c# - 如何在代码中写入 JSON 字符串值？

Question

我想将以下字符串存储在 String 变量中

{"Id":"123","DateOfRegistration":"2012-10-21T00:00:00+05:30","状态":0}

这是我使用的代码..

String str="{"Id":"123","DateOfRegistration":"2012-10-21T00:00:00+05:30","Status":0}";

..但它显示错误..

Answer 1

你必须这样做

String str="{\"Id\":\"123\",\"DateOfRegistration\":\"2012-10-21T00:00:00+05:30\",\"Status\":0}";

请参阅此内容以供参考
也来自 msdn :)

Short Notation  UTF-16 character    Description
\'  \u0027  allow to enter a ' in a character literal, e.g. '\''
\"  \u0022  allow to enter a " in a string literal, e.g. "this is the double quote (\") character"
\\  \u005c  allow to enter a \ character in a character or string literal, e.g. '\\' or "this is the backslash (\\) character"
\0  \u0000  allow to enter the character with code 0
\a  \u0007  alarm (usually the HW beep)
\b  \u0008  back-space
\f  \u000c  form-feed (next page)
\n  \u000a  line-feed (next line)
\r  \u000d  carriage-return (move to the beginning of the line)
\t  \u0009  (horizontal-) tab
\v  \u000b  vertical-tab

Answer 2

我更喜欢这个，只要确保字符串中没有单引号

 string str = "{'Id':'123','DateOfRegistration':'2012 - 10 - 21T00: 00:00 + 05:30','Status':0}"
              .Replace("'", "\"");

Answer 3

有另一种方法可以使用 Expando 对象或 XElement 编写这些复杂的 JSON，然后进行序列化。

https://blogs.msdn.microsoft.com/csharpfaq/2009/09/30/dynamic-in-c-4-0-introducing-the-expandoobject/

dynamic contact = new ExpandoObject
{    
    Name = "Patrick Hines",
    Phone = "206-555-0144",
    Address = new ExpandoObject
    {    
        Street = "123 Main St",
        City = "Mercer Island",
        State = "WA",    
        Postal = "68402"
    }
};

//Serialize to get Json string using NewtonSoft.JSON
string Json = JsonConvert.SerializeObject(contact);

Answer 4

微调sudhAnsu63 的答案，这是一个单行：

使用.NET 核心：

string str = JsonSerializer.Serialize(
  new {    
    Id = 2,
    DateOfRegistration = "2012-10-21T00:00:00+05:30",
    Status = 0
  }
);

使用Json.NET：

string str = JsonConvert.SerializeObject(
  new {    
    Id = 2,
    DateOfRegistration = "2012-10-21T00:00:00+05:30",
    Status = 0
  }
);

无需实例化dynamic ExpandoObject.

Answer 5

您必须像这样转义字符串中的引号：

String str="{\"Id\":\"123\",\"DateOfRegistration\":\"2012-10-21T00:00:00+05:30\",\"Status\":0}";

Answer 6

您需要像这样转义内部引号：

String str="{\"Id\":\"123\",\"DateOfRegistration\":\"2012-10-21T00:00:00+05:30\",\"Status\":0}";

Answer 7

使用 Verbatim String Literals ( @"...") 您可以通过用双引号对交换双引号来编写内联多行 json - "" 而不是 "。示例：

string str = @"
{
    ""Id"": ""123"",
    ""DateOfRegistration"": ""2012-10-21T00:00:00+05:30"",
    ""Status"": 0
}";

Answer 8

对于一个开箱即用的解决方案，我将 JSON 编码为 base64，以便可以将其作为字符串值导入一行。

这可以保留行格式，而无需手动编写动态对象或转义字符。格式与从文本文件中读取 JSON 相同：

var base64 = "eyJJZCI6IjEyMyIsIkRhdGVPZlJlZ2lzdHJhdGlvbiI6IjIwMTItMTAtMjFUMDA6MDA6MDArMDU6MzAiLCJTdGF0dXMiOjB9";
byte[] data = Convert.FromBase64String(base64);
string json = Encoding.UTF8.GetString(data);

//using the JSON text
var result = JsonConvert.DeserializeObject<object>(json);