databricks-david-lewis commented on code in PR #40947:
URL: https://github.com/apache/spark/pull/40947#discussion_r1177391819
##########
sql/core/src/main/scala/org/apache/spark/sql/execution/datasources/FileFormat.scala:
##########
@@ -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`
--
This is an automated message from the Apache Git Service.
To respond to the message, please log on to GitHub and use the
URL above to go to the specific comment.
To unsubscribe, e-mail: [email protected]
For queries about this service, please contact Infrastructure at:
[email protected]
---------------------------------------------------------------------
To unsubscribe, e-mail: [email protected]
For additional commands, e-mail: [email protected]
