Question

我为预订航班编写了以下代码，但效果很好。

我唯一的问题是我无法更改每个select的每个选项的值。

例如：我想将0到12的值设置为每个option。但它只重复了内部选项标签的值。

如何专门为选项设置值，不仅要将其中的文本重复为值？我怎样才能做到这一点？这是我的片段：

$(function() {

  // Function to create child dropdown
  var createChildDropdown = function(i) {
  
    // Create a div in the following format:
    // <div class="childs">
    //    <label for="...">Child (number)</label>
    //    <select id="..."></select>
    // </div>
    var $childDropdown = $('<div />', {
      'class': 'childs'
    });
    $childDropdown.append($('<label />', {
      'for': 'childDropdown-' + i
    }).text('Child ' + i));
    $childDropdown.append($('<select />', {
      'id': 'childDropdown-' + i
    }));
    
    // Define options made available for each child
    var options = ['a', 'b', 'c'];
    options.forEach(function(option) {
      // Create new option element and append to <select>
      $childDropdown.find('select').append($('<option />').text(option).attr('value', option));
    });
    
    // Return documentFragment so that we can append it
    return $childDropdown;
  };
  
  // Function to destroy child dropdown
  var destroyChildDropdown = function($el, i) {
    $el.find('div.childs').get(i).remove();
  };

  $(".button-click a").on("click", function() {
    var button = $(this);
    var oldVal = parseInt(button.closest("ul").prev().val());
    var newVal = (button.text() == "+") ? oldVal + 1 : (oldVal > 0) ? oldVal - 1 : 0;
    var total_value = "";

    button.closest("ul").prev().val(newVal);

    $(".travel").each(function() {
      var cat = $(this).prev('span').text();
      total_value += cat + ": " + $(this).val() + ", ";
    });

    if (oldVal < newVal) {
      $('.childDropdowns').append(createChildDropdown(newVal));
    } else if (oldVal > newVal) {
      destroyChildDropdown($('.childDropdowns'), newVal);
    }

    total_value = total_value.substring(0, total_value.length - 2);
    $(".main").val(total_value);
  })
})

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<label>
Count all1 Traveller(s)
<input type="text" name="no_travellers" class="main" value="Adults: 1" placeholder="" />
</label>
<br/>
<label>
   <span>Children</span>
   <input type="text" class="travel" id="CAT_Custom_410672" name="CAT_Custom_410672"  value="0" />
   <ul class="button-group button-click">
      <li><a href="#" class="small button secondary"><i class="fa fa-plus"><span class="hide">+</span></i></a></li>
      <li><a href="#" class="small button secondary"><i class="fa fa-minus"><span class="hide">-</span></i></a></li>
   </ul>
</label>
<div class="childDropdowns"></div>

Answer 1

forEach将循环索引作为第二个参数传递给回调函数 - 所以你需要做的就是让你的回调接受第二个参数，然后用它作为值：

$(function() {

  // Function to create child dropdown
  var createChildDropdown = function(i) {
  
    // Create a div in the following format:
    // <div class="childs">
    //    <label for="...">Child (number)</label>
    //    <select id="..."></select>
    // </div>
    var $childDropdown = $('<div />', {
      'class': 'childs'
    });
    $childDropdown.append($('<label />', {
      'for': 'childDropdown-' + i
    }).text('Child ' + i));
    $childDropdown.append($('<select />', {
      'id': 'childDropdown-' + i
    }));
    
    // Define options made available for each child
    var options = ['a', 'b', 'c'];
    options.forEach(function(option, index) {
//-> added parameter name index here ^^^^^ 
      // Create new option element and append to <select>
      $childDropdown.find('select').append($('<option />')
             .text(option).attr('value', index));
//-> and using it here ...               ^^^^^
    });
    
    // Return documentFragment so that we can append it
    return $childDropdown;
  };
  
  // Function to destroy child dropdown
  var destroyChildDropdown = function($el, i) {
    $el.find('div.childs').get(i).remove();
  };

  $(".button-click a").on("click", function() {
    var button = $(this);
    var oldVal = parseInt(button.closest("ul").prev().val());
    var newVal = (button.text() == "+") ? oldVal + 1 : (oldVal > 0) ? oldVal - 1 : 0;
    var total_value = "";

    button.closest("ul").prev().val(newVal);

    $(".travel").each(function() {
      var cat = $(this).prev('span').text();
      total_value += cat + ": " + $(this).val() + ", ";
    });

    if (oldVal < newVal) {
      $('.childDropdowns').append(createChildDropdown(newVal));
    } else if (oldVal > newVal) {
      destroyChildDropdown($('.childDropdowns'), newVal);
    }

    total_value = total_value.substring(0, total_value.length - 2);
    $(".main").val(total_value);
  })
})

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<label>
Count all1 Traveller(s)
<input type="text" name="no_travellers" class="main" value="Adults: 1" placeholder="" />
</label>
<br/>
<label>
   <span>Children</span>
   <input type="text" class="travel" id="CAT_Custom_410672" name="CAT_Custom_410672"  value="0" />
   <ul class="button-group button-click">
      <li><a href="#" class="small button secondary"><i class="fa fa-plus"><span class="hide">+</span></i></a></li>
      <li><a href="#" class="small button secondary"><i class="fa fa-minus"><span class="hide">-</span></i></a></li>
   </ul>
</label>
<div class="childDropdowns"></div>

如何使用jquery更改选项的值

1 个答案: