Convert object with two string parameters to comma separated string

Question

This is my object and i'm getting list of this object:

public class MeetingTypeModel
{
    public string MeetingTypeId { get; set; }
    public string MeetingTypeName { get; set; }
}

List<MeetingTypeModel> MeetingTypeList

I want to convert it into format:

({ "1": "Bangladesh", "2": "Belgium"})

just asking all of you guys can he achieve this by overriding the String.Join() method? — Lucifer, May 29 '18 at 05:16
@Lucifer what do you mean, by overriding `String.Join()` method? — Yurii N., May 29 '18 at 05:18
we can have a list in a CSV format by using String.Join() what if user uses that logic and creates his own method to achieve `({ "1": "Bangladesh", "2": "Belgium"})` .I am just asking can we or can we not? c# — Lucifer, May 29 '18 at 05:20
@Lucifer you can't override `string.Join`, it is a static method on a `Type`. you could create a different type `StringExtensions` and an Extension method called `public static string Join(this string, bool myMethodSignatureIsDifferent, params string[] args)` as some sort of overload concept to `string.Join` more or less. but that would still not be static method on `string`. and it would only serve to confuse people. — Brett Caswell, May 29 '18 at 05:26
Do you need to effectively generate a JSON representation of your list or is it just a coincidence that you need the output to be formatted this way ? — Spotted, May 29 '18 at 05:41

TheGeneral · Accepted Answer · 2018-05-29T05:54:35.783

4

The way you shouldn't create a json Dictionary

var temp = MeetingTypeList.Select(x => $"\"{x.MeetingTypeId}\" : \"{x.MeetingTypeName}\"");
var json = $"{{ {string.Join(", ", temp)} }}";

They way you should create a json Dictionary

var dict = MeetingTypeList.ToDictionary(x => x.MeetingTypeId, x => x.MeetingTypeName); 
string json = JsonConvert.SerializeObject(dict, Formatting.Indented);

Update from Brett Caswells comments

Some considerations to this.

We cannot guarantee Distinction with MeetingTypeId MeetingTypeName pairing (it doesn't likely though).

Whether "(" is prepended and ")" appended to the result of this serialization.

If you need the extra parenthesises

json = $"({json})";

edited May 29 '18 at 05:54

answered May 29 '18 at 05:38

TheGeneral

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this is likely the best answer, though there are some considerations to this. the first is that we cannot guarantee Distinction with `MeetingTypeId` `MeetingTypeName` pairing (it doesn't likely though). – Brett Caswell May 29 '18 at 05:50
Second consideration is based on whether "(" is prepended and ")" appended to the result of this serialization. I'm not familiar with parenthesis in Json myself. – Brett Caswell May 29 '18 at 05:51
@BrettCaswell both valid points – TheGeneral May 29 '18 at 05:51
@BrettCaswell updated and attributed – TheGeneral May 29 '18 at 05:54

score 1 · Answer 2 · answered May 29 '18 at 05:19

1

Simply add reference to Newtonoft.Json and add using Newtonsoft.Json;

var MeetingTypeList = new List<MeetingTypeModel>();
// add data to list
string json = JsonConvert.SerializeObject(MeetingTypeList);

answered May 29 '18 at 05:19

Awais Mahmood

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you are also going to have to decorate the properties with `[JsonProperty("1")]` and `[JsonProperty("2")]`... – Zohar Peled May 29 '18 at 05:29
this is probably incorrect. the result output OP put forward uses value of `MeetingTypeId ` property as a field – Brett Caswell May 29 '18 at 05:29
to clarify, wouldn't the result of this serialization be: `[{ "MeetingTypeId" : "1", "MeetingTypeName" : "Bangladesh" }, {"MeetingTypeId" : "2", "MeetingTypeName" : "Belgium" }]` – Brett Caswell May 29 '18 at 06:02

PepitoSh · Answer 3 · 2018-05-29T06:38:03.300

If you would like to avoid using external libraries, you could do it item by item like the following:

    StringBuilder SB = new StringBuilder();
    SB.Append("({");
    for(int i=0; i<MeetingTypeList.Count; i++) {
        MeetingTypeModel mtm = MeetingTypeList[i];

        SB.AppendFormat("\"{0}\":"\"{1}\", mtm.MeetingTypeId, mtm.MeetingTypeName);
        if {i<MeetingTypeList.Count -1) {
          SB.Append(",");
        }
    }
    SB.Append("})");

Prasad Telkikar · Answer 4 · 2018-05-29T05:38:50.887

0

You can iterate through list and by using StringBuilder, you can generate expected output

public string GetString(List<MeetingTypeModel> MeetingTypeList)
{
StringBuilder sb = new StringBuilder();
sb.Append("({");
int i = 1;
foreach(MeetingTypeModel item in MeetingTypeList){

    sb.AppendFormat("\"{0}\" : \"{1}\"" , item.MeetingTypeId, item.MeetingTypeName);
    //To not append last comma
    if(i != MeetingTypeList.Count)
        sb.Append(",");
    i++;
}
sb.Append("})");
return sb.ToString();
}

POC: DotNetFiddler

edited May 29 '18 at 05:38

answered May 29 '18 at 05:21

Prasad Telkikar

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Your code will have an extra comma – SSS May 29 '18 at 05:26
@SSS I updated my answer, why downvote, I can work on your suggestions – Prasad Telkikar May 29 '18 at 05:39
your answer is correct, and doesn't warrant a downvote; but, I wouldn't upvote this, since I consider it to be low quality by virtue of using `StringBuilder`. `StringBuilder` does have it's role, but the better answers here are going to involve serialization. – Brett Caswell May 29 '18 at 06:23
I didn't downvote, someone else did – SSS May 29 '18 at 06:24
@BrettCaswell, I agreed that serialization would be elegant approach over function which I wrote. I used `StringBuilder` as it is mutable and having advantages over `string` – Prasad Telkikar May 29 '18 at 06:30

Convert object with two string parameters to comma separated string

4 Answers4