C# Json：如何使用 json.net 正确去除转义字符

Json: how to properly strip the escape characters with json.net

提问by Sharon C Lawrence

I have json response in the below format.

我有以下格式的 json 响应。

"[{\\"JobID\\":\\"1\\",\\"BillGenerationDate\\":\\"4/29/2013 2:53:34 PM\\",\\"BillID\\":\\"115743\\",\\"BillNo\\":\\"115743\\",\\"CustomerID\\":\\"4041705\\",\\"PayStatus\\":\\"0\\",\\"PaymentRequiredStatus\\":\\"True\\",\\"ProductName\\":\\"Epic FBO test\\",\\"Description\\":\\"Epic Automation 2\\r\\n\\",\\"ProductType\\":\\"eBill \\",\\"DueType\\":\\"-1\\",\\"DueDate\\":\\"2013-03-15\\",\\"Amount\\":\\"63.70\\",\\"Cost\\":\\"\\"},
{\\"JobID\\":\\"9\\",\\"BillGenerationDate\\":\\"5/2/2013 10:21:39 AM\\",\\"BillID\\":\\"115743\\",\\"BillNo\\":\\"115743\\",\\"CustomerID\\":\\"4041705\\",\\"PayStatus\\":\\"0\\",\\"PaymentRequiredStatus\\":\\"True\\",\\"ProductName\\":\\"FBO Test Product\\",\\"Description\\":\\"FBO Product Test\\",\\"ProductType\\":\\"eBill \\",\\"DueType\\":\\"-1\\",\\"DueDate\\":\\"2013-05-01\\",\\"Amount\\":\\"150.70\\",\\"Cost\\":\\"\\"}]

I believe json.net handles the escape characters and I used the below code to deserialize it to a dictionary collection.

我相信 json.net 处理转义字符，我使用以下代码将其反序列化为字典集合。

var billList = JsonConvert.DeserializeObject<List<Dictionary<string, string>>>(contentCorrected);

But this json parsing throws exception "Invalid property identifier character: . Path '[0]', line 1, position 2." Could we solve this by manipulating the json response string?

但是这个json解析抛出异常“无效的属性标识符字符：路径'[0]'，第1行，位置2。” 我们可以通过操作 json 响应字符串来解决这个问题吗？

采纳答案by Bijoy K Jose

Try string contentCorrected = contentCorrected.Replace(@"\", "");before deserialization process.

string contentCorrected = contentCorrected.Replace(@"\", "");在反序列化过程之前尝试。

回答by Murugan

1. Remove all the "\" character before you deserialize it. Use replace function.

   yourJsonString.Replace("\\\\\", "");

2. Your Json string is incomplete or doesnot seems to be of type List<Dictionary<string, string>>". Correct the type you want the json to be converted. I modified your json a little as follows and it worked.

   newJson = "{ \"array\":" + yourJsonString + "}"

1. 在反序列化之前删除所有“\”字符。使用替换功能。

   yourJsonString.Replace("\\\\", "");

2. 您的 Json 字符串不完整或似乎不是 类型List<Dictionary<string, string>>"。更正您希望转换 json 的类型。我对您的 json 进行了如下修改，它起作用了。

   newJson = "{ \"array\":" + yourJsonString + "}"

回答by Prabhat

For me the code below works

对我来说，下面的代码有效

string contentCorrected = contentCorrected.Replace(**@"\""", ""**);

回答by Jos R.

The problem occurs when valid double quotes are used within the answer. Removing and/or Replacing won't solved this in all cases. It frustrated me too until I found a simple solution:

在答案中使用有效的双引号时会出现问题。在所有情况下，删除和/或替换不会解决这个问题。它也让我感到沮丧，直到我找到了一个简单的解决方案：

var billList = JsonConvert.DeserializeObject<List<Dictionary<string, string>>>(@contentCorrected);

回答by Johan Alzate

THE SHORT ANSWER:first you need to deserialize the escaped string, but not to the target CLR type, but deserialize to another string:

简短回答：首先您需要反序列化转义字符串，但不是反序列化到目标 CLR 类型，而是反序列化到另一个字符串：

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


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


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


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

LONG ANSWER (but interesting)Using string.Replace(...could generate an invalid string, because it could damage certain special characters that needed the backslash to be deserialized correctly .

长答案（但很有趣）Usingstring.Replace(...可能会生成无效字符串，因为它可能会损坏某些需要正确反序列化反斜杠的特殊字符。

This type of escaped stringsare usually generated when a string that was already a json string, its serialized again (or even more times). This causes something like "various levels of serialization" (it really is a serialization of a string with reserved characters), and the result is backshash characters (or groups of one, two or more backslash followed: \, \\, \\\) scattered all over the string. So, to remove them correctly is not enough to replace them by empty.

这种类型的转义字符串通常是在一个已经是 json 字符串的字符串再次序列化（甚至更多次）时生成的。这会导致类似“各种级别的序列化”（它实际上是带有保留字符的字符串的序列化），结果是反斜杠字符（或一组、两个或多个反斜杠后跟：\、\\、\\\ ) 散落在整个弦上。因此，正确删除它们不足以用空替换它们。

THE RIGHT WAY:A better way to get a unescaped stringwould be to do a first deserialization to string type (Repeat this several times if necessary), And then do a final deserialization to target CLR type:

正确的方法：获得未转义字符串的更好方法是对字符串类型进行第一次反序列化（如有必要，重复几次），然后对目标 CLR 类型进行最终反序列化：

// -- SERIALIZATION --

// Initial object
MyClass originObj = new MyClass { Property1 = 1988, Property2 = "Some data :D" };

// "First level" Of serialization.
string jsonString = JsonConvert.SerializeObject(originObj);
Debug.WriteLine(jsonString);
// Prints: 
// {"Property1":1988,"Property2":"Some data :D"}


// "Second level" of serialization.
string escapedJsonString = JsonConvert.SerializeObject(jsonString);
Debug.WriteLine(escapedJsonString);            
// "{\"Property1\":1988,\"Property2\":\"Some data :D\"}"
// Note the initial and final " character and de backslash characters

// ...
// at this point you could do more serializations ("More levels"), Obtaining as a result more and more backslash followed,
// something like this:
// "\"{\\"Property1\\":1988,\\"Property2\\":\\"Some data :D\\"}\""
// Note that is... very very crazy :D
// ...

// -- DESERIALIZATION --

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

// ...
// at this point you could repeat more deserializations to string, if necessary. For example if you have many backslash \\
// ...

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