Replacing escape characters from JSON

Question

I want to replace the "\" character from a JSON string by a empty space. How can I do that?

Answer 1

I have found that the easiest and best way to remove all escape characters from your JSON string, is to pass the string into Regex.Unescape() method. This method returns a new string with no ecapes, even \n \t etc, are removed.

See this MSDN article for more details: Regex.Unescape Method (String) (System.Text.RegularExpressions)

Answer 2

If the json object is a string, in .Net the escape "\" characters are added, should you want to clean up the json string, JObject.Parse({string}) as demonstrated in the following code snippet cleans up nicely:

var myJsonAsString = "{\"Name\": \"John\", \"LastName\": \"Doe\", \"Age\": 199 }";

var myCleanJsonObject = JObject.Parse(myJsonAsString);

Should give us a clean Json object with the escape characters removed.

{
"Name": "John",
"LastName": "Doe",
"Age": 199
}

Answer 3

This type of escaped strings are usually generated when you try to encode a string of JSON into JSON a second time. This causes something like "various levels of serialization"

Solution:

• First, you need to deserialize the escaped string, but not to the target CLR type, but deserialize to another string
• Repeat deserialization to string type again, if necessary.
• Then perform final deserialization to the target type:

Code:

// Initial example json string:  "\"{\\\"Property1\\\":1988,\\\"Property2\\\":\\\"Some data :D\\\"}\""    

// First, deserialize to another string (unescaped string).
string unescapedJsonString = JsonConvert.DeserializeObject<string>(escapedJsonString);
// Result: "{\"Property1\":1988,\"Property2\":\"Some data :D\"}"

// Second, deserialize to another string, again (in this case is necessary)
var finalUnescapedJsonString = JsonConvert.DeserializeObject<string>(unescapedJsonString);
// {"Property1":1988,"Property2":"Some data :D"}   This time prints a final, unescaped, json string:

// Finally, perform final deserialization to the target type, using the last unescaped string.
MyClass targetObject = JsonConvert.DeserializeObject<MyClass>(finalUnescapedJsonString);

Answer 4

Regex.Unescape() method will work just fine in most cases, but some particular cases require custom replacements. E.g. Regex.Unescape() will generate actual line breaks which are not supported in JSON.

Unescaped JSON:

{"comments_count":27,"comments":"<a name=\"comments\"><\/a>\n\n\t\n\t\n\t\t\n\t<div class=\"CommentsList\">\n\t\t<strong>COMENTARII<\/strong>\n\t\t"}

Regex.Unescape

{"comments_count":27,"comments":"<a name="comments"></a>

Replacements
    <div class="CommentsList">
        <strong>Comments</strong>
"}

Custom replacements

private string SanitizeReceivedJson(string uglyJson)
{
    var sb = new StringBuilder(uglyJson);
    sb.Replace("\\\t", "\t");
    sb.Replace("\\\n", "\n");
    sb.Replace("\\\r", "\r");
    return sb.ToString();
}

{"comments_count":27,"comments":"<a name=\"comments\"><\/a>\n\n\t\n\t\n\t\t\n\t<div class=\"CommentsList\">\n\t\t<strong>COMENTARII<\/strong>"}

Answer 5

C# string assignment does that for you, although if name or value contains \ it will be double escaped. The proper way is to use variable.Replace("\\\\","\\"); or variable.Replace(@"\\",@"\"); This will remove double escaped \ character, Leaving REQUIRED \ in value. for example if JSON contains "Domain\Username" this will be returned as \"Domain\\\\Username\" assigning that to string will result you will have Domain\\Username

Answer 6

deserialize to another string then to your class

string resultString = JsonConvert.DeserializeObject<string>(content);
var result = JsonConvert.DeserializeObject<T>(resultString);

This worked for me.

Answer 7

In c# you have only one way to create standard JSON result:

You must be add one class with your custom property name and then return Json(myClassForJsonResult) as bottom code:

public ActionResult testJsonData()
{
    var myClassForJsonResult=new YourClassOfYourCustomProperties();
    myClassForJsonResult.FirstPropertyStringType="first";
    myClassForJsonResult.SecondPropertyIntType=2;

    return Json(myClassForJsonResult);
}

class Defination:

    public class YourClassOfYourCustomProperties
    {
        public string FirstPropertyStringType{ get; set; }
        public int SecondPropertyIntType{ get; set; }
    }

Answer 8

You can use Regex.Unescape(yourEscapedString);

Answer 9

var json = System.Text.Json.JsonSerializer.Deserialize<JsonElement> (escapedJson)