How to pass Json string with variable in c#?

Question

string a;

string b;

string jSonContent = @"{""fields"":{""summary"":""summary"" , ""description"": ""modified.""}}";

I want a instead of summary and b instead of modified. It works if I replace summary but gives errors if I replace both.

The edited code is, string jSonContent = @"{""fields"":{""summary"":"+ "\""+ a+ "\"" +" , ""description"": "+ "\""+ b+ "\"" +"}}"; It's not giving any error, summary field is getting updated, but the description field is not.

Please revert. Thanks!

I strongly recommend that you use a JSON library such as Json.NET to construct the JSON, rather than trying to use string operations. — Jon Skeet
– Jon Skeet, Commented Mar 30, 2017 at 11:04
Just to clarify: you want the contents of variables a and b as values of those JSON properties? — Hans Keﬆing
– Hans Keﬆing, Commented Mar 30, 2017 at 11:06
Also please show the code you tried with that replacing, and what those "errors" were exactly. — Hans Keﬆing
– Hans Keﬆing, Commented Mar 30, 2017 at 11:07
@HansKesting Yes, I want the values of a and b to be passed to the server. — Bhakti Kulkarni
– Bhakti Kulkarni, Commented Mar 30, 2017 at 11:08
So please edit the question, instead of adding it as a comment - and it's very confusing when you initially say "but gives errors" and now you're saying there aren't any errors. — Jon Skeet
– Jon Skeet, Commented Mar 30, 2017 at 11:14

Darjan Bogdan · Accepted Answer · 2017-03-30 11:23:01Z

5

I highly suggest you to use JSON library, for example Json.Net.

First of all, you'll be able to work with strongly typed objects, in addition you will avoid typos and similar bugs since the serializer will do the serialization for you.

public void Test()
{
    string a = ""; //content of 'a' variable
    string b = ""; //content of 'b' variable

    var obj = new RootObject();
    obj.Fields = new Fields();

    obj.Fields.Summary = a;
    obj.Fields.Description = b;

    var jsonOutput = Newtonsoft.Json.JsonSerializer.Serialize(obj, typeof(RootObject));
}

public class Fields
{
    [JsonProperty("summary")]
    public string Summary { get; set; }

    [JsonProperty("description")]
    public string Description { get; set; }
}

public class RootObject
{
    [JsonProperty("fields")]
    public Fields Fields { get; set; }
}

Note: If you don't want to create unnecessary types, you can work directly with JObject, this is one of possible uses:

var jobj = Newtonsoft.Json.Linq.JObject.FromObject(new {
    fields = new {
        summary = a,
        description = b
    }
});
var jsonOutput = jobj.ToString();

edited Mar 30, 2017 at 11:23

answered Mar 30, 2017 at 11:12

Darjan Bogdan

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4 Comments

Jon Skeet Over a year ago

Not that you need to create extra types like this - I'd just use JObject unless this is needed in multiple places.

Bhakti Kulkarni Over a year ago

I have them all.. The only problem I am facing is with passing these variables in Json string. Also, it works fine if I pass only summary(with variable), it also works fine if I pass both of them, but with static data, the problem is with making both of them dynamic.

Ole Albers Over a year ago

That "<content>" really was irritating. Thought you want to insert xml-tags :)

Darjan Bogdan Over a year ago

@OleAlbers here you go :)

Haresh Kola · Accepted Answer · 2017-03-30 12:53:29Z

-1

string a = "a";
string b = "b";

string jSonContent = (@"{""fields"":{""summary"":""summary"" , ""description"": ""modified.""}}").Replace(@":""summary""", a).Replace("modified.", b);

edited Mar 30, 2017 at 12:53

answered Mar 30, 2017 at 11:25

Haresh Kola

284 bronze badges

6 Comments

Bhakti Kulkarni Over a year ago

BTW, why are there these many quotes in the replace part? What do they mean? The colon and the single quotes?

Haresh Kola Over a year ago

actually in C# we have to replace " we use previously '''

Bhakti Kulkarni Over a year ago

@HareshKola it suddenly isn't working for summary. description is working fine.

Jon Skeet Over a year ago

This is a really bad idea. Use a JSON library instead to handle JSON. This won't escape anything in a or b.

Haresh Kola Over a year ago

use this string jSonContent = (@"{""fields"":{""summary"":""summary"" , ""description"": ""modified.""}}").Replace(@":""summary""", a).Replace("modified.", b);

|

Damith Asanka · Accepted Answer · 2017-03-30 11:10:19Z

-1

You can use string.Format or string.concate Method

string jSonContent = string.Concat("{\"fields\":{\"summary\":","\"",a,"\" , \"description\": \"",b,"\"}}");

edited Mar 30, 2017 at 11:10

answered Mar 30, 2017 at 11:04

Damith Asanka

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5 Comments

Jon Skeet Over a year ago

Those are bad ideas though, as they won't perform any escaping etc required for JSON. It's much better to use a dedicated JSON library.

Damith Asanka Over a year ago

Thanks for you comments, but here only two field so string Concat do same thing

Jon Skeet Over a year ago

And what if the contents of a contains a double quote, for example? Then you've got invalid JSON. The number of fields is irrelevant - this solution is the JSON equivalent of building a SQL string dynamically instead of using parameterized SQL. Just don't do it.

Damith Asanka Over a year ago

I assume a is server side variable

Jon Skeet Over a year ago

I have no idea what you think that has to do with it.

bmvr · Accepted Answer · 2017-03-30 11:52:19Z

-1

Use string fromat $""

String format allow you to replace values from your variables into a string. E.g.

string name = "Bruno";
string result = $"My name is {name}";
Console.WriteLine(result);

The output will be

My name is Bruno

In your case you just have to escape the json brackets with a double bracket {{

string a = "this is a sumary";
string b = "this is a description";
string jSonContent = $"{{\"fields\":{{\"summary\":\"{a}\" , \"description\": \"{b}\"}}}}";

Console.WriteLine(jSonContent);

the output is

{"fields":{"summary":"this is a sumary" , "description": "this is a description"}}

However I strongly advise to not build your json strings manually, use Json.NET or any other NuGet Packages.

edited Mar 30, 2017 at 11:52

answered Mar 30, 2017 at 11:47

bmvr

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3 Comments

Jon Skeet Over a year ago

No, don't build JSON like this. Use a JSON library instead. Consider what will happen if name contains backslashes, or quotes, or anything else that needs escaping.

bmvr Over a year ago

@JonSkeet I've just answered to what he asked for. Also I advised to use a JSON lib instead. He didn't specified for what purpose he needed this.

Jon Skeet Over a year ago

Would you do the same if someone asked you how to build up a SQL string with values? This is basically the equivalent. The OP appears to want a JSON document with two values which are specified using strings. Using string formatting is a very dangerous way of approaching this.

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