bash bash中函数参数和for循环的问题
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Problem with function arguments and for loop in bash
提问by devoured elysium
Why doesn't this print all the passed arguments, in bash?
为什么这不在 bash 中打印所有传递的参数？
function abc() {
echo "" #prints the correct argument
for x in `seq 1 $#`; do
echo "$x" #doesn't print the 1st, 2nd, etc arguments, but instead 1, 2, ..
done
}
It is printing
它正在打印
1
2
3
4
...
instead.
反而。
回答by Gordon Davisson
I'll just add a couple more options to what everyone else has given. The closest to the way you're trying to write this is to use bash indirect expansion:
我只会在其他人给出的内容中添加更多选项。最接近您尝试编写此代码的方式是使用 bash 间接扩展：
function abc() {
for x in `seq 1 $#`; do
echo "${!x}" # the ! adds a level of indirection
done
}
...another option if you want to operate on only some of the arguments, is to use array slicing with $@:
...如果您只想对某些参数进行操作，另一种选择是使用 $@ 的数组切片：
function def() {
for arg in "${@:2:3}"; do # arguments 2 through 4 (i.e. 3 args starting at number 2)
echo "$arg"
done
}
similarly, "${@:2}"will give you all arguments starting at number 2, "${@:$start:$((end-start+1))}"will give you arguments $start through $end (the $((expression calculates how many arguments there are between $start and $end), etc...
同样，"${@:2}"会给你从数字 2 开始的所有参数，"${@:$start:$((end-start+1))}"会给你从 $start 到 $end 的参数（$((表达式计算 $start 和 $end 之间有多少个参数），等等......
回答by DarkDust
The seqcommand returns all numbers from start to stop. What you are calling here is seq 1 <number_of_arguments_to_abc>. For example, if you call abc alpha beta gamma, then the arguments would be seq 1 3, thus you get the numbers 1, 2 and 3.
该seq命令返回从开始到停止的所有数字。你在这里叫的是seq 1 <number_of_arguments_to_abc>. 例如，如果您调用abc alpha beta gamma，则参数将为seq 1 3，因此您会得到数字 1、2 和 3。
If you want the arguments to abc instead, the expression is for x in "$@".
如果您希望参数为 abc，则表达式为for x in "$@".
回答by jcubic
If you want to print all arguments try this
如果你想打印所有参数试试这个
function abc() {
for arg in $@; do
echo "$arg"
done
}
回答by Chen Levy
Actually there is a special short-hand for this case:
其实这种情况有一个特殊的简写：
function abc() {
for arg ; do
echo "$arg"
done
}
That is, if the in ...part is omitted, argloops over the function's argument $@.
也就是说，如果该in ...部分被省略，则arg循环遍历函数的参数$@。
Incidentally if you have for arg ; ...outside of a function, it will iterate over the arguments given on the command line.
顺便说一句，如果您for arg ; ...在函数之外，它将迭代命令行上给出的参数。
回答by Paused until further notice.
You should use the for argform that others have shown. However, to address some things in your question and comments, see the following:
您应该使用for arg其他人展示的表格。但是，要解决您的问题和评论中的某些问题，请参阅以下内容：
In Bash, it's not necessary to use seq. You can use C-style forloops:
在 Bash 中，没有必要使用seq. 您可以使用 C 风格的for循环：
for ((i = 2; i <= $#; i++))
do
echo "${@:i:1}"
done
Which demonstrates array slicing which is another technique you can use in addition to direct iteration (for arg) or using shift.
它演示了数组切片，这是除了直接迭代 ( for arg) 或使用shift.
An advantage of using either version of foris that the argument array is left intact, while shift modifies it. Also, with the C-style form with array slicing, you could skip any arguments you like. This is usually not done to the extent shown below, because it would rely on the arguments following a strict pattern.
使用任一版本的优点for是参数数组保持不变，而 shift 修改它。此外，使用带有数组切片的 C 样式形式，您可以跳过您喜欢的任何参数。这通常不会在下面显示的范围内完成，因为它会依赖于遵循严格模式的参数。
for ((i = 2; i < $# - 2; i+=2))
That bit of craziness would start at the second argument, process every other one and stop before the last two or three (depending on whether $#is odd or even).
那种疯狂会从第二个参数开始，每隔一个处理一个，并在最后两个或三个之前停止（取决于$#是奇数还是偶数）。
回答by Robert Wohlfarth
The variable Xholds the literal numbers. You're trying to do indirection- substitute $1where there's a $x. Indirection warps the brain. $@provides a simpler mechanism for looping over the arguments - without any adverse effects on your psyche.
变量X保存文字数字。您正在尝试进行间接访问-在有$x 的地方替换$1。间接扭曲大脑。$@提供了一种更简单的循环参数机制 - 不会对您的心理产生任何不利影响。
for x in "$@"; do
echo $x
done
See the bash man pagefor more details on $@.
有关$@ 的更多详细信息，请参阅bash 手册页。
