如何使用Ajax和JSON制作下拉菜单？

Question

这是我用来在不同级别的OpenCart中显示类别菜单的代码.它有效,但每次点击后它会产生越来越多的s XHR finished loading: POST和XHR finished loading: GETs,它会通过点击有时停止页面:

<script type="text/javascript">
 _url = '';

 $(document).ready(function(){                
    $('#mnav a').on('click', function() {
        var cat = $(this).attr('id');
        _url = '&category_id=' + cat;

        $.post('index.php?route=test/category/child' + _url,
            function(data) {
               if(data.length>10){
                    $('#mnav #sub').remove();
                    $(data).insertAfter($('#mnav #' + cat));
               }
            });
    });
 });

$.ajaxPrefilter(function( options, original_Options, jqXHR ) {
    options.async = true;
});
</script>

HTML代码:

<div id="mnav" class="list-group">
  <?php foreach ($categories as $category) { ?>
  <a id="<?php echo $category['category_id']; ?>" class="list-group-item active"><?php echo $category['name']; ?></a>
  <?php } ?>
</div>

控制器代码:

<?php
class ControllerTestCategory extends Controller {
    public function index() {
        if (isset($this->request->get['path'])) {
            $parts = explode('_', (string)$this->request->get['path']);
        } else {
            $parts = array();
        }

        $data['category_id'] = 0;
        if (isset($parts[0])) {
            $data['category_id'] = $parts[0];
        } else {
            $data['category_id'] = 0;
        }

        if (isset($parts[1])) {
            $data['child_id'] = $parts[1];
        } else {
            $data['child_id'] = 0;
        }

        $this->load->model('catalog/cat');

        $data['categories'] = array();

        $categories = $this->model_catalog_cat->getCategories(0);

        foreach ($categories as $category) {
            $children_data = array();

            $filter_data = array(
                'filter_category_id'  => $category['category_id'],
                'filter_sub_category' => true
            );

            $data['categories'][] = array(
                'category_id' => $category['category_id'],
                'name'        => $category['name'],
                'children'    => $category['children'],
                'products'    => $category['products'],
                'href'        => $this->url->link('product/category', 'path=' . $category['category_id'])
            );
        }

        $this->response->setOutput($this->load->view('test/category', $data));
    }
    public function child() {
        if (isset($this->request->get['category_id'])) {
            $this->load->model('catalog/cat');

            $data['categories'] = array();

            $categories = $this->model_catalog_cat->getCategories($this->request->get['category_id']);

            $data['x'] = '<div id="sub">';

            foreach ($categories as $category) {
                $data['x'] .= '<li>' . $category['name'] . '</li>';
            }
            $data['x'] .= '</div>';
        } else {
            $data['x'] = 'NA';
        }
        $this->response->setOutput($this->load->view('test/category', $data));
    }
}

SQL代码:

public function getCategories($parent_id = 0) {
    $sql = "SELECT c.category_id, c.parent_id, cd.name,
        (SELECT COUNT(DISTINCT ch.category_id) from category ch where ch.parent_id = c.category_id and cd.language_id = '" . (int)$this->config->get('config_language_id') . "') as children";

    $sql .= " , (SELECT COUNT(DISTINCT p.product_id) 
FROM product p  
    LEFT JOIN product_description pd ON (p.product_id = pd.product_id) 
    LEFT JOIN product_to_category p2c ON (p2c.product_id = p.product_id) 
    LEFT JOIN category_path cp ON (cp.category_id = p2c.category_id) 
WHERE 
    pd.language_id = '" . (int)$this->config->get('config_language_id') . "' AND 
    p.status = '1' AND 
    p.date_available <= NOW()) AS items";

    $sql .= " FROM category c LEFT JOIN category_description cd ON (c.category_id = cd.category_id) WHERE c.parent_id = '" . (int)$parent_id . "' AND cd.language_id = '" . (int)$this->config->get('config_language_id') . "' AND c.status = '1' ORDER BY c.sort_order, LCASE(cd.name)";

    $query = $this->db->query($sql);
    return $query->rows;
}

如果您通过提供所有必要的JavaScript,jQuery和JSON代码来帮助我,我将非常感激,因为我知道这些主题很少:-(

Answer 1

您可以将post请求的结果存储在javascript数组中,以便重用它,请参阅以下内容:

var cachedObj = [];

