jasperreports cant locate subreport

jasperreports cant locate subreport  using -'java,spring,jasper-reports,ireport'

It's drives me crazy.. Last hour I am trying to figure out why my report stopped to worked right after I added a subreport to it..
I already checked and tried all similar issues here, here and here.
Without success..  Subreport is in the same directory for sure..
The definition looks like:

<subreport isUsingCache="false">
                <reportElement uuid="db816b3c-a13d-440f-a6a2-f899762e61e4" x="0" y="89" width="555" height="100"/>
                <subreportParameter name="footerAddress">
                <subreportParameter name="footerManager">
                <subreportParameter name="footerContact">
                <subreportParameter name="footerBank">
                <subreportExpression class="net.sf.jasperreports.engine.JasperReport"><![CDATA[$P{SUBREPORT_DIR}]]></subreportExpression>

Where $P{SUBREPORT_DIR} is "/valid/path/sub.report.jasper"

And I still getting an error: Resource not found at /valid/path/sub.report.jasper


asked Sep 14, 2015 by RayY36myreyj
0 votes

3 Answers

0 votes

In first message author says:
Where $P{SUBREPORT_DIR} is "/valid/path/sub.report.jasper"

$P{SUBREPORT_DIR} is a directory, but not filename. It is first error.

The second: sometimes jasper can't find subreport, even path is correct.

The way for solving this problem:

  1. Pass $P{SUBREPORT_DIR} into main report (In this case: "/valid/path/"), put this parameter into HashMap - standard way pass parameters into jasper
  2. Pass $P{SUBREPORT_DIR} from main report into subreport as parameter: $P{SUBREPORT_DIR} main report -> $P{SUBREPORT_DIR} of subreport (this way used when exist subsubreport, which calling from subreport)
  3. In main report set expression to subreport as $P{SUBREPORT_DIR} + "sub.report.jasper"

Good luck

answered Sep 14, 2015 by GKELeandrokl
0 votes

I had this issue earlier. My approach was - i was sending the absolute folder in a parameter say folder_path .And then in the sub-report expression i was using
new File(($P{folder_path} + "*.jasper" ) .
this *.jasper file can be replace by "\\inner_folder\\*.jasper". And this worked perfectly

answered Sep 14, 2015 by JackieWeathe
0 votes

This happens to me from time-to-time, and its always been a platform OS that is the key deciding factor of why some code works, and others doesnt. However what has always worked was relative references instead of absolute references.

My suggestion would be if cjava's solution doesnt work, is to find out what JasperReports is doing on Disk. Depending on your platform, *nix, Windows, You may need to figure out what system calls JasperReports is running to find the path it is using. In my application I found that even though I was specifying a path-less reference that worked before, some platforms would path search differently and fail to find the report.

Some other questions on how to troubleshoot System calls:

*nix - Strace

strace java applet

Windows - Process Monitor

On Windows, how can I find what files a given process is using? Is there software that does this?

Mac - DTruss

ltrace equivalent for osx?

Good Luck.

answered Sep 14, 2015 by SERSus