将 JSON 个数组值合并到单个 CSV 列中

Merge JSON Array values into a single CSV column

我有一个 JSON 文件是这样的:

{
        "id": 2,
        "name": "I.1.A.2",
        "activeFlag": true,
        "recipients": [
            {
                "id": 3,
                "identityName": "idenity1",
                "fullName": "FullName1"
            },
            {
                "id": 4,
                "identityName": "identity2",
                "fullName": "FullName2"
            }
        ]
    }

我需要使用 C# 和 dotnet Core 将其转换为类似于此的 CSV 输出。

"id","name","activeFlag","identityName"
"2","I.1.A.2","true","identity1;identity2"

但是,我只能将 CSV 输出为:

"id","name","activeFlag","recipients_0", "recipients_1"
"2","I.1.A.2","true","identity1","identity2"

这是生成上述输出的代码:

    using (var csv = new ChoCSVWriter(".\temp\csvoutput.csv").WithFirstLineHeader()
    )
    {
        using (var json = new ChoJSONReader(".\temp\tmpjson.json")
        .Configure(c => c.ConvertToFlattenObject(arrayIndexSeparator: ';'))
        .Configure(c => c.ArrayValueSeparator = ';')
        .Configure(c => c.ArrayValueSeparator = ';')
        .WithField("id", jsonPath: "$..id", isArray: false)
        .WithField("recipients", jsonPath: "$..recipients[*]..identityName", isArray: true, fieldName: "recipients")
)
        {
            csv.Write(json);
        }
    }

现在,我正在使用 ChoEtl 库,但对其他 options/suggestions 开放。一直在寻找这个问题的答案,但还没有找到。对不起,如果我还没有找到一些解决方案。我确实在这里尝试了类似的解决方案: 但并没有完全满足我的需求。

就个人而言,我发现“投影”方法比基于配置的方法更易于使用和推理。参见 https://www.codeproject.com/Articles/1193650/Cinchoo-ETL-Quick-Start-Converting-JSON-to-CSV-Fil

这对我有用:

using (var csv = new ChoCSVWriter(output).WithFirstLineHeader()) {
    using (var json = new ChoJSONReader(input))
    {
        csv.Write(json.Select(jsonItem =>
        {
            var recipientList = new List<dynamic>(jsonItem.recipients);
            string recipientString = string.Join(';', recipientList.Select(r => r.identityName));
            return new
            {
                id = jsonItem.id,
                name = jsonItem.name,
                activeFlag = jsonItem.activeFlag,
                identityName = recipientString
            };
        }));
    }
}

可能有更优雅的方法来构造 identityName 的值,但由于值是动态的,因此很难直接使用 Linq 而不会 运行 出现 CS1977 错误。

但是,如果您更喜欢基于配置的方法,则可以将 valueConverterjsonPath 结合使用,如下所示:

using (var csv = new ChoCSVWriter(output).WithFirstLineHeader())
{
    using (var json = new ChoJSONReader(input)
        .WithField("id")
        .WithField("name")
        .WithField("activeFlag")
        .WithField("recipients", jsonPath: "$.recipients[*].identityName"
            , valueConverter: o => string.Join(';', ((object[])o).Select(x => x.ToString())))
    )
    {
        csv.Write(json);
    }
}

我找不到 ConvertToFlattenObject 的任何文档,所以我不确定它应该如何工作。

以上@Jack.A 的回答涵盖了产生预期输出的场景。

这是另一种方法,通过在 CSV 编写器中将 UseNestedKeyFormat 设置为 false 来生成预期的 CSV 输出

string json = @"
{
        ""id"": 2,
        ""name"": ""I.1.A.2"",
        ""activeFlag"": true,
        ""recipients"": [
            {
                ""id"": 3,
                ""identityName"": ""idenity1"",
                ""fullName"": ""FullName1""
            },
            {
                ""id"": 4,
                ""identityName"": ""identity2"",
                ""fullName"": ""FullName2""
            }
        ]
    }";


StringBuilder csv = new StringBuilder();

using (var r = ChoJSONReader.LoadText(json)
    .WithField("id")
    .WithField("name")
    .WithField("activeFlag")
    .WithField("recipients", jsonPath: "$..recipients[*]..identityName")
    )
{
    using (var w = new ChoCSVWriter(csv)
        .Configure(c => c.ArrayValueSeparator = ';')
        .WithFirstLineHeader()
        .QuoteAllFields()
        .UseNestedKeyFormat(false)
        )
    {
        w.Write(r);
    }
}
Console.WriteLine(csv.ToString());

输出:

id,name,activeFlag,recipients
"2","I.1.A.2","True","idenity1,identity2"

示例 fiddle:https://dotnetfiddle.net/O1fKQA