Question

鉴于此类数据集

records = [
  {id: 1, name: "John Smith", age: 49},
  {id: 2, name: "Jane Brown", age: 45},
  {id: 3, name: "Tim Jones",  age: 60},
  {id: 4, name: "Blake Owen", age: 78}    
]

如何过滤记录以返回刚刚超过50岁的简化数组。

假设返回数组是

over50s = // something

我看过很多相似的代码，但它对我来说很合适。

感谢您的帮助！

Answer 1

这个示例脚本怎么样？使用WebView检索结果over50s。

示例脚本：

filter()

结果：

records = [
  {id: 1, name: "John Smith", age: 49},
  {id: 2, name: "Jane Brown", age: 45},
  {id: 3, name: "Tim Jones",  age: 60},
  {id: 4, name: "Blake Owen", age: 78}
]
over50s = records.filter (e) -> e.age >= 50

如果我误解了你的问题，请告诉我。我想修改。

Answer 2

答案很完美，但我想使用reference添加“Coffeescript方式”：

 over50s = (record for record in records when record.age > 50)

说明：

for record in records
  console.log(record)
# This would loop over the records array, 
#   and each item in the array will be accessible 
#   inside the loop as `record`

console.log(record) for record in records
# This is the single line version of the above

console.log(record) for record in records when record.age > 50
# now we would only log those records which are over 50

over50s = (record for record in records when record.age > 50)
# Instead of logging the item, we can return it, and as coffeescript
#   implicitly returns from all blocks, including comprehensions, the 
#   output would be the filtered array
# It's shorthand for:

over50s = for record in records
  continue unless record.age > 50
  record

# This long hand is better for complex filtering conditions. You can
#   just keep on adding `continue if ...` lines for every condition
# Impressively, instead of ending up with `null` or `undefined` values
#   in the filtered array, those values which are skipped by `continue`
#   are completely removed from the filtered array, so it will be shorter.

过滤coffeescript中的记录数组，以返回所有带有年龄的记录＆gt; 50

2 个答案:

示例脚本：

结果：

说明：