运行 python 来自 Node.js (child_process) 的带有命名参数的脚本

Question

我有一个 python 脚本，可以运行在命令行上使用此参数：

python2 arg1 --infile abc.csv --encrypt true --keyfile xyz.bin 1234 WOW path

但是，如果我尝试从 Node.js 子进程做同样的事情，我会得到一个错误：

const spawn = require("child_process").spawn;

const process = spawn("python2", [
  path.join(rootDir, "public", "python", "script.py"),
  "arg1",
  "--infile abc.csv",
  "--encrypt true",
  "--keyfile xyz.bin",
  "1234",
  "WOW",
  "path",
]);

不是运行宁并给出错误。但是，如果我运行没有命名参数（--encrypt true）等，它运行s 成功：

const process = spawn("python2", [
  path.join(rootDir, "public", "python", "script.py"),
  "arg1",
  "1234",
  "WOW",
  "path",
]);

我认为我传递命名参数的方式可能不正确。请帮忙！

Answer 1

这篇文章可能对您有用：

https://medium.com/swlh/run-python-script-from-node-js-and-send-data-to-browser-15677fcf199f

Answer 2

您需要拆分参数的每个部分：

const process = spawn("python2", [
  path.join(rootDir, "public", "python", "script.py"),
  "arg1",
  "--infile",
      "abc.csv", // indentation for clarity, it's not necessary
  "--encrypt",
      "true",
  "--keyfile",
      "xyz.bin",
  "1234",
  "WOW",
  "path",
]);

您的原始脚本类似于运行在命令提示符下执行此操作：

python script.py arg1 "--infile abc.csv" "--encrypt true" "--keyfile xyz.bin" 1234 WOW path

基本上，您传递的参数名为 --infile abc.csv，值为 --encrypt true。这不是您想要的运行。你想要的是：

python script.py arg1 --infile abc.csv --encrypt true --keyfile xyz.bin 1234 WOW path

运行 python 来自 Node.js (child_process) 的带有命名参数的脚本

Run python script from Node.js (child_process) with named arguments

python

spawn

child-process

node.js

python-2.7