jasperserver

HTTP Status code = 400

我是研究僧i 提交于 2020-01-06 07:44:58
问题 I am currently doing a Rest call with the jasper rest api and I have this response : trailers: {}, readable: false, url: '', method: null, statusCode: 400, client: So this is a bad request Bad request 400 The request had bad syntax or was inherently impossible to be satisfied. And here is my nodejs request with request module : var request = require("request"); request.post({url: "http://localhost:8080/jasperserver/rest/login", qs: {j_username:"jasperadmin", j_password: "jasperadmin"}},

How to make query based on logged in user

好久不见. 提交于 2020-01-05 07:24:57
问题 I am new to JasperReports Server . I am having different campaigns in my DB. And users assigned to these campaigns. I want my JR Server user to provide campaign IDs to run a report. For this, I have create a query and an input control to show the list of campaigns in a list to take input from user. The query I created shows all the campaigns in the DB. I want to show only those campaigns in which the logged in user is assigned. Is there any way to do this in JasperReports Server ? 回答1: It

Interactive features of JasperReports without involving JasperReports Server

家住魔仙堡 提交于 2020-01-05 06:02:38
问题 I have a few reports that I developed, compiling jrxmls using Java, filling reports and displaying report output in IFrame (Jsp). Even though I used table components I can't see any interactive features in report, Question I have: Are there any other alternatives that I can acheive interactive features like Sorting, Filtering etc.., without using JasperReports Server? 回答1: The same question was asked on the jaspersoft community site. Duplicating my answer here: There is a sample - webapp-repo

Retrieve resource from jasperserver repository using server java api

拜拜、爱过 提交于 2020-01-01 17:27:51
问题 I'm trying to retrieve resource from Jasperserver repository using its java API, according to jasper report server ultimate guide, I should get an instance of the ExecutionContext interface: ExecutionContext context = JasperServerUtil.getExecutionContext(); then, get an instance of the RepositoryService interface: RepositoryService repositoryService = ...; //how?? Now I can get the file using the following code: FileResourceData fileResourceData = repositoryService.getContentResourceData

JasperReport Server REST 2 api - Creating a new report

余生长醉 提交于 2019-12-25 06:13:13
问题 I was not able to find any api that can actually create a report on the JasperReports Server. There is one to create resource. http://community.jaspersoft.com/documentation/jasperreports-server-web-services-guide/v550/creating-resource To an extend it tells about how to create a folder etc. but talks nothing about creating a report in all. Any idea on how a report (jrxml) can be done in a programmatic way? Thanks. 回答1: To post your reports manually to the JasperServer repository, use the same

MDX query from PHP

丶灬走出姿态 提交于 2019-12-24 09:27:43
问题 Is it possible to create an MDX query for JasperServer using PHP as a host language, given that my underlying relational database is MySQL? What I would like to do is send the query to JasperServer in order to create an OLAP View and receive the view as a result (preferably in some text-based format, as oposed to an image or PDF). I'm using JasperServer CE 4.2.1. As a side note, is it even correct to think about making MDX queries from you application, just like you would make SQL queries? If

How is the JasperServer REST client path?

Deadly 提交于 2019-12-24 06:31:24
问题 I'm working to make client rest service with jasperserver to generate reports. I'm using the following code to make that: I have problem in setting server url and report path, for server url I put http://localhost:8081/jasperserver/ and as shown in next image I put report path rest/report/mytest/my_report but I get 404 not found error in line File remoteFile = resource.get(File.class); So how can I get the proper report path from jasperserver? public class App2 { private final static String

JasperServer proxy CSRF error

元气小坏坏 提交于 2019-12-23 10:56:15
问题 I have a new installation of JasperReports Server 6.2 using the bundled Tomcat on Ubuntu 14.04 LTS with an Nginx proxy so I can access https://mydomain.xyz/jasperserver. It mostly works, but I'm not able to manage users and roles. If I bypass Nginx and go straight to Tomcat http://123.123.123.123:8080/jasperserver, everything works perfectly. The log indicates: 2016-02-28 19:44:08,024 ERROR CsrfGuard,http-nio-8080-exec-3:44 - potential cross-site request forgery (CSRF) attack thwarted (user:,

Encrypting Parameters in URLs

寵の児 提交于 2019-12-23 02:53:22
问题 I have a URL that looks like this: http://localhost:8001/jasperserver&reportUnit=somereport&username=cghan& password=somepasswrd As you can see, the password is in clear site, along with some other parameters. What's the best way to secure this? 回答1: Set the method of the FORM (in HTML) to POST. That way attributes won't be shown in the URL (they can be retrieved exactly the same way). Anyway, for anyone with a sniffer, it will make little difference. If it is really important, force the

Encrypting Parameters in URLs

爷,独闯天下 提交于 2019-12-23 02:53:07
问题 I have a URL that looks like this: http://localhost:8001/jasperserver&reportUnit=somereport&username=cghan& password=somepasswrd As you can see, the password is in clear site, along with some other parameters. What's the best way to secure this? 回答1: Set the method of the FORM (in HTML) to POST. That way attributes won't be shown in the URL (they can be retrieved exactly the same way). Anyway, for anyone with a sniffer, it will make little difference. If it is really important, force the