Saving the output of a child process in a variable in the parent in NodeJS

血红的双手。 提交于 2020-12-29 13:15:20

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

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