databricks-david-lewis commented on code in PR #40947:
URL: https://github.com/apache/spark/pull/40947#discussion_r1177391819
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sql/core/src/main/scala/org/apache/spark/sql/execution/datasources/FileFormat.scala:
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@@ -265,8 +265,8 @@ object FileFormat {
* fields of the [[PartitionedFile]], and do have entries in the file's
metadata map.
*/
val BASE_METADATA_EXTRACTORS: Map[String, PartitionedFile => Any] = Map(
- FILE_PATH -> { pf: PartitionedFile => pf.toPath.toString },
- FILE_NAME -> { pf: PartitionedFile => pf.toPath.getName },
+ FILE_PATH -> { pf: PartitionedFile => pf.filePath.urlEncoded },
+ FILE_NAME -> { pf: PartitionedFile => new
Path(pf.filePath.urlEncoded).getName },
Review Comment:
Just using `Path.getName` returns the unencoded string (i.e. `some
file.txt`).
The new way will return the encoded path name: `some%20file.txt`.
It's not exactly right with URI parameters though... I just updated it to
`new Path(pf.filePath.toUri.getRawPath).getName`
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