大家好：我经常遇到以下关于咖啡脚本的问题，想知道是否有人知道解决方案。请考虑以下代码：
k=for x in [0...3]
{foo:{bar:x,baz:3},qux:5}
我有很多代码都有这种基本布局。然而，它很难阅读。以以下方式编写此代码会更干净：
#Gives Error- PARSE ERROR ON LINE 5: UNEXPECTED TERMINATOR
k=for x in [0...3]
foo:
bar:x
baz:3
qux:5
这个错误可以通过下面的破解来规避，看起来真的很难看：
k=for x in [0...3]
g=
foo:
bar:x
baz:3
qux:5
g
有人知道在理解中使用Coffeescapet的多行assoc数组语法而不会遇到这个错误的干净方法吗？谢谢你的帮助！
所以
k=for x in [0...3]
foo:
bar:x
baz:3
qux:5
未能编译，但是
func
foo:
bar:x
baz:3
qux:5
（例如）编译正确。我相信这是CoffeeScript解析器中的一个已知错误。不幸的是，还有几个类似的问题悬而未决，因为解析YAML样式的对象变得异常棘手。所以现在，我会使用显式花括号，就像c3rin建议的那样。
[编辑：请参阅特别发布981。]
.map（）是您的好友：
k = [0...3].map (i) ->
foo:
bar: "#{i}"
baz: i
qux: i*3
（我知道你的问题实际上是一个bug，但这更有意义。列表理解更适合简单的任务）
我已经更改了几次答案，但我认为第一个例子的问题是，coffeescript编译器认为foo：是你想要构建的对象，并且在它到达qux时很担心：因为它认为它与foo完全不同。有趣的是，您可以混合JSON样式和YAML样式的声明，使用JSON样式的大括号显式地声明对象定义的边界，并在边界内部使用YAML以提高可读性。
{
foo:
bar:x
baz:3
qux:5
}
我通常的解决方案如下：
k = for x in [0...3]
g =
foo:
bar:x
baz:3
qux:5
设置变量会返回其设置的值。这仍然有点麻烦，但比设置后显式返回g的版本稍微好一点。不过，这绝对是解决coffeescript错误的方法。
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