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尝试定义函数时出现jQuery错误_Jquery_Select - Fatal编程技术网

尝试定义函数时出现jQuery错误

尝试定义函数时出现jQuery错误,jquery,select,Jquery,Select,我有一个javaScript函数,可以更新选择输入选项。该函数工作正常,但我必须重复两次更新字段的代码,以便在页面加载和父选择更改时完成更新。这是有效的代码: jQuery.fn.filterOn = function(radio, values) { var firsttime = true; return this.each(function() { var select = this; var options = []; $(select).find('op

我有一个javaScript函数,可以更新选择输入选项。该函数工作正常,但我必须重复两次更新字段的代码,以便在页面加载和父选择更改时完成更新。这是有效的代码:

jQuery.fn.filterOn = function(radio, values) {
  var firsttime = true;
  return this.each(function() {
    var select = this;
    var options = [];
    $(select).find('option').each(function() {
      options.push({
        value: $(this).val(), 
        text: $(this).text()
      });
    });
    $('#upload_range_manufacturer_category').data('options', options);
    $(radio).change(function() {
      var options = $(select).empty().data('options');
      var haystack = values[$(radio).val()];
      $.each(options, function(i) {
        var option = options[i];
        if($.inArray(option.value, haystack) !== -1) {
          $(select).append(
            $('<option>').text(option.text).val(option.value)
            );
        }
      });
    });
    if(firsttime){
      $(function() {
        var options = $(select).empty().data('options');
        var haystack = values[$(radio).val()];
        firsttime = false;
        $.each(options, function(i) {
          var option = options[i];
          if($.inArray(option.value, haystack) !== -1) {
            $(select).append(
              $('<option>').text(option.text).val(option.value)
              );
          }
        });
      })
    } 
  });
};
以及选择和函数调用的代码:

<td><select name="upload_range[product_category]" id="upload_range_product_category"> 
<option value="2">Products</option> 
<option value="3">Manufacturers</option> 
</select></td> 
</tr> 
<tr> 
  <th><label for="upload_range_manufacturer_category">Manufacturer category</label></th> 
  <td><select name="upload_range[manufacturer_category]" id="upload_range_manufacturer_category"> 
<option value="4">Sports Equipment</option> 
<option value="6">Bricks</option> 
</select></td> 


<script type="text/javascript"> 
  $(document).ready(function () {
    var data = {"2":["4"],"3":["6"]};
    $('#upload_range_manufacturer_category').filterOn('#upload_range_product_category', data);
  }); 
</script>
为了避免代码重复,我想我可以编写一个可以从当前函数内部调用的函数。但是,我的新代码给了我以下错误:

未捕获类型错误:对象函数a,b{返回新的e.fn.inita,b,h} 没有“filterSelect”方法

这是新代码:

jQuery.fn.filterOn = function(field, values) {
  var firsttime = true;
  return this.each(function() {
    var select = this;
    var options = [];
    $(select).find('option').each(function() {
      options.push({
        value: $(this).val(), 
        text: $(this).text()
      });
    });
    $('#upload_range_manufacturer_category').data('options', options);
    $(field).change(function() {
      $.filterSelect(field,select,values);
    });
    if(firsttime){
      firsttime = false;
      $.filterSelect(field,select,values);
    } 
  });
};
jQuery.fn.filterSelect = function(field,select,values) {
  var options = $(select).empty().data('options');
  var haystack = values[$(field).val()];
  $.each(options, function(i) {
    var option = options[i];
    if($.inArray(option.value, haystack) !== -1) {
      $(select).append(
        $('<option>').text(option.text).val(option.value)
        );
    }
  });
};
如果有人知道我可能犯了什么学生错误,我们将非常感谢你的意见。我曾尝试在filterOn函数的上方和下方声明filterSelect函数,但两者都不起作用

谢谢

Luke

尝试删除过滤器选择声明中的.fn,使其看起来像

jQuery.filterSelect = function(field,select,values) {
  var options = $(select).empty().data('options');
  var haystack = values[$(field).val()];
  $.each(options, function(i) {
    var option = options[i];
    if($.inArray(option.value, haystack) !== -1) {
      $(select).append(
        $('<option>').text(option.text).val(option.value)
        );
    }
  });
};

它限定了您的功能范围,比如jQuery.fn.filterOn如何处理$'upload\u range\u manufacturer\u category'。filterOn'upload\u range\u product\u category',data;