c# 科学计数法值转换成正常值，返回字符串

/// <summary>///  科学计数法值转换成正常值/// </summary>/// <param name="value"></param>/// <returns></returns>public string ValueScientificNotationConvert(JToken value){if (value == null) return "";var s = value.ToString(Formatting.Indented).ToString();string num = String.Empty;// s = s.Trim();string[] arr = s.Split('E');int LastZeroCountInPart1 = 0;foreach (char ch in arr[0].Reverse()){if (ch == '0')LastZeroCountInPart1++;else{break;}}bool isPart1Double = double.TryParse(arr[0], out var part1);int decimalDigitCountInPart1 = arr[0].IndexOf('.') == -1 ? 0 : arr[0].Length - 1 - arr[0].IndexOf('.');string newPart2 = arr.ElementAtOrDefault(1);bool isPart2Int = int.TryParse(newPart2, out var part2);if (arr.Length == 1){if (isPart1Double)num = part1.ToString();}if (arr.Length == 2){if (part2 < 0 || part2 - decimalDigitCountInPart1 < 0){if (isPart1Double && isPart2Int)num = Decimal.Parse((part1 * Math.Pow(10, part2)).ToString(), System.Globalization.NumberStyles.Float).ToString(); // 此时Math.Pow(10, part2)并不会越界，故可使用Decimal.Parse从scientific num转换为real num
                }else if (part2 - decimalDigitCountInPart1 >= 0){num = part1.ToString().Replace(".", "") + new String('0', part2 - decimalDigitCountInPart1);  // 避免越界，此处使用字符串拼接代替Math.Pow(10, part2)，拼接还可用StringBuilder的append方法
                }}return num + new String('0', LastZeroCountInPart1); // 此处拼接还可使用StringBuilder的append方法}