合并两个无名的名字和姓氏列表

Question

我试图合并两个列表。一个是名字列表，第二个是姓氏列表。我想弄清楚jQuery的最佳方法是什么

我有这个：

<ul class="first-names">
    <li>John</li>
    <li>Betty</li>
</ul>

<ul class="last-names">
    <li>Doe</li>
    <li>White</li>
</ul>

我喜欢这个：

<ul class="full-names">
    <li>John Doe</li>
    <li>Betty White</li>
</ul>

到目前为止，我已经尝试了这一点，它甚至没有接近工作：

$( "ul.last-names li" ).text(function( index ) {
    $('u.first-names li').append(this).text();
});
$("ul.first-names").removeClass("first-names").addClass("full-names");

最终：

<ul class="full-names">
    <li>John</li>
    <li>Doe</li>
    <li>White</li>
    <li>White</li>
    <li>Betty</li>
    <li>Doe</li>
    <li>White</li>
    <li>White</li>
</ul>

谢谢！

Answer 1

假设两个列表长度相等，您可以使用类似以下的函数循环遍历每个<ul>元素，并将名称附加到生成的<ul>中：

<script>
    $(function(){
        MergeNames();
    });

    function MergeNames(){
      // Assumes that the number of elements are equal
      var firstNames = $('.first-names li');
      var lastNames = $('.last-names li');
      // Ensure they are equal (optional)
      if(firstNames.length == lastNames.length){
        // Loop through and build your new list
        for(var i = 0; i < firstNames.length; i++){
           $('.full-names').append('<li>' + $(firstNames[i]).text() + ' ' + $(lastNames[i]).text() + '</li>');
        }
      }
    }
  </script>

您可以see an example of this in action here以及下面的结果输出：

Answer 2

我建议，假设您只想留下一个列表，全名和，相关的名字和姓氏按顺序排列：

// caching the relevant last-name <li> elements, and then
// removing them from the HTML:
var lastNames = $('.last-names li').remove();

// selecting the <ul> with the class-name of 'first-names',
// using .toggleClass() to remove the classes that the
// <ul> currently has ('first-names') and adding those
// that it does not currently have ('full-names').
$('ul.first-names').toggleClass('first-names full-names')

  // finding the <li> elements in that <ul>:
  .find('li')

  // updating the text of those elements, via the anonymous
  // function of the text() method; the first argument ('i')
  // is the index of the current element in the jQuery collection
  // returned by the selector:
  .text(function(i){

    // here we trim the text content of the current Node (not
    // a jQuery object) using String.prototype.trim()
    // and concatenating that with a blank space and the
    // <li> of the equal index to the current node (here we
    // use get() to retrieve the DOM node from the cached
    // collection):
    return this.textContent.trim() + ' ' + lastNames.get( i ).textContent.trim();
});

var lastNames = $('.last-names li').remove();

$('ul.first-names').toggleClass('first-names full-names').find('li').text(function(i) {
  return this.textContent.trim() + ' ' + lastNames.get(i).textContent.trim();
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<ul class="first-names">
  <li>John</li>
  <li>Betty</li>
</ul>

<ul class="last-names">
  <li>Doe</li>
  <li>White</li>
</ul>

使用纯JavaScript也比较简单，但更详细：

// caching a reference to the last-names <ul>:
var lastNamesList = document.querySelector('ul.last-names'),

// caching references to the descendant <li> elements of the
// last-names <ul>, converting the collection to an Array,
// using Array.from():
    lastNames = Array.from( lastNamesList.querySelectorAll('li') ),

// caching the first-names <ul>:
    firstNameList = document.querySelector('ul.first-names');

// removing the 'last-names' class from the element:
firstNameList.classList.remove('last-names');

// adding the 'full-names' class-name:
firstNameList.classList.add('full-names');

// using querySelectorAll() to find the child <li> nodes of the
// first-names <ul>, and converting that collection to an Array
// in order to use the Array.prototype.forEach() method:
Array.from( firstNameList.querySelectorAll('li') ).forEach(function(li, index) {
  // in the anonymous function:
  // 'li':     a reference to the current <li> element of the
  //           array of nodes over which we're iterating,
  // 'index' : the index of the current array-element within the
  //           array over which we're iterating

  // here we update the textContent of the current <li> to a
  // combination of the trimmed current textContent, plus a
  // space plus the trimmed textContent of the <li> of the
  // same index held within the lastNames array:
  li.textContent = li.textContent.trim() + ' ' + lastNames[ index ].textContent.trim();
});

// here we find the parent of the lastNamesList node, and
// use the cached reference to remove that Node from the
// document:
lastNamesList.parentNode.removeChild(lastNamesList);

var lastNamesList = document.querySelector('ul.last-names'),
  lastNames = Array.from(lastNamesList.querySelectorAll('li')),
  firstNameList = document.querySelector('ul.first-names');

firstNameList.classList.remove('last-names');
firstNameList.classList.add('full-names');

Array.from(firstNameList.querySelectorAll('li')).forEach(function(li, index) {
  li.textContent = li.textContent.trim() + ' ' + lastNames[index].textContent.trim();
});

lastNamesList.parentNode.removeChild(lastNamesList);

<ul class="first-names">
  <li>John</li>
  <li>Betty</li>
</ul>

<ul class="last-names">
  <li>Doe</li>
  <li>White</li>
</ul>

参考文献：

• JavaScript的：
  • Array.from()。
  • Array.prototype.forEach()。
  • document.querySelector()。
  • document.querySelectorAll()。
  • Element.classList API。
  • Node.textContent。
  • String.prototype.trim()。
• jQuery的：
  • find()。
  • get()。
  • remove()。
  • toggleClass()。

Answer 3

将来，请展示您的尝试，并在适当的时候展示演示。这就是我要做的事情：

<script>
$('#first-names li').each(function() {
    var idx = $(this).index();
    var firstName = $(this).text();
    var lastName = $('#last-names li').eq(idx).text();

    $('#full-names').append('<li>' + firstName + ' ' + lastName + '</li>');
});
</script>

<强> Demo

请注意，我已将您的选择器转换为ID。我认为这是合适的。此外，还不清楚您是否打算以编程方式创建组合列表，以及是否应该替换原始列表之一。