将类似Excel列字母转换为整数的最简单方法是什么?
例如:
AB --> 27
AA --> 26
A --> 0
Z --> 25
def excel_col_index( str )
value = Hash[ ('A'..'Z').map.with_index.to_a ]
str.chars.inject(0){ |x,c| x*26 + value[c] + 1 } - 1
end
或者
def excel_col_index( str )
offset = 'A'.ord - 1
str.chars.inject(0){ |x,c| x*26 + c.ord - offset } - 1
end
def column_name_to_number(column_name)
multipliers = ('A'..'Z').to_a
chars = column_name.split('')
chars.inject(-1) { |n, c| multipliers.index(c) + (n + 1) * 26 }
end
以下是被接受答案的版本,带有规范:
RSpec.describe "#excel_col_index" do
def excel_col_index(str)
value = Hash[('A'..'Z').map.with_index.to_a]
str.chars.inject(0) { |x, c| x * 26 + value[c] + 1 } - 1
end
{ "A" => 0, "Z" => 25, "AB" => 27, "AA" => 26 }.each do |col, index|
it "generates #{index} from #{col}" do
expect(excel_col_index(col)).to eq(index)
end
end
end
- 1
)def cell2num col
val = 0
while col.length > 0
val *= 26
val += (col[0].ord - 'A'.ord + 1)
col = col[1..-1]
end
return val - 1
end
"AB1".each_codepoint.reduce(0) do |sum, n|
break sum - 1 if n < 'A'.ord # reached a number
sum * 26 + (n - 'A'.ord + 1)
end # => 27
excel_col_index('b') #=> 34
)。如果你需要改变它,可以使用...str.upcase.chars.inject
... - Arye Eidelman