blocks|key|43891|text|在newer+versions(>=1.8.0)+of+docker中，您可以这样做|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|43892|docker+build+-f+Dockerfile.db+.
docker+build+-f+Dockerfile.web+.|code-block|syntax|javascript|43893|一次很大的节省。|43894|编辑:根据raksja的评论更新版本|43895|编辑:来自@vsevolod的评论:通过赋予文件.Dockerfile扩展名(而不是名称)，例如Prod.Dockerfile，Test.Dockerfile等，可以在VS代码中突出显示语法。|43896|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/docker/docker/wiki^0|1|X|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|11|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|C|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|C|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

In <a href="https://github.com/docker/docker/wiki" rel="noreferrer">newer versions(>=1.8.0) of docker</a>, you can do this

<pre class="lang-bash prettyprint-override"><code>docker build -f Dockerfile.db .
docker build -f Dockerfile.web .
</code></pre>

A big save.

EDIT: update versions per raksja's comment

EDIT: comment from @vsevolod: it's possible to get syntax highlighting in VS code by giving files .Dockerfile extension(instead of name) e.g. Prod.Dockerfile, Test.Dockerfile etc.

blocks|key|4550548|text|在处理需要使用多个dockerfile的项目时，只需在单独的目录中创建每个dockerfile即可。例如,|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4550549|应用程序/+db/|4550550|上述每个目录都将包含它们的dockerfile。在构建应用程序时，docker将搜索所有目录并构建所有dockerfile。|4550551|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

When working on a project that requires the use of multiple dockerfiles, simply create each dockerfile in a separate directory. For instance,

app/
db/

Each of the above directories will contain their dockerfile. When an application is being built, docker will search all directories and build all dockerfiles.

blocks|key|43954|text|添加一个抽象层，例如，在这个项目https://github.com/larytet/dockerfile-generator中添加一个YAML文件，如下所示|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|43955|centos7:
++++base:+centos:centos7
++++packager:+rpm
++++install:
++++++-+$build_essential_centos+
++++++-+rpm-build
++++run:
++++++-+$get_release
++++env:
++++++-+$environment_vars|code-block|syntax|javascript|43956|一个简短的Python脚本/make可以从配置文件生成所有Dockerfile。|43957|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/larytet/dockerfile-generator^0|G|1B|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

Add an abstraction layer, for example, a YAML file like in this project <a href="https://github.com/larytet/dockerfile-generator" rel="nofollow noreferrer">https://github.com/larytet/dockerfile-generator</a> which looks like 

<pre class="lang-yaml prettyprint-override"><code>centos7:
 base: centos:centos7
 packager: rpm
 install:
 - $build_essential_centos 
 - rpm-build
 run:
 - $get_release
 env:
 - $environment_vars
</code></pre>

A short Python script/make can generate all Dockerfiles from the configuration file.

blocks|key|44009|text|在Intellij中，我简单地将docker文件的名称更改为*.Dockerfile，并将文件类型*.Dockerfile与docker语法关联起来。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|44010|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

In Intellij, I simple changed the name of the docker files to *.Dockerfile, and associated the file type *.Dockerfile to docker syntax.

How do you organize the Dockerfiles belonging to a project when you have one Dockerfile for the database, one for the application server, and so on? Do you create some sort of hierachy in the source?
A big enterprise project can't consist of only one Dockerfile?

Docker: Multiple Dockerfiles in project

数据库

服务器

Docker

Python

当数据库和应用服务器各有一个Dockerfile时，如何组织属于项目的Dockerfile？您是否在源代码中创建了某种层次结构？一个大型企业项目不能只由一个Dockerfile组成？

问Docker:项目中有多个Dockerfile
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