Question

我的字符串就像这个"SHN::test::cg1","RDC::TEST::CONFIG 1"，但是将它存储在变量

中

new_string="SHN::test::cg1","RDC::TEST::CONFIG 1","MDC::test::cg1"

我插入像这样的“＃{new_string}”。值类似于

""SHN::test::cg1","RDC::TEST::CONFIG 1""

更新1：这是我需要使用那个

的场景

$redis.subscribe("RDC::TEST::CONFIG 1","SHN::test::cg1") do |on|
 #hardcoded values inside braces
end

但我需要传递一个变量来形成像上面那样的字符串（在大括号内），我试过

$redis.subscribe("#{new_string}") do |on|
 #interpolated value inside braces
end

更新2：回答我只需要将new_string作为数组传递，不需要转换为字符串这是我的场景

的答案

$redis.subscribe(new_string) do |on|
     #interpolated value inside braces
 end

提前致谢....

Answer 1

你的new_string实际上是一个数组，而不是一个字符串！原因是因为你实际上有三个用逗号分隔的字符串，所以Ruby将它解析为一个字符串数组：

irb> new_string="SHN::test::cg1","RDC::TEST::CONFIG 1","MDC::test::cg1"
=> ["SHN::test::cg1", "RDC::TEST::CONFIG 1", "MDC::test::cg1"]

当你将它插入到一个新字符串中时，Ruby会在数组上调用.to_s，这会（根据你的Ruby版本）产生你所看到的内容：

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，只需将new_string设为实际字符串（通过修复定义中的语法）。或者，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"

如何避免在rails字符串中开始和结束双引号？

1 个答案: