如何使用记录将JSON弦乐转换为二进制?Web.Contents Post Power查询

Question

web.contents方法将内容作为二进制

我使用此代码。它有效

query = "{
    ""field1"" : ""value1"",
    ""field2"" : ""value2"",
    ""field3"" : {
        ""sub_field_3_1"" : [""value_3_1_1"", ""value_3_1_2"", ""value_3_1_1""],
        ""sub_field_3_2"" : [""value_3_2_1"", ""value_3_2_2"", ""value_3_2_1""]
        }
    }",
content = Text.ToBinary(query),
Web.Contents("https://my_url", [
    Headers = [#"Content-Type"="text/xml; charset=utf-8"],
    Content=content
])

我明白，这不是一个很好的解决方法，因为没有理由进行双重转换。但是我找不到如何应用记录的方法，应该看起来像这样：

record = [
    field1 = value1,
    field2 = value2,
    field3 = [
        sub_field_3_1 = {value_3_1_1, value_3_1_2, value_3_1_1},
        sub_field_3_2 = {value_3_2_1, value_3_2_2, value_3_2_1}
    ]
],
content = SOME_CONVERTER(record),
Web.Contents("https://my_url", [
    Headers = [#"Content-Type"="text/xml; charset=utf-8"],
    Content = content
])

尝试使用 uri.buildquilderystring （如何使用Power Query的web.contents发布multipart/form-data），但它不能正确形成二进制

record = [
    field1 = value1,
    field2 = value2,
    field3 = [
        sub_field_3_1 = {value_3_1_1, value_3_1_2, value_3_1_1},
        sub_field_3_2 = {value_3_2_1, value_3_2_2, value_3_2_1}
    ]
],
content = Text.ToBinary(Uri.BuildQueryString(record)),
Web.Contents("https://my_url", [
    Headers = [#"Content-Type"="text/xml; charset=utf-8"],
    Content=content
]

有更好的解决方法吗？

Answer 1

目前，您的硬编码JSON字符串是更好的解决方案之一。

它不是理想的，但是您可以像toJson一样滚动自己的价值与json转换功能：

let
    record = [
        field1 = "value1",
        field2 = "value2",
        field3 = [
            sub_field_3_1 = {"value_3_1_1", null, 3.2},
            sub_field_3_2 = {"value_3_2_1", "value_3_2_2", "value_3_2_1"}
        ]
    ],
    toJson = (v as any) as text =>
      if v is null then "null" else 
      if v is logical or v is number then Text.From(v) else
      if v is text then """" & Text.Replace(Text.Replace(v, "", "\"), """", """") & """" else
      if v is list then "[" & Text.Combine(List.Transform(v, @toJson), ", ") & "]" else
      if v is record then "{" & 
        Text.Combine(List.Transform(
          Record.FieldNames(v),
          (n) => @toJson(n) & ": " & @toJson(Record.Field(v, n))), ", ")
        & "}" else
      error "not implemented",  
    jsonText = toJson(record)
in
    jsonText

与真实的Json.FromValue库功能相比，一些缺陷应该做什么：

• 只有原始文本逃脱
  • 有关您需要逃脱的所有特殊字符，请参见JSON.org
• 不处理环状M值，特殊非数字或其他类型的值类型
• 会在非常大的值（字符串concat都会使用很多模因）
上扼杀

Answer 2

some_converter == json.fromvalue

let
    record = [
        field1 = "value1",
        field2 = "value2",
        field3 = [
            sub_field_3_1 = {"value_3_1_1", "value_3_1_2", "value_3_1_1"},
            sub_field_3_2 = {"value_3_2_1", "value_3_2_2", "value_3_2_1"}
        ]
    ],
    content = Json.FromValue(record),
    web=Web.Contents("https://my_url", [
        Headers = [#"Content-Type"="text/xml; charset=utf-8"],
        Content = content
    ])
in
    web