在有条件防止此类情况发生时，两个console.log（）如何执行？

Question

I am trying to automate the MySQL server startup once I start my application with the following code:

 var exec = require('child_process').exec; let check = new Promise((resolve, reject) => exec('mysql-ctl status', function(error, stdout, stderr) { if (error) reject(error); else resolve(stderr); }) ); let start = new Promise((resolve, reject) => exec('mysql-ctl start', function(error, stdout, stderr) { if (error) reject(error); else resolve(console.log("MySQL server is up!")); }) ); async function check_status() { return await check; } async function start_server(status) { if (status.trim() == 'MySQL is stopped') { console.log("Starting MySQL server..."); await start; } else console.log("MySQL server was up..."); } module.exports = async function() { check_status() .then((status) => start_server(status)) .catch((error) => console.log("Error: " + error)); }; 

However, if the server is already up, I get the following output:

 App Started. Listening on PORT 8080 MySQL server was up... MySQL server is up! 

How come both console.log() are executing when there is a condition to prevent precisely that?

Answer 1

When you are calling new Promise(resolve, reject => ()) you are starting execution of function given to promise. That function is executed asynchronously.

So in your case you start checking if MySQL is running and next you start script for starting MySQL, but before check is finished.

module.exports = async function() {
  check_status()
    .then((status) => start_server(status))
    .catch((error) => console.log("Error: " + error));
};

When this function is executed you are just awaiting result from check in check_status() . But that check was started when check variable is created.

Result of this is in your then method => MySQL server is up! . When check is being executed start is also being executed and because MySQL server is not started you are getting Starting MySQL server...