node.js fs.readFileSync 不是文件相关的吗？节点.js
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fs.readFileSync is Not File Relative? Node.js
提问by Arman
-Suppose I have a file in the root of my project called "file.xml"
- 假设我的项目根目录中有一个名为“file.xml”的文件
-Suppose I have a test file in tests/ called "test.js" and it has
- 假设我在 tests/ 中有一个名为“test.js”的测试文件，它有
const file = fs.readFileSync("../file.xml");
If I now run node ./tests/test.jsfrom the root of my project it says ../file.xml does not exist. If I run the same command from within the tests directory, then it works.
如果我现在node ./tests/test.js从我的项目的根目录运行，它会说 ../file.xml 不存在。如果我从测试目录中运行相同的命令，那么它就可以工作。
It seems fs.readFileSync is relative to the directory where the script is invoked from, instead of where the script actually is. If I wrote fs.readFileSync("./file.xml")in test.js it would look more confusing and is not consistent with relative paths in a require statement which are file relative.
似乎 fs.readFileSync 与调用脚本的目录有关，而不是脚本实际所在的目录。如果我fs.readFileSync("./file.xml")在 test.js 中编写它会看起来更混乱，并且与文件相关的 require 语句中的相对路径不一致。
Why is this?
为什么是这样？
How can I avoid having to rewrite the paths in my fs.readFileSync?
如何避免在 fs.readFileSync 中重写路径？
回答by cartant
You can resolve the path relative the location of the source file - rather than the current directory - using path.resolve:
您可以解析相对于源文件位置的路径 - 而不是当前目录 - 使用path.resolve：
const path = require("path");
const file = fs.readFileSync(path.resolve(__dirname, "../file.xml"));
