F#数据类型提供程序-使用字符串变量创建 [英] F# Data Type Provider - Create with string variable

问题描述

在F#中，我具有以下代码

In F# I have the following bit of code

[<Literal>]
let fname=fstFile.FullName.ToString()
type genomeFile = CsvProvider<fname>

其中fstFile是FileInfo类的实例.但是，此操作失败:

Where fstFile is an instance of the FileInfo class. However, this fails with:

This is not a valid constant expression or custom attribute value

如果没有实际的硬编码字符串，我似乎无法创建类型.有人知道如何解决这个问题吗?

And I can't seem to create the type without an actual hard coded string. Does anyone know how to fix this?

推荐答案

CSV类型提供程序的参数必须为常量，以便可以在编译时生成类型(而无需实际评估程序.)，您可以在运行时加载具有相同架构的其他文件.

The parameter to the CSV type provider needs to be a constant, so that the types can be generated at compile-time (without actually evaluating the program. However, you can load a different file with the same schema at runtime.

因此，处理此问题的最佳方法是将实际输入之一复制到某个知名位置(例如，与脚本位于同一目录中的sample.csv)，然后在运行时使用实际路径:

So, the best way to handle this is to copy one of your actual inputs to some well-known location (e.g. sample.csv in the same directory as your script) and then use the actual path at runtime:

// Generate type based on a statically known sample file
type GenomeFile = CsvProvider<"sample.csv">

// Load the data from the actual input at runtime
let actualData = GenomeFile.Load(fstFile.FullName.ToString())

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