无法从shell脚本调用Openscad -D命令行

Question

我在Ubuntu linux上使用openscad命令从命令行生成模型.我最终成功运行了openscad命令,使用-D参数覆盖变量值:

$ openscad -o output.stl -D 'param1="name1"' -D 'param2="name2"' openscad-script.scad

编辑:注意我们必须传递-D参数的方式,单引号和双引号必须根据因为Openscad文档.

但是当我从shell脚本生成并执行相同的命令时,openscad失败并出现错误:

$ ./myscript.sh value1 value2 

错误:第XX行中的分析器错误:语法错误无法解析文件'openscad-script.scad'!

其中XX =文件的最后一行.

这是bash脚本

#!/bin/bash
# run openscad command
param1="-D 'param1=\"$1\"'"
param2="-D 'param2=\"$2\"'"
echo "openscad -o $1-$2.stl $param1 $param2 openscad-script.scad"
openscad -o $1-$2.stl $param1 $param2 openscad-script.scad

这看起来很简单,我仍然无法弄清楚opencad在运行命令时失败的原因.

谢谢你的帮助,

编辑:我发现了一种让它工作的方法,可能不是最好的

#!/bin/bash
# run openscad command
param1="-D 'param1=\"$1\"'"
param2="-D 'param2=\"$2\"'"
command = "openscad -o $1-$2.stl $param1 $param2 openscad-script.scad"
eval $command

Answer 1

如果您的命令行是:

openscad -o name1-name2.stl -D 'param1="name1"' -D 'param2="name2"' openscad-script.scad

...然后执行此操作的正确脚本将是:

#!/bin/bash
openscad \
  -o "$1-$2.stl" \
  -D "param1=\"$1\"" \
  -D "param2=\"$2\"" \
  openscad-script.scad

...或者,如果你真的想因为某种原因在多行上建立东西:

#!/bin/bash

args=( -o "$1-$2.stl" )
args+=( -D "param1=\"$1\"" )
args+=( -D "param2=\"$2\"" )

openscad "${args[@]}" openscad-script.scad

openscad字面上无法知道命令行中是否使用了单引号或双引号,因此没有强制要求单引号的方法.而且,shell引用是逐个字符的属性!也就是说:

'param1="name1"' # becomes the C string "param1=\"name1\""

导致完全相同的字符串传递为:

param1='"name1"' # becomes the C string "param1=\"name1\"", same as above

要么

param1='"'name1'"' # becomes the C string "param1=\"name1\"", same as above

......当这些值都是常数时.然而,当name1被替换为类似的东西$1时,根据使用的引用类型,含义会变得非常不同:

set -- name1; IFS=0123456789
"param1=\"$1\"" # becomes the C string "param1=\"name1\"", as above
'param1="$1"'   # becomes the C string "param1=\"$1\"", not substituting name1

# ...and, as an example of something to look out for:
param1='"'$1'"'  # becomes TWO C strings, "param1=\"name" and "\"", due to bad quoting
# ...the IFS above makes it split on numbers; by default this risk would happen with spaces