My name is ${firstName} ${lastName} and I am ${age} years old."
echo "My favorite hobbies include: ${favoriteHobbies[*]}"
}
换句话说，您不仅可以通过名称调用参数（这样可以构成更具可读性的核心），您实际上可以传递数组（以及对变量的引用 - 这个功能仅适用于bash 4.3）！另外，映射变量都在本地范围内，就像$ 1（和其他）一样。
使这项工作的代码非常轻松，并且在bash 3和bash 4中都有效（这些是我用它测试过的唯一版本）。如果您对这样的更多技巧感兴趣，那么使用bash开发更好更容易，您可以查看我的Bash Infinity Framework，下面的代码可用作其功能之一。
shopt -s expand_aliases
function assignTrap {
local evalString
local -i paramIndex=${__paramIndex-0}
local initialCommand="${1-}"
if [[ "$initialCommand" != ":" ]]
then
echo "trap - DEBUG; eval \"${__previousTrap}\"; unset __previousTrap; unset __paramIndex;"
return
fi
while [[ "${1-}" == "," || "${1-}" == "${initialCommand}" ]] || [[ "${#@}" -gt 0 && "$paramIndex" -eq 0 ]]
do
shift # first colon ":" or next parameter's comma ","
paramIndex+=1
local -a decorators=()
while [[ "${1-}" == "@"* ]]
do
decorators+=( "$1" )
shift
done
local declaration=
local wrapLeft='"'
local wrapRight='"'
local nextType="$1"
local length=1
case ${nextType} in
string | boolean) declaration="local " ;;
integer) declaration="local -i" ;;
reference) declaration="local -n" ;;
arrayDeclaration) declaration="local -a"; wrapLeft= ; wrapRight= ;;
assocDeclaration) declaration="local -A"; wrapLeft= ; wrapRight= ;;
"string["*"]") declaration="local -a"; length="${nextType//[a-z\[\]]}" ;;
"integer["*"]") declaration="local -ai"; length="${nextType//[a-z\[\]]}" ;;
esac
if [[ "${declaration}" != "" ]]
then
shift
local nextName="$1"
for decorator in "${decorators[@]}"
do
case ${decorator} in
@readonly) declaration+="r" ;;
@required) evalString+="[[ ! -z \$${paramIndex} ]] || echo \"Parameter '$nextName' ($nextType) is marked as required by '${FUNCNAME[1]}' function.\"; " >&2 ;;
@global) declaration+="g" ;;
esac
done
local paramRange="$paramIndex"
if [[ -z "$length" ]]
then
# ...rest
paramRange="{@:$paramIndex}"
# trim leading ...
nextName="${nextName//\./}"
if [[ "${#@}" -gt 1 ]]
then
echo "Unexpected arguments after a rest array ($nextName) in '${FUNCNAME[1]}' function." >&2
fi
elif [[ "$length" -gt 1 ]]
then
paramRange="{@:$paramIndex:$length}"
paramIndex+=$((length - 1))
fi
evalString+="${declaration} ${nextName}=${wrapLeft}\$${paramRange}${wrapRight}; "
# continue to the next param:
shift
fi
done
echo "${evalString} local -i __paramIndex=${paramIndex};"
}
alias args='local __previousTrap=$(trap -p DEBUG); trap "eval \"\$(assignTrap \$BASH_COMMAND)\";" DEBUG;'
答案 3 :(得分：26)
错过了parens和逗号：
myBackupFunction ".." "..." "xx"
并且函数应该如下所示：
function myBackupFunction() {
# here $1 is the first parameter, $2 the second etc.
}
答案 4 :(得分：6)
我希望这个例子可以帮到你。它从用户那里获取两个数字，将它们提供给名为add的函数（在代码的最后一行），add将它们相加并打印出来。
#!/bin/bash
read -p "Enter the first value: " x
read -p "Enter the second value: " y
add(){
arg1=$1 #arg1 gets to be the first assigned argument (note there are no spaces)
arg2=$2 #arg2 gets to be the second assigned argument (note there are no spaces)
echo $(($arg1 + $arg2))
}
add x y #feeding the arguments
答案 5 :(得分：4)
我想提到另一种将命名参数传递给bash的方法......通过引用传递。从bash 4.0开始支持此功能
#include <stdio.h>
#include <string.h>
#include <dirent.h>
#include <stdlib.h>
//Prototyping
int fileNameBegin(const char *a, const char *b);
void returner(char directory[256], char string[32]);
void print();
//Array of Node Pointer
struct node* arrayOfLinkedLists[26];
//Main
int main() {
printf("Enter Directory Address:\n");
char str[256];
gets(str);
char letter[32];
do {
printf("Enter letters to search by:\n");
letter[0] = '\0';
gets(letter);
returner(str, letter);
print();
} while (letter[0] != '\0');
return 0;
}
//Constructing the Node Struct
struct node{
char fileName[50];
struct node *next;
};
//Narrowing Down Search
int fileNameBegin(const char *a, const char *b)
{
if(strncasecmp(a, b, strlen(b)) == 0) return 1; //not case sensitive, string comparing var a and b with String length
return 0;
}
#define DATA_MAX_LEN 50
//Adding the node (Files) to the LinkedList in Array
void addFileName(struct node **pNode, const char *c)
{
while (*pNode)
pNode = &(*pNode)->next; //It equals the address of the pointer
*pNode = malloc( sizeof **pNode );
strncpy((*pNode)->fileName,c,DATA_MAX_LEN-1); //Copying characters from String
(*pNode)->fileName[ DATA_MAX_LEN-1] = 0;
(*pNode)->next = NULL;
}
//Opening the Directory. Reading from Directory. Comparing File Name to String and Adding if there's a match
void returner(char directory[256], char string[32])
{
DIR *pDir = opendir (directory);
if (pDir)
{
struct dirent *pent;
while ((pent = readdir(pDir)))
{
if (pent->d_name[0] == '.' && (pent->d_name[1] == 0 || (pent->d_name[1] == '.' && pent->d_name[2] == 0)))
continue;
if(fileNameBegin(pent->d_name, string))
addFileName(arrayOfLinkedLists + ((int) strlwr(string)[0] - 97), pent->d_name);
}
closedir (pDir);
}
}
//I have no idea what this does.... oh, it displays it, duh.
void print(){
int i;
struct node *temp;
for(i=0 ; i < 26; i++){
temp = arrayOfLinkedLists[i];
while(temp != NULL){
printf("%s\n",temp->fileName);
temp = temp->next;
}
}
free(temp);
}
bash 4.3的另一种语法是使用nameref
虽然nameref更方便，因为它无缝解除引用，但是一些较旧的受支持的发行版仍然发布了older version，所以我还是不推荐它。
答案 6 :(得分：4)
一个简单的例子，在调用函数时将在执行脚本或内部脚本期间清除。
#!/bin/bash
echo "parameterized function example"
function print_param_value(){
value1="${1}" # $1 represent first argument
value2="${2}" # $2 represent second argument
echo "param 1 is ${value1}" #as string
echo "param 2 is ${value2}"
sum=$(($value1+$value2)) #process them as number
echo "The sum of two value is ${sum}"
}
print_param_value "6" "4" #space sparted value
#you can also pass paramter durign executing script
print_param_value "$1" "$2" #parameter $1 and $2 during executing
#suppose our script name is param_example
# call like this
# ./param_example 5 5
# now the param will be $1=5 and $2=5