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java - JHipster：使用附加信息注册用户[帮助]

问题描述

我一直在按照 本指南尝试向用户添加信息，一切都很顺利，直到我到达将输入字段添加到 register.component.html 的部分，当我添加 ngModel 指令时页面停止工作，密码强度不再有效，当我尝试注册用户时，我注册失败并出现以下错误：

2019-12-10 16:31:18.875 ERROR 912 --- [ XNIO-1 task-14] o.h.engine.jdbc.spi.SqlExceptionHelper   : Column "USEREXTRA0_.USER_ID" not found; SQL statement`select userextra0_.user_id as user_id8_25_, userextra0_.activo as activo1_25_, userextra0_.cod_p as cod_p2_25_, userextra0_.morada as morada3_25_, userextra0_.nif as nif4_25_, userextra0_.nome as nome5_25_, userextra0_.permiss_change as permiss_6_25_, userextra0_.telef as telef7_25_, userextra0_.user_permissions_id as user_per9_25_, userextra0_.user_profile_id as user_pr10_25_ from user_extra userextra0_ [42122-200]`
2019-12-10 16:31:18.876 ERROR 912 --- [ XNIO-1 task-14] dw.aop.logging.LoggingAspect             : Exception in dw.web.rest.UserExtraResource.getAllUserExtras() with cause = 'org.hibernate.exception.SQLGrammarException: could not prepare statement' and exception = 'could not prepare statement; SQL [select userextra0_.user_id as user_id8_25_, userextra0_.activo as activo1_25_, userextra0_.cod_p as cod_p2_25_, userextra0_.morada as morada3_25_, userextra0_.nif as nif4_25_, userextra0_.nome as nome5_25_, userextra0_.permiss_change as permiss_6_25_, userextra0_.telef as telef7_25_, userextra0_.user_permissions_id as user_per9_25_, userextra0_.user_profile_id as user_pr10_25_ from user_extra userextra0_]; nested exception is org.hibernate.exception.SQLGrammarException: could not prepare statement'
2019-12-10 16:31:18.878 ERROR 912 --- [ XNIO-1 task-14] o.z.problem.spring.common.AdviceTraits   : Internal Server Error

当我尝试查看我的 UserExtra 实体时，我得到一个内部服务器错误。

这是我的 UserExtra 表，我意识到 ID 字段应该称为 User_Id 但我按照指南做了所有事情，不知道出了什么问题

标签： javaangularspringjhipster

解决方案

我建议您检查您的 Liquibase 脚本。很可能该字段未在用户额外表定义中定义。

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