JsonConvert.DeserializeObject特殊字符未终止字符串。预期分隔符：

问题描述

出于某种原因，当我在我的淘汰模型中有一个特殊字符并将其转换为json对象时，字符串将结束特殊字符所在的位置，并且在反序列化时出现错误：

For some reason, when I have a special character in my knockout model and convert it to a json object, the string ends where the special character is supposed to be and I get an error when deserializing it:

$.ajax({
    url: "/Admin/Forms/Convert",
    type: "post",
        //contentType: "application/json",
        dataType: "text",
        data: "modelData=" + ko.toJSON(theModel),
        success: function (data) {

            // window.open("/Admin/Forms/DisplayClient");
            var win = getFullWindow('/Admin/Forms/DisplayClient');
            win.open();
        },
        error: function (xhr, status, msg) { alert(msg); }
    });

当我使用这种方法时：

public void Convert(string modelData)
{
    Form form = JsonConvert.DeserializeObject<Form>(modelData);
}

我收到错误消息：

Unterminated string. Expected delimiter: ". Path 'Name', line 1, position 178.

推荐答案

如果JSON字符串包含特殊字符，如双引号，则反斜杠 \ 或斜杠 / ，需要使用反斜杠 \ 进行转义。没有JSON解析器能够处理首先没有正确格式化的JSON字符串。

If a JSON string contains special characters like double quotes ", backslashes \ or slashes /, they need to be escaped with backslashes \. There is no JSON parser that will be able to deal with a JSON string that isn't properly formatted in the first place.

所以你需要确保你的 JSON.org 标准，code> theModel 的格式正确。

So you need to make sure that your theModel is formatted appropriately and according to JSON.org standards.

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