Generate string out of a combination of input strings

Question

I've a little messy method, which generates a name out of three strings. The three strings can be null. I've to check each possible combination and create a string, based on the given available values. There isn't a fix position / layout for each variable in the resulting name, it depends on the availability of all strings.

My method looks like this right now:

    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;
    }

Any suggestions on how to improve this method (Get rid of the if / else construct)?

I think you missed case of else if (!String.IsNullOrEmpty(dimension)) — Nikhil Agrawal
– Nikhil Agrawal, Commented Jun 25, 2014 at 7:40

Hassan Nisar · Accepted Answer · 2014-06-25 08:17:06Z

5

check the nulls at the top, change to strings if need be, then create a final result

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

edited Jun 25, 2014 at 8:17

Hassan Nisar

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answered Jun 25, 2014 at 7:41

Sayse

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6 Comments

Hassan Nisar Over a year ago

Adding to List<string>() and String.Join is simple and elegant solution.

Sayse Over a year ago

@gildor if any are null they arent included in the join

Gildor Over a year ago

@Sayse sorry I misunderstood. Vote up!

cb_dev Over a year ago

Good hint, but there would be some format errors, when only dimension and variable are available. It should be "Variable [Dimension]", with the List/String.Join method it would lead to "[Dimension] (Variable)" (Wrong order + variable in braces)

Sayse Over a year ago

@cb_dev - are you saying that the second parameter needs square brackets and the third needs parenthesis? if so then tim's answer is correct.

|

Tim Schmelter · Accepted Answer · 2014-06-25 07:50:11Z

0

If i've understood your logic correctly, you could also shorten it in this way:

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;
}

answered Jun 25, 2014 at 7:50

Tim Schmelter

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5 Comments

Sayse Over a year ago

This was my first thought too but this would give the variable square brackets instead of ( as per the second else if

Tim Schmelter Over a year ago

@Sayse: good catch. I haven't noticed that "variable" must have parenthesis since OP mentioned that "there isn't a fix position / layout for each variable".

Hassan Nisar Over a year ago

Just for my understanding. Is it possible to determine by using notNull.Length that which string is out of all three strings? What if dimension variable parameter is given and other two are null?

Tim Schmelter Over a year ago

@HassanNisar: In my approach it is not necessary to determine which string was null, but - as Sayse has already noticed - i've probably misunderstood OP requirement a little bit.

cb_dev Over a year ago

Intresting solution, but Sayse is right, in some cases it would lead to a wrong result

Prinshul Jain · Accepted Answer · 2014-06-25 08:12:41Z

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);
    }

Eugene Podskal · Accepted Answer · 2014-06-25 13:54:36Z

Taking into account that your can't just use empty strings in templates or use simple composition, your best bet is to use some sort of Func table(dictionary) that will contain formatting function:

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);

It could be made more generalized(I mean not just for 3 strings), but in the end such solution will be able to handle any possible combination or formatting idiosyncrasies.

Daniel · Accepted Answer · 2014-06-25 07:46:19Z

-1

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();
}

answered Jun 25, 2014 at 7:46

Daniel

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