bash: calling a scripts with double-quote argument

问题

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"

Within the script, there are two select statements:

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

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.

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).

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

**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. **

Basically, I'm looking for way to accept an argument within double quotes to a wrapper bash script, and pass it - with the double quotes - to an inner script.

CLARIFICATIONS:

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

Any ideas?

Thanks,

Adam

回答1:

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\""

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"

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

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 "$@"

And invoking it:

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

来源：https://stackoverflow.com/questions/1994928/bash-calling-a-scripts-with-double-quote-argument

标签

bash

quotes