Question

如果item小于1，是否有人知道如何从以下代码中list删除item.profit？

Splice似乎不起作用。

让我们说这会返回如下内容：

item 1, profit = 10
item 2, profit = 5
item 3, profit = -3
item 4, profit = 5
item 5, profit = -2

我正试图以列表结束：

item 1, profit = 10
item 2, profit = 5
item 3, profit = 5

var i = 0;
async.forEach(list, function (item, callback) {

    var url = 'http://www.website.com';
    request(url, function (err, response, html) {
        if (err) {
            console.log(err)
        } else {
            /// some code before....

            item.profit = (some calculation here);

            if (item.profit < 1) {
                console.log("Profit less than 1, so removing from list...");
                list.splice(i, 1); // This doesn't seem to work
            }

            // some code after...

        }
        callback();
    });
}, function (err) { //This is the final callback
    callback(err, list);
});

谢谢，

安东尼

Answer 1

请注意，async.forEach（）并行处理项目。因此，修改原始列表的条目是一个问题。您可能希望将所有想要的条目保存在新列表中。这是代码：

var i = 0, 
    newList = [];
async.eachSeries(list, function (item, callback) {  // <<<<< done is series to keep
                                                    // the same order as the original
                                                    // list, if you don't care about
                                                    // the order, then use
                                                    // async.each()

  var url = 'http://www.website.com';
  request(url, function (err, response, html) {
    if (err) {
        console.log(err)
    } else {
        /// some code before....

        item.profit = (some calculation here);

        if (item.profit > 0) {
            newList.push(item);
        }

        // some code after...

    }
    callback();
  });
}, function (err) { //This is the final callback
  callback(err, newList);
});

从列表Node.JS中删除项目

1 个答案: