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seanccc17-Peridot
17-Peridot
March 28, 2019
Solved

ThingWorx on Postgresql cannot launch:Improper value found for model provider configuration password

  • March 28, 2019
  • 1 reply
  • 1726 views

Hi,

I'm installing ThingWorx8.4 on Postgresql, but got error "Improper value found for model provider configuration password, system cannot start without valid password"  when launch tomcat.  I've executed 

 thingworxPostgresDBSetup.bat and thingworxPostgresSchemaSetup.bat without errors , so I'm sure that the password in the platform-settings.json is correct for the postgresql user "twadmin".   

 

If switch to H2 database , then tomcat can launch successfully and Thingworx application works well. 

 

Any suggestions ?   The tomcat error log and platform-settings for H2 and PostgreSQL are attached. 

==================Environment Information===============

ThingWorx version: 8.4

OS: Windows server 2012 R2

Tomcat: 9.0 

Database: PostgreSql 10

JDK: Oracle 1.8.0_192/64bit

 

 

Best answer by posipova

Can you try with just this in platform-settings for postgres?

{
 "PlatformSettingsConfig": {
 "AdministratorUserSettings": {
 "InitialPassword": "Adminxxxxxx"
 },
		"LicensingConnectionSettings": {
			"username": "******",
			"password": "******",
			"timeout":"60"
		}
 },
 "PersistenceProviderPackageConfigs": {
"PostgresPersistenceProviderPackage": {
"ConnectionInformation": {
"jdbcUrl": "jdbc:postgresql://localhost:5432/thingworx",
 "password": "Adminxxxxxx",
 "username": "twadmin"
 }
 }
 }
}

  

1 reply

P
posipova20-TurquoiseAnswer
20-Turquoise
March 28, 2019

Can you try with just this in platform-settings for postgres?

{
 "PlatformSettingsConfig": {
 "AdministratorUserSettings": {
 "InitialPassword": "Adminxxxxxx"
 },
		"LicensingConnectionSettings": {
			"username": "******",
			"password": "******",
			"timeout":"60"
		}
 },
 "PersistenceProviderPackageConfigs": {
"PostgresPersistenceProviderPackage": {
"ConnectionInformation": {
"jdbcUrl": "jdbc:postgresql://localhost:5432/thingworx",
 "password": "Adminxxxxxx",
 "username": "twadmin"
 }
 }
 }
}

  

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