I am developing a jasper report in PHP I have tried embedded jasper library 0.9d -- PhpJasperLibrary-master After fixing part of the code in the library, It works normally but it cannot display table format which is very important for me to develop Jasper report so I give up in this method and build up an jasper server.
I have installed the PHP Client(v2.0) by composer and using the sample code in the following link
http://community.jaspersoft.com/wiki/php-client-sample-code http://community.jaspersoft.com/project/php-client/releases
I am not sure why there is a lot of bugs in the sample code like getjob(s) function and passing variable to class client.
Anyway, I fixed it and the server info can be called. In the last stage, I cannot run the report and the error message shown that the resource of my report not found.
Jaspersoft\Exception\RESTRequestException
Resource /20150604 not found.
I could like you to help me and I hope I can release a prefect version after all parts done.Thanks
And the below is my report information.
I have put my report under the folder /reports and it can run in the jasper server
I have checked the URL that calling the report is like below
localhost:9090/jasperserver/rest_v2/reports/20150604.html
I have concern about the URL but I cant find any reference to it. So please support me to fix the problem. Thanks
additional: here is the code I call the report
$report = $c->reportService()->runReport('/20150604', 'html');
Hmm the rest endpoint to "run" a report is js/rest_v2/?path to report?.html
Unless your report is in the root of your organization, then you need a path.
Check out this (line 20): https://github.com/jasonhuber/Jaspersoft_InputControlsFromCSV/blob/master/index.html#L20
and check in the web services guide page 71...