使用Regex C＃解析字符串

Question

I am trying to customise a DevExpress grid filter.

Currently I return the data from my api, and so I need to parse the built in string which is returned from the grid.

An example of the filter string is;

StartsWith([name], 'test') And StartsWith([quantity], '12') And StartsWith([id], '1') And StartsWith([date], '01/10/2015')

I would like to convert this to a Dictionary in the most efficient way?

Answer 1

You could use Regex for filtering the key/value pair outta your string and a simple foreach to prepare the dictionary.

This could be a solution:

public static Dictionary<string, object> FilterAPIData(string data)
    {
        var r = new Regex(@"\[\w+\], \'[\w/]+\'");

        var result = r.Matches(data);
        var dict = new Dictionary<string, object>();

        foreach (Match item in result)
        {
            var val = item.Value.Split(',');
            dict.Add(val[0], val[1]);
        }

        return dict;
    }

Answer 2

Regex might be the best option for this, but I'll show you how to do it without Regex as it can be a bit difficult to understand.

Assuming your string will always be in this format you can do this:

string str = "StartsWith([name], 'test') And StartsWith([quantity], '12') And StartsWith([id], '1') And StartsWith([date], '01/10/2015')";
var strArray = str.Split(new string[]{"And "}, StringSplitOptions.None);

var dict = new Dictionary<string, string>();
foreach(var value in strArray)
{
    dict.Add(GetStringBetween(value, "[", "]"), GetStringBetween(value, "'", "'"));
}

private string GetStringBetween(string value, string startDelim, string endDelim)
{
     int first = value.IndexOf(startDelim) + startDelim.Length;
     int last = value.LastIndexOf(endDelim);

     return value.Substring(first, last - first);
}

//Output : 
//name test 
//quantity 12 
//id 1 
// date 01/10/2015 

If there other formats the string can be in you can adjust as needed. I would also consider adding in more validation/error handling, but I will let you figure that out ;)