英文:
Go: Inline string concatenation
问题
我需要使用os.Open来打开一个文件。我有路径./XML/和文件名foo.xml，每次都会变化。也就是说，我有一个从目录中读取的xml文件数组，我想逐个打开它们（或者使用线程，无所谓）。
我需要简单地将./XML/添加到SOMETHING.xml。在Java中，这很简单：
String a = "whatever", b = "whatever";
doSomething(a + b);
在Go语言中如何实现这个功能？谷歌没有给我找到答案。
file, err := os.Open(????????????)
英文:
I need to call os.Open to open a file. I have the path ./XML/ and the filename foo.xml, which changes each time. That is I have an array of xml files read from a dir which I am trying to open one at a time (or threaded, doesn't matter).
I need to simple add ./XML/ to SOMETHING.xml. In Java this is trivial,
String a = "whatever", b = "whatever";
doSomething(a + b);
How is this done in Go? Google has failed me.
file, err := os.Open(????????????)
答案1
得分: 9
最便携的路径拼接方法是使用filepath.Join：
import "path/filepath"
file, err := os.Open(filepath.Join("XML", fileinfo.Name()))
英文:
The most portable way to do path concatenation is by using filepath.Join:
import "path/filepath"
file, err := os.Open(filepath.Join("XML", fileinfo.Name()))
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