如果长度超过90个字符,如何将字符串拆分为2个部分

时间:2011-11-11 13:37:43

标签: ruby-on-rails ruby string

我有一个动态字符串(在Ruby程序中),如果它超过91个字符,我需要拆分它。我需要将它分成关闭逗号','char。

的部分

字符串示例:

"1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41"

谢谢!

3 个答案:

答案 0 :(得分:6)

这就是你需要的:

string.split(',').in_groups_of(90, false)

结果:

> str.split(',').in_groups_of(10, false)
=> [["1", "2", "3", "4", "5", "6", "7", "8", "9", "10"],
 ["11", "12", "13", "14", "15", "16", "17", "18", "19", "20"],
 ["21", "22", "23", "24", "25", "26", "27", "28", "29", "30"],
 ["31", "32", "33", "34", "35", "36", "37", "38", "39", "40"],
 ["41"]]

或具有连接值的数组:

str.split(',').in_groups_of(10, false).map {|s| s.join(',')}

结果:

> str.split(',').in_groups_of(10, false).map {|s| s.join(',')}
=> ["1,2,3,4,5,6,7,8,9,10",
 "11,12,13,14,15,16,17,18,19,20",
 "21,22,23,24,25,26,27,28,29,30",
 "31,32,33,34,35,36,37,38,39,40",
 "41"]

<强>更新

使用纯Ruby(不是Rails):

str.split(',').each_slice(10).to_a

或加入:

str.split(',').each_slice(10).map {|s| s.join(',')}

答案 1 :(得分:1)

试试这个

  str.split(',') if str.length>91

答案 2 :(得分:0)

我假设你需要一些带String的东西并返回一个Array,其中Array的每个元素都是一个长度为91或更短的字符串。但是,在分割分隔符之前,您希望生成的字符串尽可能长。

a = "1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41"

def my_split(str, delimiter, max_length)
    return [str] if str.length <= max_length
    i = str.rindex(delimiter, max_length)
    the_rest = my_split(str[i+1..-1], delimiter, max_length)
    return [str[0..i-1]] + the_rest
end

for r in my_split(a, ',', 90)
    puts "#{r}:Length: #{r.length.to_s}"
end

结果:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33:Length: 89
34,35,36,37,38,39,40,41:Length: 23