我的字符串是这样的，"SHN::test::cg1","RDC::TEST::CONFIG 1"但是当将其存储在变量中时
new_string="SHN::test::cg1","RDC::TEST::CONFIG 1","MDC::test::cg1"
我像这样的“＃{new_string}”进行插值。值就像
""SHN::test::cg1","RDC::TEST::CONFIG 1""
更新1：这是我需要使用该场景的情况
$redis.subscribe("RDC::TEST::CONFIG 1","SHN::test::cg1") do |on|
#hardcoded values inside braces
end
但是我需要传递一个变量来形成上面尝试过的字符串（在大括号内）
$redis.subscribe("#{new_string}") do |on|
#interpolated value inside braces
end
更新2：答案我只需要传递new_string作为数组本身，不需要转换为字符串，这就是我的情况的答案
$redis.subscribe(new_string) do |on|
#interpolated value inside braces
end
提前致谢....
您new_string实际上是一个数组，而不是字符串！原因是因为您实际上有三个用逗号分隔的字符串，所以Ruby将其解析为字符串数组：
irb> new_string="SHN::test::cg1","RDC::TEST::CONFIG 1","MDC::test::cg1"
=> ["SHN::test::cg1", "RDC::TEST::CONFIG 1", "MDC::test::cg1"]
然后将其插值到新字符串中时，Ruby会调用.to_sArray，该数组（取决于Ruby的版本）会产生您所看到的内容：
irb> ["SHN::test::cg1","RDC::TEST::CONFIG 1","MDC::test::cg1"].to_s
=> "[\"SHN::test::cg1\", \"RDC::TEST::CONFIG 1\", \"MDC::test::cg1\"]"
irb> puts ["SHN::test::cg1","RDC::TEST::CONFIG 1","MDC::test::cg1"].to_s
["SHN::test::cg1", "RDC::TEST::CONFIG 1", "MDC::test::cg1"]
要解决此问题，只需制作new_string一个实际的字符串（通过在定义中修正语法）。或者，您自己将数组转换为字符串，例如：
irb> "#{new_string.join('')}"
=> "SHN::test::cg1RDC::TEST::CONFIG 1MDC::test::cg1"
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