将列索引转换为相应的列字母 [英] Convert column index into corresponding column letter

问题描述

我需要将Google Spreadsheet列索引转换为其对应的字母值，例如，给定电子表格： .stack.imgur.com / DyaPs.pngalt =

我需要这样做（这个函数显然不存在，它是一个例如）：

  getColumnLetterByIndex（4）; //这应该返回D
 getColumnLetterByIndex（1）; //这应该返回A
 getColumnLetterByIndex（6）; //这应该返回F
  

现在，我不记得索引是否从 0 或从 1 ，无论如何这个概念应该是清楚的。

我没有发现关于气体文件的任何内容..我是否失明？任何想法？

谢谢

解决方案

（将返回列号> 26的双字母列名称）：

 函数columnToLetter（column）
 {
 var temp，letter =''; 
 while（column> 0）
 {
 temp =（column  -  1）％26; 
 letter = String.fromCharCode（temp + 65）+ letter; 
 column =（column  -  temp  -  1）/ 26; 
} 
返回信; 
} 
 
函数letterToColumn（字母）
 {
 var column = 0，length = letter.length; 
 for（var i = 0; i  {
 column + =（letter.charCodeAt（i） -  64）* Math.pow（26，length  -  i -  1）; 
} 
返回列; 
} 
  

I need to convert a Google Spreadsheet column index into its corresponding letter value, for example, given a spreadsheet:

I need to do this (this function obviously does not exist, it's an example):

getColumnLetterByIndex(4);  // this should return "D"
getColumnLetterByIndex(1);  // this should return "A"
getColumnLetterByIndex(6);  // this should return "F"

Now, I don't recall exactly if the index starts from 0 or from 1, anyway the concept should be clear.

I didn't find anything about this on gas documentation.. am I blind? Any idea?

Thank you

解决方案

I wrote these a while back for various purposes (will return the double-letter column names for column numbers > 26):

function columnToLetter(column)
{
  var temp, letter = '';
  while (column > 0)
  {
    temp = (column - 1) % 26;
    letter = String.fromCharCode(temp + 65) + letter;
    column = (column - temp - 1) / 26;
  }
  return letter;
}

function letterToColumn(letter)
{
  var column = 0, length = letter.length;
  for (var i = 0; i < length; i++)
  {
    column += (letter.charCodeAt(i) - 64) * Math.pow(26, length - i - 1);
  }
  return column;
}

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