grant我需要给当前用户(" userA ")以允许他更改由另一个用户(" userC ")所属的对象所有者的选项/技巧是什么?
更确切地说,联系表由userC拥有,当我执行以下查询以将所有者更改为userB时,与userA连接:
alter table contact owner to userB;
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我收到此错误:
ERROR: must be owner of relation contact
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但是userA通常都需要权限(" 在模式上创建 "授权选项应该足够了):
grant select,insert,update,delete on all tables in schema public to userA;
grant select,usage,update on all sequences in schema public to userA;
grant execute on all functions in schema public to userA;
grant references, trigger on all tables in schema public to userA;
grant create on schema public …Run Code Online (Sandbox Code Playgroud) 我想从命令行中检索groupId,artifactId和 Maven项目的版本.
本主题中提出的解决方案" 如何将Maven项目版本添加到bash命令行 "是使用以下插件:
mvn org.apache.maven.plugins:maven-help-plugin:2.2:evaluate -Dexpression=project.artifactId
它的工作原理很好,但我想不出如何设置,在同一时间内,project.groupId, project.artifactId & project.version到-Dexpression说法.
我会避免每次使用不同的-Dexpression参数启动3次Maven命令...
THKS
local pom_groupid=`mvn org.apache.maven.plugins:maven-help-plugin:2.2:evaluate -Dexpression=project.groupId |grep -Ev '(^\[|Download\w+:)'`
local pom_artifactid=`mvn org.apache.maven.plugins:maven-help-plugin:2.2:evaluate -Dexpression=project.artifactId |grep -Ev '(^\[|Download\w+:)'`
local pom_version=`mvn org.apache.maven.plugins:maven-help-plugin:2.2:evaluate -Dexpression=project.version |grep -Ev '(^\[|Download\w+:)'`
Run Code Online (Sandbox Code Playgroud) 我已经定义了$resource一个API端点,它返回一个包含多个响应的响应,headers但在transformResponse配置函数中,headersGetter函数参数中缺少大多数标头.
我该如何解决?
API的响应标头
HTTP/1.1 201 Created
Server: Apache-Coyote/1.1
x-content-type-options: nosniff
x-xss-protection: 1; mode=block
Cache-Control: no-cache, no-store, max-age=0, must-revalidate
Pragma: no-cache
Expires: 0
x-frame-options: DENY
Access-Control-Allow-Origin: http://localhost:9000
access-control-allow-methods: POST, PUT, GET, DELETE, OPTIONS
access-control-allow-headers: Content-Type
Access-Control-Allow-Credentials: true
Content-Disposition: attachment; filename="testCall.pcap"
FileName: testCall.pcap
Content-Type: application/pcap
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transformResponse的标题
Pragma: no-cache
Content-Type: application/pcap
Cache-Control: no-cache, no-store, max-age=0, must-revalidate
Expires: 0
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app.factory("MyService", function ($resource, ENV, _) {
return {
testCall: $resource(ENV.apiEndpoint + "/test-call", {}, {
launch: { …Run Code Online (Sandbox Code Playgroud) 我在项目上使用Liquibase,并且每次更新Liquibase后都需要始终执行一个过程...
当前,changeSet如下所示:
<changeSet id="liquibase-0" author="liquibase" runAlways="true">
<sqlFile relativeToChangelogFile="true" path="procedure/owner-changer.sql"/>
</changeSet>
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如何确保此changeSet将作为上一次更新操作运行?
谢谢
如何使用 Maven Assembly 插件将生成的 zip 文件包含到主zip 文件中?
例如,我有以下文件夹结构:
./folderA/
./file1
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我想生成一个 zip 文件,其中的内容是这样的:
folderA.zip
file1
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这是我的简化配置':
pom.xml
<?xml version="1.0" encoding="UTF-8"?>
<project >
<build>
<plugins>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-assembly-plugin</artifactId>
<configuration>
<appendAssemblyId>false</appendAssemblyId>
<descriptor>${basedir}/assembly-zipFolderA.xml</descriptor>
<finalName>folderA.zip</finalName>
</configuration>
</plugin>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-assembly-plugin</artifactId>
<configuration>
<appendAssemblyId>false</appendAssemblyId>
<descriptor>${basedir}/assembly.xml</descriptor>
<finalName>${module-zipFinalName}</finalName>
</configuration>
</plugin>
</plugins>
</build>
</project>
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程序集zipFolderA.xml
<assembly>
<id>folderA-zip</id>
<includeBaseDirectory>false</includeBaseDirectory>
<formats>
<format>zip</format>
</formats>
<fileSets>
<fileSet>
<directory>folderA</directory>
<outputDirectory>/</outputDirectory>
</fileSet>
</fileSets>
</assembly>
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程序集.xml
<assembly>
<id>main</id>
<includeBaseDirectory>false</includeBaseDirectory>
<formats>
<format>zip</format>
</formats>
<files>
<file>
<source>file1</source>
</file>
<file>
<source>folderA.zip</source>
</file>
</files>
</assembly>
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===>使用这个配置,Maven 抱怨它找不到 …