$(document).ready(function(){                
  $('#mnav a').on('click', function() {
    var cat = $(this).attr('id');
    _url = '&category_id=' + cat;
    getData(cat, _url); //<-- Get data from ajax or cache
  });
});

//This function replaces the $.post call (just for example) 
function dummyPost(id, url){
  //url should be used to make the post call
  var data = "<span class='sub'>Test " + id + "</span>";
  return data;
}

function getData(id, url){
  //Try to get data from cache
  var data;
  if(cachedObj[url]) {
    data = cachedObj[url];
    console.log("Data retrived from cache");
  }
  else {
    data = dummyPost(id, url);
    cachedObj[url] = data;
    console.log("Data retrived from post");
  }
  
  $('#mnav .sub').remove();
  //$(data).insertAfter($('#mnav #' + id));
  $('#mnav #' + id).append($(data));
}

.sub{
  color: red;
  font-weight: bold;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="mnav" class="list-group">
  <a id="1" class="list-group-item active">One</a>
  <a id="2" class="list-group-item active">Two</a>
  <a id="3" class="list-group-item active">Three</a>
</div>

我已经制作了dummyPost应该修改的函数来执行post请求.

您可以在我的示例日志中看到,当您第一次单击链接时,它会使用"post"检索其子菜单,而下一次,它会从缓存的数组中获取数据cachedObj.

我希望它对你有所帮助.再见.

更新: 应用于您的代码应该像:

<script type="text/javascript">

  var cachedObj = []; //<-- Add an array to cache submenus

  //Add a function to retrieves data from cache or REST
  function getData(url){
    //Try to get data from cache
    if(cachedObj[url]) {
      console.log("Data retrived from cache");
    }
    else {
      $.ajax({
         type: 'GET',
         url: 'index.php?route=test%2Fcategory%2Fchild' + url,
         success: function(data) {
               cachedObj[url] = data;
               console.log("Data retrived from post");
            }),
         async:false
      });
    }
    return cachedObj[url];
 }

 $(document).ready(function(){                
    $('#mnav a').on('click', function() {
        var cat = $(this).attr('id');
        var url = '&category_id=' + cat;

        var data = getData(url); //<-- Call the new function to get data
        if(data.length>10){
           $('#mnav #sub').remove();
           $(data).insertAfter($('#mnav #' + cat));
        }
    });
 });
</script>

我无法测试它,所以它可能包含一些错误.

Answer 2

问题可能是您没有阻止锚标记的默认操作.尝试添加event.preventDefault.这样浏览器将触发POST请求,而不是那个GET.

此外,如果您#mnav a通过ajax 添加新元素,最好将事件绑定到文档而不是元素.

 $(document).ready( function() {                
    $(document).on('click', '#mnav a', function( event ) {
        event.preventDefault();
        // some code
    });
 });

Answer 3

我在我的ajax驱动项目中遇到过这个问题.我希望我的解决方案可以帮到你.这可能会阻止创建更多呼叫.

// DOM ready
$(function() {

    $(document).off('click', '#mnav a').on('click', '#mnav a', function(e){

      e.preventDefault();

      // your stuff here

    });
});

Answer 4

在编写任何代码之前,您似乎需要制定有关您希望客户端和服务器进行通信的策略.

我对您的问题的理解是"菜单会在每次点击时生成ajax请求,而我不希望这样做."

但这就是你构建代码的方式:jQuery设置的菜单的onclick处理程序中有一个$.post()调用.

另一种方法是将预先填充菜单所需的所有数据发送到客户端.然后点击每个菜单,从已经在内存中的数据中提取,而不是发送ajax请求.

我能想到有几种方法可以做到这一点.您选择哪种策略取决于您对系统的舒适度和/或控制,以及页面速度的优先级.

1. 让服务器准备这些数据,并将其写入服务器端模板,以便数据立即可用,例如<script><?php echo 'var menuData = '.json_encode($my_data).'; ?></script>.这是最快的选择.
2. 使用现有的ajax URL,但在页面加载时触发一系列ajax请求.将所有结果发送到某个中央数据对象,如选项1中所示.一旦所有数据都返回,您的菜单将起作用.这里的优点是您不必更改任何服务器端代码,但很多调用需要时间.
3. 创建一个新的ajax URL,它能够在一个批处理中返回所有菜单数据.这将比选项2更快,但您必须使用您的服务器API.