schema

Create trigger on Oracle from a user that listens to events for another user

由 做~自己de王妃 提交于 2020-01-16 00:43:10
问题 I have two Oracle schemas A & B. Can I create a trigger in schema A that will listen to Create Tables on Schema B? If yes, which grants do I need to do? 回答1: Yes, you can. The user a needs to have CREATE ANY TRIGGER : As DBA: CREATE USER a IDENTIFIED BY a; GRANT CREATE SESSION, CREATE ANY TRIGGER TO a; CREATE USER b IDENTIFIED BY b; GRANT CREATE SESSION TO b; As a: CREATE OR REPLACE TRIGGER a.create_trigger BEFORE CREATE ON b.SCHEMA BEGIN raise_application_error(num => -20000, msg=> 'create

SpringMVC中RestFul风格

由 落花浮王杯 提交于 2020-01-15 20:09:39
先说一下什么是RestFul风格，以一个链接为例子，如果我们访问一个网页，想要给a和b传参数，传统的方式是?a=1&b=2，而RestFul就是改变了传统的方式，用/a/1/2的形式，达到了简洁、安全、高效（支持缓存）。 这里我们以一个简单的SpringMVC例子来进行演示，首先我们写一个 Controller控制器（核心） package com.zhiying.controller4; import org.springframework.stereotype.Controller; import org.springframework.ui.Model; import org.springframework.web.bind.annotation.PathVariable; import org.springframework.web.bind.annotation.RequestMapping; @Controller public class ControllerTest4 { @RequestMapping("/add/{a}/{b}") public String test1(@PathVariable int a, @PathVariable int b, Model model) { int result = a + b; model.addAttribute(

Selecting Nodes in XSD schema using Xpath

由 倾然丶 夕夏残阳落幕 提交于 2020-01-15 18:52:07
问题 I have the following code that I wish to use to select all the elements I will need in a certain sequence. Here's the snippet: XmlDocument schema = new XmlDocument(); schema.Load(SchemaFileName); XmlNamespaceManager xnm = new XmlNamespaceManager(schema.NameTable); xnm.AddNamespace("xs", "http://www.w3.org/2001/XMLSchema"); XmlNodeList list = schema.SelectNodes(Path); However, I'm not sure what I should write as the path. Ideally I want to select all the child nodes of the "sequence" tag, but

Selecting Nodes in XSD schema using Xpath

由 自古美人都是妖i 提交于 2020-01-15 18:47:31
问题 I have the following code that I wish to use to select all the elements I will need in a certain sequence. Here's the snippet: XmlDocument schema = new XmlDocument(); schema.Load(SchemaFileName); XmlNamespaceManager xnm = new XmlNamespaceManager(schema.NameTable); xnm.AddNamespace("xs", "http://www.w3.org/2001/XMLSchema"); XmlNodeList list = schema.SelectNodes(Path); However, I'm not sure what I should write as the path. Ideally I want to select all the child nodes of the "sequence" tag, but

Selecting Nodes in XSD schema using Xpath

由 柔情痞子 提交于 2020-01-15 18:47:03
问题 I have the following code that I wish to use to select all the elements I will need in a certain sequence. Here's the snippet: XmlDocument schema = new XmlDocument(); schema.Load(SchemaFileName); XmlNamespaceManager xnm = new XmlNamespaceManager(schema.NameTable); xnm.AddNamespace("xs", "http://www.w3.org/2001/XMLSchema"); XmlNodeList list = schema.SelectNodes(Path); However, I'm not sure what I should write as the path. Ideally I want to select all the child nodes of the "sequence" tag, but

使用注解开发

由 拜拜、爱过 提交于 2020-01-15 17:47:40
使用注解开发 使用注解所需要的包: org.springframework:spring-aop:5.2.2.RELEASE 添加注解的支持 <?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:context="http://www.springframework.org/schema/context" xsi:schemaLocation="http://www.springframework.org/schema/beans https://www.springframework.org/schema/beans/spring-beans.xsd http://www.springframework.org/schema/context https://www.springframework.org/schema/context/spring-context.xsd"> <context:annotation-config/> </beans> bean //放在类上,等价于注册了<bean id="user"

Using Java schemacrawler, why is it scanning every table in my database?

由 て烟熏妆下的殇ゞ 提交于 2020-01-15 15:23:19
问题 Using Java schemacrawler, why is it scanning every table in my database? Shouldn't it just be scanning the database I specified on the command line: -database=openfire ??? :: schemacrawler batch launcher @echo off C:\JDK\bin\java.exe -classpath jtds-1.2.4.jar;schemacrawler-8.8.jar; \ schemacrawler-sqlserver-8.8.jar schemacrawler.tools.sqlserver.Main \ -user=sa -password=password -database=openfire -port=1433 -host=localhost \ -table_types=TABLE -command=schema -schemas=.*\.dbo.* -infolevel

Using Java schemacrawler, why is it scanning every table in my database?

由 廉价感情. 提交于 2020-01-15 15:22:53
问题 Using Java schemacrawler, why is it scanning every table in my database? Shouldn't it just be scanning the database I specified on the command line: -database=openfire ??? :: schemacrawler batch launcher @echo off C:\JDK\bin\java.exe -classpath jtds-1.2.4.jar;schemacrawler-8.8.jar; \ schemacrawler-sqlserver-8.8.jar schemacrawler.tools.sqlserver.Main \ -user=sa -password=password -database=openfire -port=1433 -host=localhost \ -table_types=TABLE -command=schema -schemas=.*\.dbo.* -infolevel

Spring+SpringMVC+MyBatis整合Redis

由 折月煮酒 提交于 2020-01-15 11:44:09
Spring+SpringMVC+MyBatis整合 项目目录结构如下 代码如下： pom.xml <?xml version="1.0" encoding="UTF-8"?> <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd"> <modelVersion>4.0.0</modelVersion> <groupId>top.xz.hand</groupId> <artifactId>ssmdemo</artifactId> <version>1.0-SNAPSHOT</version> <packaging>war</packaging> <properties> <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding> <maven.compiler.source>1.8</maven.compiler.source> <maven.compiler

jaxbcontext generating incomplete schemas?

由 这一生的挚爱 提交于 2020-01-15 08:52:06
问题 I've run into a strange problem with JAXB. I've used xjc to generate my java classes from my XSD and all looks good. If I use schemagen, it produces a proper schema that matches my original xsd. However, if I use JAXBContext.generateSchema(), then the generated schema is incomplete. I'm using Oracle Java 1.6.0_29 and jaxb-2.2.4-1.jar as the implementation. I'm enclosing the java code (which generates the schema), and the xsd below as well as the output of the jaxb call.
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