Ruby:OptionParser:String Arg&;散列赋值
使用OptionParser进行字符串参数输入和哈希赋值。为一个参数读入多个变量的最佳方法是什么?然后如何将它们分配给要引用的哈希?以下是我到目前为止的情况:Ruby:OptionParser:String Arg&;散列赋值,ruby,variables,split,variable-assignment,optionparser,Ruby,Variables,Split,Variable Assignment,Optionparser,使用OptionParser进行字符串参数输入和哈希赋值。为一个参数读入多个变量的最佳方法是什么?然后如何将它们分配给要引用的哈希?以下是我到目前为止的情况: large_skus = Hash.new small_skus = Hash.new OptionParser.new do |opts| opts.on("-b", "--brands bName1,bName2,bNameN", String, "Check specific brands by name") do |b| o
large_skus = Hash.new
small_skus = Hash.new
OptionParser.new do |opts|
opts.on("-b", "--brands bName1,bName2,bNameN", String, "Check specific brands by name") do |b|
options[:brands] = b.split(",")
end
opts.on("-l", "--large lSku1,lSku2,lSkuN", String, "Large SKUs - List CSVs") do |l|
options[:large_skus] = l.split(",")
##For each sku given
brandName = options[:brands]
large_skus[brandName] = l[$sku].to_i
##
end
opts.on("-s", "--small sSku1,sSku2,sSkuN", String, "Small SKUs - List CSVs") do |s|
options[:small_skus] = s.split(",")
##For each sku given
brandName = options[:brands]
small_skus[brandName] = s[$sku].to_i
##
end
end.parse!(ARGV)
如果输入:
ruby test.rb --brands bName1,bName2,bNameN --large lSku1,lSku2,lSkuN --small wQueue1,wQueue2,wQueueN
此代码
#!/usr/bin/env ruby
require 'ap'
require 'optparse'
options = {}
OptionParser.new do |opts|
opts.on("-b", "--brands bName1,bName2,bNameN", Array, "Check specific brands by name") do |b|
options[:brands] = b
end
opts.on("-l", "--large lSku1,lSku2,lSkuN", Array, "Large SKUs - List CSVs") do |l|
options[:large_skus] = l
end
opts.on("-s", "--small wQueue1,wQueue2,wQueueN", Array, "Small SKUs - List CSVs") do |s|
options[:small_skus] = s
end
end.parse!(ARGV)
ap options
生成此输出:
{
:brands => [
[0] "bName1",
[1] "bName2",
[2] "bNameN"
],
:large_skus => [
[0] "lSku1",
[1] "lSku2",
[2] "lSkuN"
],
:small_skus => [
[0] "wQueue1",
[1] "wQueue2",
[2] "wQueueN"
]
}
请注意,我使用的不是每个选项的字符串类型,而是数组。这让OptionParser能够完成将元素解析为数组的繁重工作。从这一点上看,如何处理数组元素取决于您。我认为您的做法是错误的。您希望您的用户必须跟踪他们输入的参数的顺序,但您不希望自己在代码中这样做 你不要求任何人跟踪什么与什么相关,并使其明确:
ruby test.rb --input bName1,lSku1,wQueue1 --input bName2,lSku2,wQueue2 --input bNameN,lSkuN,wQueueN
代码:
要知道,阵列是一个很好的选择。散列赋值呢?我更新了我原来的帖子,我注意到一些打字错误(队列与sku)。我的想法是,我不必按索引号迭代大小sku阵列以找到它们的匹配/关联品牌。根据示例代码,您试图做的事情并不明显。您的代码缺少变量($sku),并且您没有解释使用大sku和小sku散列的意图。
opts.on("--input <brand,Large_Skus,Small_Skus>", "input description",
"NOTE: Can be used more than once.") do |opt|
list = opt.split(',')
unless list.lenght == 3
raise "some error because you didn't place all arguments"
end
options[:input].push list
end
[ [ 'bName1', 'lSku1', 'wQueue1' ],
[ 'bName2', 'lSku2', 'wQueue2' ],
[ 'bNameN', 'lSkuN', 'wQueueN' ] ]