Question

是否可以使用闭包从值列表中创建一组变量？问这个的原因是基于（例如）两个三个四个或五个部分的列表创建一些递归功能这里的代码当然不起作用，但任何指针都会有所帮助。然后

def longthing = 'A for B with C in D on E'
//eg shopping for 30 mins with Fiona in Birmingham on Friday at 15:00
def breaks = [" on ", " in ", "with ", " for "]
def vary = ['when', 'place', 'with', 'event']

i = 0
line = place = with = event = ""
breaks.each{
shortline = longthing.split(breaks[i])
longthing= shortline[0]
//this is the line which obviously will not work
${vary[i]} = shortline[1]
rez[i] = shortline[1]
i++
 }
return place + "; " + with + "; " + event
// looking for answer of D; C; B

编辑＆gt;＆GT;

是的我正在尝试找到一种更加清晰的方法来清理它，我必须在每次循环之后做这件事

len = rez[3].trim()
if(len.contains("all")){
len = "all"
} else if (len.contains(" ")){
len = len.substring(0, len.indexOf(" ")+2 )
}
len = len.replaceAll(" ", "")
with = rez[2].trim()
place = rez[1].trim()
when = rez[0].trim()
event = shortline[0]

如果我决定在列表中添加另一个项目（我刚刚做了），我必须记住哪个[i]成功提取它

这是解析日期/时间的工作部分，然后使用jChronic将自然文本转换为公历日历信息，以便我可以在Google日历中设置事件

Answer 1

怎么样：

def longthing = 'A for B with C in D on E'
def breaks = [" on ", " in ", "with ", " for "]
def vary = ['when', 'place', 'with', 'event']
rez = []
line = place = with = event = ""

breaks.eachWithIndex{ b, i ->
  shortline = longthing.split(b)
  longthing = shortline[0]
  this[vary[i]] = shortline[1]
  rez[i] = shortline[1]
}
return place + "; " + with + "; " + event

Answer 2

当你使用带有List和“each”的闭包时，groovy循环遍历List中的元素，将值放在列表中的“it”变量中。但是，由于您还想跟踪索引，因此还有一个groovy eachWithIndex，它也传入索引

http://groovy.codehaus.org/GDK+Extensions+to+Object

类似

breaks.eachWithIndex {item, index ->
   ... code here ...
}

groovy闭包实例化变量

2 个答案: