如何在CoffeeScript中

Question

我試圖從經典的書重寫例如遍歷兩個數組「的JavaScript：好的部分」中的CoffeeScript通過列表解析：

var parse_url = /^(?:([A-Za-z]+):)?(\/{0,3})([0-9.\-A-Za-z]+) 
(?::(\d+))?(?:\/([^?#]*))?(?:\?([^#]*))?(?:#(.*))?$/; 

var url = 「http://www.ora.com:80/goodparts?q#fragment」; 
var result = parse_url.exec(url); 
var names = ['url', 'scheme', 'slash', 'host', 'port', 'path', 'query', 'hash']; 
var blanks = '  '; 
var i; 
for (i = 0; i < names.length; i += 1) { 
    document.writeln(names[i] + ':' + 
     blanks.substring(names[i].length), result[i]); 
} 

我不能怎麼寫循環，堅持這一代碼：

console.log "#{name}: " for name in names 

Answer 1

我想象你正在試圖做一些事情Like this

parse_url = /^(?:([A-Za-z]+):)?(\/{0,3})([0-9.\-A-Za-z]+)(?::(\d+))?(?:\/([^?#]*))?(?:\?([^#]*))?(?:#(.*))?$/ 

url = 'http://www.ora.com:80/goodparts?q#fragment' 

result = parse_url.exec(url) 

names = ['url', 'scheme', 'slash', 'host', 'port', 'path', 'query', 'hash'] 

alert "#{names[i]}: #{result[i]}" for i in [0..names.length-1] 

Answer 2

萬一有人來這裏實際上需要在同一時間進行迭代兩個數組，這個工程：

for [course_event, remote_event] in zip(course.events, remote.events) 
    course_event.lessons = remote_event.lessons if event.lessons_need_updating 

zip()爲here：

zip =() -> 
    lengthArray = (arr.length for arr in arguments) 
    length = Math.min(lengthArray...) 
    for i in [0...length] 
    arr[i] for arr in arguments