Send application name when connecting to an Oracle DB using HikariCP in Spring

Question

I have a Spring Boot 1.4.7 application that I am currently updating to version 2.0.5. The application connects to an Oracle DB using JDBC using the following configuration:

spring:
  jpa:
    database-platform: org.hibernate.dialect.Oracle12cDialect
  datasource:
    url: jdbc:oracle:thin:@<db_server>
    username: ${credentials.database.username}
    password: ${credentials.database.password}
    driver-class: oracle.jdbc.OracleDriver.class
    platform: oracle
    tomcat:
      connection-properties: v$session.program=${spring.application.name}

After updating the application to Spring Boot 2.0.5 the application name sent to the server is JDBC Thin Client instead of ${spring.application.name}. The reason for this seems to be the switch to HikariCP as the default connection pool in Spring 2.x. How would I migrate this configuration to Hikari in a way that allows me to send a custom property for v$session.program to the db?

What I have tried:

Appending ?ApplicationName=<name> to the JDBC url.
Solutions mentioned in this Stackoverflow question
Setting System.setProperty("oracle.jdbc.v$session.program", <name>)
Setting spring.datasource.hikari.data-source-properties.v$session.program: <name> in the application.yml

Thanks a lot ! This helped my a lot when I tried to set `defaultRowPrefetch` which makes Oracle ojdbc8.jar connections run much more efficiently ! — Marged, Oct 28 '18 at 10:08

score 4 · Answer 1 · answered Oct 25 '18 at 06:37

4

In yaml, the dollar sign is escaped.

spring.datasource.hikari.data-source-properties.v$session.program: <name>

com.zaxxer.hikari.HikariConfig : dataSourceProperties............{password=<masked>, vsession.program=<name>}

Try this.

spring:
  datasource:
    hikari:
      data-source-properties: v$session.program=name

answered Oct 25 '18 at 06:37

yeonguk kim

41
4

This way it works. Which means everything in yaml is escaped. – Pankaj Jan 18 '19 at 23:27
1

This does not appear to work. You mention "escaping", but don't show anything in your answer that is escaped. – trevorsky Apr 22 '19 at 21:01

score 2 · Answer 2 · answered Dec 29 '20 at 15:42

2

If you want to reference to spring.application.name you have to use it like this:

spring:
  datasource:
    hikari:
      data-source-properties:
        "[v$session.program]": ${spring.application.name}

answered Dec 29 '20 at 15:42

twonky

106
3

Karol Dowbecki · Answer 3 · 2018-10-15T09:21:28.917

1

Using HikariCP pool properties (default pool in Spring Boot 2)

spring:
  datasource:
    hikari:
      data-source-properties:
        v$session.program: MyAppName

edited Oct 15 '18 at 09:21

answered Oct 15 '18 at 09:16

Karol Dowbecki

43,645
9
78
111

6

This does not seem to have any effect. – Joba Oct 23 '18 at 06:20

Send application name when connecting to an Oracle DB using HikariCP in Spring

3 Answers3