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Support Article

Archiving generated Zips are not readable



File listener unable to process the zip file through PRFile handle, which is created in the activity.

Error Messages

2017-09-12 17:17:22,433 [ WebContainer : 4] ( internal.archive.PRJarFile) ERROR Unable to create PRJarFile error in opening zip file
at Method)
at java.util.jar.JarFile.<init>(
at java.util.jar.JarFile.<init>(
at com.pega.pegarules.deploy.internal.archive.PRJarFile.<init>(
at com.pega.pegarules.deploy.internal.archive.ParUtilsImpl.getArchiveType(
at com.pega.pegarules.deploy.internal.PRRuleMgmtUtilsImpl.readArchive(
at com.pega.pegarules.deploy.internal.PRRuleMgmtUtilsImpl.readArchive(
at com.pega.pegarules.deploy.internal.PRRuleMgmtUtilsImpl.createArchiveForRead(
at com.pega.pegarules.deploy.internal.PRRuleMgmtUtilsImpl.createArchiveForRead(

Steps to Reproduce

1.Turn on the File listener to process the zip file.
2.Create PRFile handle in activity to access the file.

Root Cause

User created the following PRFile handle:

ClipboardPage logServiceFilePage = tools.findPage("LogServiceFile");
String renamedFileName= logServiceFilePage .getString("pyWorkFileName");
PRFile prfile = new PRFile(<FileListener_defaultlocation> +RenamedFileName);

After picking file, file service renames it and then starts processing.

Once this processing is done it copies that file to completed folder inside work_<fileListener>.

File is present in work_<listenername> but as this file was renamed, PRFile handle is not able to find it with the original name.


As file listener renames and moves the files to work_<listenername> folder, a new name must be provided to PRFile handle.

For example file listener name is test, then PRFile handle must be created like this below:

PRFile prfile = new PRFile(file://default:/work_test+renamedFileName);
Suggest Edit

Published March 14, 2018 - Updated October 8, 2020

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