Question

我有点凌乱的方法，它会从三个字符串中生成一个名字。这三个字符串可以为null。我根据给定的可用值检查每个可能的组合并创建一个字符串。结果名称中的每个变量都没有固定位置/布局，这取决于所有字符串的可用性。

我的方法现在看起来像这样：

    private String GetName(string variable, string label, string dimension)
    {
        string result = String.Empty;
        if (!String.IsNullOrEmpty(label) && !String.IsNullOrEmpty(variable) && !String.IsNullOrEmpty(dimension))
        {
            result = String.Format("{0} [{1}] ({2})", label, dimension, variable);
        }
        else if (!String.IsNullOrEmpty(label) && !String.IsNullOrEmpty(dimension))
        {
            result = String.Format("{0} [{1}]", label, dimension);
        }
        else if (!String.IsNullOrEmpty(label) && !String.IsNullOrEmpty(variable))
        {
            result = String.Format("{0} ({1})", label, variable);
        }
        else if (!String.IsNullOrEmpty(label))
        {
            result = String.Format("{0}", label);
        }
        else if (!String.IsNullOrEmpty(variable) && !String.IsNullOrEmpty(dimension))
        {
            result = String.Format("{0} [{1}]", variable, dimension);
        }
        else if (!String.IsNullOrEmpty(variable))
        {
            result = String.Format("{0}", variable);
        }
        return result;
    }

有关如何改进此方法的任何建议（摆脱if / else结构）？

Answer 1

检查顶部的空值，如果需要则更改为字符串，然后创建最终结果

var stringsToInclude = new List<string>();
if(!String.IsNullOrEmpty(label))
    stringsToInclude.Add(label);
if(!String.IsNullOrEmpty(dimension))
    stringsToInclude.Add(string.Format("[{0}]", dimension));
if(!String.IsNullOrEmpty(variable))
    stringsToInclude.Add(string.Format("({0})", variable));


result = string.Join(" ", stringsToInclude);

String.Join

Answer 2

考虑到你不能在模板中使用空字符串或使用简单的合成，最好的办法是使用某种包含格式化功能的Func表（字典）：

public class MyStringFormatterKey : IEquatable<MyStringFormatterKey>
{
    private Boolean isVariableAbsent, isLabelAbsent, isDimensionAbsent;
    public MyStringFormatterKey (String variable, String  label, String dimension) : 
        this(String.IsNullOrEmpty(variable), String.IsNullOrEmpty(label), String.IsNullOrEmpty(dimension))
        {   } 

    public MyStringFormatterKey (Boolean isVariableAbsent, Boolean  isLabelAbsent, Boolean  isDimensionAbsent)
        {  this.isVariableAbsent = isVariableAbsent;  this.isLabelAbsent= isLabelAbsent; this.isDimensionAbsent = isDimensionAbsent; } 

    public override Int32 GetHashCode()
    {
        return 100 * (isVariableAbsent ? (0) : (1)) + 10 * (isLabelAbsent ? (0) : (1)) + (isDimensionAbsent ? (0) : (1));
    }

    public bool Equals(MyStringFormatterKey other)
    {
        return (this.GetHashCode() == other.GetHashCode());
    }
}

...


Dictionary<MyStringFormatterKey , Func<String, String, String, String>> formatterDictionary = 
    new Dictionary<MyStringFormatterKey , Func<String, String, String, String>>();

formatterDictionary.Add(new MyStringFormatterKey(false, true, false), 
    (variable, label, dimension) => String.Format("{0} [{1}]", variable, dimension));

...

Func<String, String, String, String> stringFormatter = formatterDictionary[new MyStringFormatterKey(variable, label, dimension)];

String result = stringFormatter(variable, label, dimension);

它可以更通用化（我的意思不仅仅是3个字符串），但最终这样的解决方案将能够处理任何可能的组合或格式化特性。

Answer 3

如果我理解了你的逻辑，你也可以用这种方式缩短它：

private String GetName(string variable, string label, string dimension)
{
    string[] all = { variable, label, dimension };
    string[] notNull = all.Where(s => !string.IsNullOrEmpty(s)).ToArray();
    string format = notNull.Length == 3 ? "{0} [{1}] ({2})" 
                  : notNull.Length == 2 ? "{0} [{1}]" : "{0}";
    string result = String.Format(format, notNull);
    return result;
}

Answer 4

public String GetName(string label, string dimension, string variable)
    {
        string result = String.Empty;
        string labelFmt = !string.IsNullOrEmpty(label) ? string.Format("{0}", label) : string.Empty;
        string dimenFmt = !string.IsNullOrEmpty(dimension) ? string.Format("[{0}]", dimension) : string.Empty;
        string varFmt = !string.IsNullOrEmpty(variable) ? string.Format("({0})", variable) : string.Empty;
        string firstSepStr = (!string.IsNullOrEmpty(label) &&
                               (!string.IsNullOrEmpty(variable) || !string.IsNullOrEmpty(dimension)))
                                  ? " "
                                  : string.Empty;
        string secondSepStr = (!string.IsNullOrEmpty(variable) && !string.IsNullOrEmpty(dimension))
                                  ? " "
                                  : string.Empty;
        return string.Format("{0}{1}{2}{3}{4}", labelFmt, firstSepStr, dimenFmt, secondSepStr, varFmt);
    }

Answer 5

private String GetName(string variable, string label, string dimension)
{
    StringBuilder result = new StringBuilder();
    bool bEmpty = true;
    if (!String.IsNullOrEmpty(label)
    {
        result.Append(label);
        result.Append(' ');
    }
    if (!String.IsNullOrEmpty(dimension)
    {
        result.Append('[');
        result.Append(dimension);
        result.Append("] ");
    }
    if (!String.IsNullOrEmpty(variable)
    {
        result.Append('(');
        result.Append(dimension);
        result.Append(") ");
    }
    if (result.Count > 0)
    {
       result.Length := result.Length - 1;
    }
    return result.ToString();
}

从输入字符串的组合生成字符串

5 个答案: