Question

我实际上是在尝试遍历LI标签的集合并插入一些文本来模拟某人编写要做的事情列表的外观。它可以工作，但它会同时写入每个列表项而不是等待。有没有一种简单的方法来实现这一目标？我在这里有一个JS小提琴设置：http://jsfiddle.net/fZpzT/但代码看起来像这样。感谢。

function addListItems() {
var str = {
    listitem1:'personal background check',
    listitem2:'look into my sketchy neighbor',
    listitem3:'look up my driving record',
    listitem4:'pick up milk',
    listitem5:'wash the car'
}

$('.list-container li').each(function(){
    var z = $(this).attr('id');
    var str2 = str[z];
    var delay = 0;
    for (var i = 0; i <= str2.length; i++) {
        (function(str2){
            delay += 100 + Math.floor(Math.random()*11)*6;
            setTimeout(function(){
                appendStr(str2);       
            },delay);
        })(str2[i])
    }
    function appendStr(str2) {
      $('#'+ z).append(str2);
    }
});
}

Answer 1

延迟累计：Demo on jsFiddle

var str = {
    listitem1: 'write the first item',
    listitem2: 'write the second item',
    listitem3: 'write the third item',
    listitem4: 'write the fourth item',
    listitem5: 'write the fifth item'
}, cumulativeDelay = 0;

$('.list-container li').each(function () {
    var z = $(this).attr('id');
    var str2 = str[z];
    var delay = cumulativeDelay;
    for (var i = 0; i <= str2.length; i++) {
        (function (str2) {
            delay += 100 + Math.floor(Math.random() * 11) * 6;
            setTimeout(function () {
                appendStr(str2);
            }, delay);
        })(str2[i])
    }
    cumulativeDelay = delay;
    function appendStr(str2) {
        $('#' + z).append(str2);
    }
    $(this).css('color', 'red');
});

Answer 2

我会改变工作方式：

您从要编写的字符串开始，

对于每个字符串：

找到相应的列表项（按ID）
以打字机样式写出整个字符串

完成后调用下一次迭代。

var strings = [
  'personal background check',
  'look into my sketchy neighbor',
  'look up my driving record',
  'pick up milk',
  'wash the car'
];

function iterate(strings, idx)
{
  if (idx >= strings.length) { return; }

  var id = 'listitem' + (idx + 1),
  el = document.getElementById(id);

  typewriter(el, strings[idx], 0, function() {
    iterate(strings, idx + 1);
  });
}

function typewriter(el, str, idx, cb)
{
  if (idx >= str.length) {
    return cb();
  }

  setTimeout(function() {
    el.innerText = str.substr(0, idx + 1);
    typewriter(el, str, idx + 1, cb);
  }, 100 + Math.floor(Math.random() * 11) * 6);
}

Demo

Answer 3

简化怎么样？使用两个变量在单个循环中迭代略微修改的数据结构的整个内容。像这样。 http://jsfiddle.net/cXykd/

var strings = 
[
    { "id":"listitem1", "text": "write the first item" },
    { "id":"listitem2", "text": "write the second item" },
    { "id":"listitem3", "text": "write the third item" },
    { "id":"listitem4", "text": "write the fourth item" },
    { "id":"listitem5", "text": "write the fifth item" },
 ]

var stringsIndex = 0;
var textIndex = 0;

AddString();

function AddString(){
    if (stringsIndex < strings.length){
        if (textIndex >= strings[stringsIndex].text.length)
        {
            stringsIndex++;
            if (stringsIndex == strings.length)
            {
                return;
            }
            textIndex = 0;
        }

        $("#" + strings[stringsIndex].id).append(strings[stringsIndex].text[textIndex]);
        textIndex++;

        var delay = 10 + Math.floor(Math.random()*11)*6;
        setTimeout(AddString, delay);
    }
}

jquery .ee等待函数在继续循环之前完成

3 个答案: