Question

问题在于，在回调中，我尝试设置article.color，但如果我在console.log(article.color)之后添加color，它仍然是＃FF0000（在color内部，它会检测到显性颜色和工作完美）。

如何在回调中更新变量？

var articles = JSON.parse( fs.readFileSync( 'src/articles/articles.json' , 'utf8') );
var totalArticles = articles.length;
var readyArticles = 0;

function testReady(){
    readyArticles++;
    if(readyArticles == totalArticles){
        fs.writeFile('src/articles/articles.build.json', JSON.stringify(articles));
    }
}

articles.forEach(function( article ) {
    var imgPath = './src/'+article.background;
    article.color = '#FF0000';

    color( imgPath , function(err, color) {
        if( !err ){
            article.color = color;
        } else {
            console.log( err, color );
        }
    });

    testReady()
});

Answer 1

提示@CherryDT工作解决方案是：

var articles = JSON.parse(fs.readFileSync('src/articles/articles.json', 'utf8'));
var totalArticles = articles.length;
var readyArticles = 0;

function testReady() {
  readyArticles++;
  if (readyArticles == totalArticles) {
    console.log('Add ' + readyArticles + ' colors');
    fs.writeFile('src/articles/articles.build.json', JSON.stringify(articles));
  }
}

articles.forEach(function(article) {
  article.color = '#FF0000';

  color('./src/' + article.background, function(err, color) {
    if (!err) {
      article.color = '#' + color;
    } else {
      console.log(err, color);
    }

    testReady();
  })
});

在回调中更新变量

1 个答案: