在根目录中创建项目，而没有项目目录
假设您有此文件结构：
/myapp/
/myapp/composer.phar
如果您在/myApp/的根中运行create-project，例如/code>
这将创建这样的文件：
/myapp/
/myapp/bar/x.txt
/myapp/bar/y.txt
/myapp/composer.phar
是否可以省略bar目录，以便看起来像这样？
/myapp/
/myapp/x.txt
/myapp/y.txt
/myapp/composer.phar
我知道我可以在运行命令之前简单地向上移动一个目录，但是我使用的是Vagrant/Docker，如果可能的话，它将更简单。
Say you have this file structure:
/myapp/
/myapp/composer.phar
If you run create-project in the root of /myapp/ e.g., php composer.phar create-project foo/bar
This will create files like this:
/myapp/
/myapp/bar/x.txt
/myapp/bar/y.txt
/myapp/composer.phar
Is it possible to omit the bar directory so that it will look like this?
/myapp/
/myapp/x.txt
/myapp/y.txt
/myapp/composer.phar
I understand that I could simply move up one directory before running the command, however I am using Vagrant/Docker and it will be simpler if the above is possible.
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正如他们所说的“当其他一切失败时，rtfm” :)，所以：
您在这里看到我们感兴趣的东西：
参数是可选的（您锻炼的），但没有任何明确提供的参数，作曲家将提供价值自己。这还包括
＆lt; directory＆gt;，然后将从＆lt; package＆gt;name中“派生”。要解决您的问题，您只需要告诉Composer您想要使用什么目录，如果应该是您当前的工作目录，只需通过。（dot），在u*ix世界中，这意味着“当前目录”：您应该很好。
As they say "When everything else fail, RTFM" :), so:
and you see that we got something interested here:
Arguments are optional (which you exercise) but not useless therefore for all non explicitly provided arguments, Composer will provide the value on its own. This also includes
<directory>which will be then "derived" from<package>name. To solve your problem you just need to tell Composer what directory your want it to use, and if that's should be your current working directory, simply pass.(dot), which in U*ix world means "current directory":and you should be good.