庆典：调用带有双引号参数的脚本

问题描述：

我有一个参数封闭双引号一个bash脚本，给定的边界值范围内创建地图的形状文件，例如

I have a bash scripts which an argument enclosed with double quotes, which creates a shape-file of map within the given boundries, e.g.

$ export_map "0 0 100 100"

在脚本中，有两个选择语句：

Within the script, there are two select statements:

select ENCODING in UTF8 WIN1252 WIN1255 ISO-8859-8;
...
select NAV_SELECT in Included Excluded;

自然地，这两个语句所需要的输入输入号码作为输入。这可以通过管道将数字，然后是一个换行符，到脚本通过旁路。

Naturally, these two statements require the input to enter a number as an input. This can by bypassed by piping the numbers, followed by a newline, to the script.

为了节省时间，我想有一个脚本，将创造8的地图 - 对编码（4个选项）和 NAV_SELECT （2个选项）。

In order to save time, I would like to have a script that would create 8 maps - for each combination of ENCODING (4 options) and NAV_SELECT (2 options).

我已经写了另外一个bash脚本， create_map ，以服务器作为包装：

I have written another bash script, create_map, to server as a wrapper:

#!/bin/bash

for nav in 1 2 3 4;
do
    for enc in 1 2;
    do
        printf "$nav\n$enc\n" | /bin/bash -c "./export_map.sh \"0 0 100 100\"" 
    done
done

*的这工作（感谢，布莱恩！），但我不能找到一种方法有数值参数0 0 100 100正在从外部脚本之外过去了。 * 的

*This works (thanks, Brian!), but I can't find a way to have the numeric argument "0 0 100 100" being passed from outside the outer script. *

基本上，我正在寻找方法来接受一个包装bash脚本双引号中的参数，并把它传递 - 用双引号 - 到内部脚本

说明：

export_map 是主脚本，正在从 create_map 所谓的8倍。

export_map is the main script, being called from create_map 8 times.

任何想法？

谢谢，

亚当

答

如果我正确地理解你的问题（这我不知道，见我的意见），你应该添加其他 \\ n 你的的printf ; 的printf 默认不添加尾随的换行符的方式，回声一样。这将确保第二个值将被正确读取选择哪些我假设在 export_map.sh 出现命令>。

If I understand your problem correctly (which I'm not sure about; see my comment), you should probably add another \n to your printf; printf does not add a trailing newline by default the way that echo does. This will ensure that the second value will be read properly by the select command which I'm assuming appears in export_map.sh.

printf "$nav\n$enc\n" | /bin/bash -c "./export_map.sh \"100 200 300 400\""

另外，我不认为你需要添加-c 的 /斌/ bash和引号。下面应该足够了，除非我失去了一些东西：

Also, I don't think that you need to add the /bin/bash -c and quote marks. The following should be sufficient, unless I'm missing something:

printf "$nav\n$enc\n" | ./export_map.sh "100 200 300 400"

修改感谢您的澄清。为了从您的包装脚本传递参数，进入内部脚本，保持它作为一个参数，你可以通过$ 1，这里的引号表明您想保持这种归纳为一个参数，而 $ 1 是你的包装脚本的第一个参数。如果你想传递到你内心的脚本从外部脚本的所有参数，每个被保留作为一个参数，你可以使用$ @代替。

edit Thanks for the clarification. In order to pass an argument from your wrapper script, into the inner script, keeping it as a single argument, you can pass in "$1", where the quotes indicate that you want to keep this grouped as one argument, and $1 is the first parameter to your wrapper script. If you want to pass all parameters from your outer script in to your inner script, each being kept as a single parameter, you can use "$@" instead.

#!/bin/bash

for nav in 1 2 3 4;
do
    for enc in 1 2;
    do
        printf "$nav\n$enc\n" | ./export_map.sh "$1"
    done
done

下面是一个如何$ @和一个简单的例子。首先， inner.bash ：

Here's a quick example of how "$@" works. First, inner.bash:

#!/bin/bash

for str in "$@"
do
    echo $str
done

outer.bash ：

#!/bin/bash

./inner.bash "$@"

和调用它：

$ ./outer.bash "foo bar" baz "quux zot"
foo bar
baz
quux zot

庆典：调用带有双引号参数的脚本

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