bash 从脚本和命令行调用时，脚本参数解析不同

Question

我正在尝试从另一个脚本调用一个脚本，但命令行参数解析不正确。这是一些错误还是我遗漏了什么？ test.sh

#!/bin/bash

set -o errexit
set -o nounset

set -x

while test $# -gt 0; do
    case "" in 
        -n) shift
            name=
            shift
            ;;
        -s)
            shift
            ssh_options=
            shift
            ;;
        *)
            echo "see --help"
            exit 1
            ;;
    esac
done

$ssh_cmd="${ssh_options}"

echo "${name} ssh_options=${ssh_options}"

call_test.sh

#!/bin/bash

set -x
cmd=$'./test.sh -n foo -s \"-foo -bar -F\"'
$($cmd)

echo done

#output

./test.sh -n foo -s "-foo -bar -F"
+ test 4 -gt 0
+ case "" in
+ shift
+ name=foo
+ shift
+ test 2 -gt 0
+ case "" in
+ shift
+ ssh_options='-foo -bar -F'
+ shift
+ test 0 -gt 0
+ echo 'foo ssh_options=-foo -bar -F'
foo ssh_options=-foo -bar -F



./call_test.sh
+ cmd='./test.sh -n foo -s "-foo -bar -F"'
++ ./test.sh -n foo -s '"-foo' -bar '-F"'
+ test 6 -gt 0
+ case "" in
+ shift
+ name=foo
+ shift
+ test 4 -gt 0
+ case "" in
+ shift
+ ssh_options='"-foo'
+ shift
+ test 2 -gt 0
+ case "" in
+ echo 'see --help'
+ exit 1
+ see --help
./call_test.sh: line 5: see: command not found
+ echo done
done

感谢任何帮助。

Answer 1

$ssh_cmd="${ssh_options}" 不是您想要的。当赋值 .

时，不要在变量上使用 $

通过https://www.shellcheck.net/-

养成运行错误的习惯

Line 26:
$ssh_cmd="${ssh_options}"
^-- SC2281: Don't use $ on the left side of assignments.
^-- SC2154: ssh_cmd is referenced but not assigned.

Did you mean: (apply this, apply all SC2281)
ssh_cmd="${ssh_options}"

再次编辑并测试。

bash 从脚本和命令行调用时，脚本参数解析不同

bash script arguments parsing differ when called from script and from command line

bash

scripting