问题描述
我想在NodeJS中启动一个子进程并将其输出保存到变量中.以下代码将其提供给stdout:
I would like to start a child process in NodeJS and save it's output into a variable. The following code gives it to stdout:
require("child_process").execSync("echo Hello World", {"stdio": "inherit"});
我想到的东西与此代码相似:
I have something in mind that is similar to this code:
var test;
require("child_process").execSync("echo Hello World", {"stdio": "test"});
console.log(test);
test 的值应该是 Hello World .
这不起作用，因为"test" 不是有效的stdio值.
Which does not work, since "test" is not a valid stdio value.
也许使用环境变量是可行的，但是我没有发现如何在子进程中修改它们，而结果仍然对父级可见.
Perhaps this is possible using environment variables, however I did not find out how to modify them in the child process with the result still being visible to the parent.
推荐答案
execSync 是一个返回所传递命令的标准输出的函数，因此您可以使用以下代码将其输出存储到变量中:
execSync is a function which returns the stdout of the command you pass in, so you can store its output into a variable with the following code:
var child_process = require("child_process");
var test = child_process.execSync("echo Hello World");
console.log(test);
// => "Hello World"
请注意，如果进程的退出代码为非零，这将引发错误.另外，请注意，您可能需要使用 test.toString()，因为 child_process.execSync 可以返回 Buffer .
Be aware that this will throw an error if the exit code of the process is non-zero. Also, note that you may need to use test.toString() as child_process.execSync can return a Buffer.
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