使用Promise在Node.js中执行一系列命令

Question

我想通过使用Q Promise通过Node.js中的串行线逐步处理一系列命令。

var cmdArr = ['cmd1', 'cmd2','cmd3'];

我不确定如何构建它。 我想到了这样的事情，但是没有用：

Q().then(function() {
  cmdArr.forEach(command) {
     //here to initialize the promise??
  }
});

重要的是要保持顺序并能够在每个步骤之间使用Q.delay。

Answer 1

假设要执行的命令是某种异步函数调用：

var Q = require('q');

// This is the function you want to perform. For example purposes, all this
// does is use `setTimeout` to fake an async operation that takes some time.
function asyncOperation(input, cb) {
  setTimeout(function() {
    cb();
  }, 250);
};

function performCommand(command) {
  console.log('performing command', command);
  // Here the async function is called with an argument (`command`),
  // and once it's done, an extra delay is added (this could be optional
  // depending on the command that is executed).
  return Q.nfcall(asyncOperation, command).delay(1000);
}

// Set up a sequential promise chain, where a command is executed
// only when the previous has finished.
var chain = Q();
[ 'cmd1', 'cmd2', 'cmd3' ].forEach(function(step) {
  chain = chain.then(performCommand.bind(null, step));
});

// At this point, all commands have been executed.
chain.then(function() {
  console.log('all done!');
});

我对q不太熟悉，所以可以做得更好。

为了完整bluebird ，这是使用bluebird的版本：

var Promise = require('bluebird');

...

var asyncOperationAsPromised = Promise.promisify(asyncOperation);

function performCommand(command) {
  console.log('performing command', command);
  return asyncOperationAsPromised(command).delay(1000);
}

Promise.each(
  [ 'cmd1', 'cmd2', 'cmd3' ],
  performCommand.bind(null)
).then(function() {
  console.log('all done!');
});

Answer 2

对数组上的一系列异步操作进行排序的常见设计模式是使用.reduce()如下所示：

var cmdArr = ['cmd1', 'cmd2','cmd3'];

cmdArr.reduce(function(p, item) {
    return p.delay(1000).then(function() {
        // code to process item here
        // if this code is asynchronous, it should return a promise here
        return someAyncOperation(item);
    });
}, Q()).then(function(finalResult) {
    // all items done here
});

注意，我还显示了您可以在其中插入Q的.delay() 。