问题
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"});
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);
The value of test
was supposed to be Hello World
.
Which does not work, since "test"
is not a valid stdio value.
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.
回答1:
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"
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
.
来源:https://stackoverflow.com/questions/41001360/saving-the-output-of-a-child-process-in-a-variable-in-the-parent-in-nodejs