C# 对每个JSON属性应用自定义格式

ChoJSONWriter或Newtonsoft是否支持对每个JSON属性应用自定义格式。从数据源中提取数据后，我希望对每个JSON记录应用以下格式

{
  "Place": "{0}",
  "SkuNumber": "SKU_{1}"
}

---

我可以通过查看每条记录并应用String.Format手动完成，但我不想重新发明轮子。

这就是使用Cinchoo ETL向每个成员添加自定义格式的方法

动态方法：

StringBuilder sb = new StringBuilder();

using (var w = new ChoJSONWriter(sb)
    .WithField("Place")
    .WithField("SkuNumber", valueConverter: (o) => String.Format("SKU_{0}", o.ToNString()))
    )
{
    dynamic o1 = new ExpandoObject();
    o1.Place = 1;
    o1.SkuNumber = 100;

    w.Write(o1);
}

Console.WriteLine(sb.ToString());

public class PlaceObj
{
    public string Place { get; set; }
    public int SkuNumber { get; set; }
}

StringBuilder sb = new StringBuilder();

using (var w = new ChoJSONWriter<PlaceObj>(sb)
    .WithField(m => m.SkuNumber, valueConverter: (o) => String.Format("SKU_{0}", o.ToNString()))
)
{
    PlaceObj o1 = new PlaceObj();
    o1.Place = "1";
    o1.SkuNumber = 100;

    w.Write(o1);
}

Console.WriteLine(sb.ToString());

POCO:

StringBuilder sb = new StringBuilder();

using (var w = new ChoJSONWriter(sb)
    .WithField("Place")
    .WithField("SkuNumber", valueConverter: (o) => String.Format("SKU_{0}", o.ToNString()))
    )
{
    dynamic o1 = new ExpandoObject();
    o1.Place = 1;
    o1.SkuNumber = 100;

    w.Write(o1);
}

Console.WriteLine(sb.ToString());

public class PlaceObj
{
    public string Place { get; set; }
    public int SkuNumber { get; set; }
}

StringBuilder sb = new StringBuilder();

using (var w = new ChoJSONWriter<PlaceObj>(sb)
    .WithField(m => m.SkuNumber, valueConverter: (o) => String.Format("SKU_{0}", o.ToNString()))
)
{
    PlaceObj o1 = new PlaceObj();
    o1.Place = "1";
    o1.SkuNumber = 100;

    w.Write(o1);
}

Console.WriteLine(sb.ToString());

POCO方法：

StringBuilder sb = new StringBuilder();

using (var w = new ChoJSONWriter(sb)
    .WithField("Place")
    .WithField("SkuNumber", valueConverter: (o) => String.Format("SKU_{0}", o.ToNString()))
    )
{
    dynamic o1 = new ExpandoObject();
    o1.Place = 1;
    o1.SkuNumber = 100;

    w.Write(o1);
}

Console.WriteLine(sb.ToString());

public class PlaceObj
{
    public string Place { get; set; }
    public int SkuNumber { get; set; }
}

StringBuilder sb = new StringBuilder();

using (var w = new ChoJSONWriter<PlaceObj>(sb)
    .WithField(m => m.SkuNumber, valueConverter: (o) => String.Format("SKU_{0}", o.ToNString()))
)
{
    PlaceObj o1 = new PlaceObj();
    o1.Place = "1";
    o1.SkuNumber = 100;

    w.Write(o1);
}

Console.WriteLine(sb.ToString());

StringBuilder sb=新建StringBuilder（）；
使用（var w=新的ChoJSONWriter（sb）
.WithField（m=>m.SkuNumber，valueConverter:（o）=>String.Format（“SKU{0}”，o.ToNString（））
)
{
PlaceObj o1=新的PlaceObj（）；
o1.Place=“1”；
o1.1编号=100；
w、 写入（o1）；
}
Console.WriteLine（sb.ToString（））；

您能告诉我们您正在使用什么方法（动态/POCO）来创建JSON文件吗？示例代码很有帮助。感谢您的回答，但解决方案需要通用，因为我不知道将返回哪些字段，并且字段列表仅在运行时可用，因为数据源是CSV文件。您可以发布包含输入和预期输出的示例代码吗？