如何刷新或重新加載數據

Question

我想僅刷新窗口小部件內以下代碼中給出的值，而不刷新60秒，而無需重新加載整個頁面：

    <?php
    // display Who's Online
    if (class_exists("VisitorMaps")) {
        $visitor_maps = new VisitorMaps();
    }
    if (isset($visitor_maps)) {
       $visitor_maps->visitor_maps_manual_sidebar();
    }
    ?>

任何幫助將不勝感激。

Answer 1

讓我更多地了解您的問題，然后我可能會改善我的回答，但是您可以使用此js從json格式的服務器獲取響應，並僅更新html頁面的某些元素/ *為數據中每個名為“選擇器”的鍵刷新DOM字典* /

/* both data and data['selectors'] need to be json objects
*/

var getJSON, refreshPage,
  __slice = [].slice;

refreshPage = function() {
  var callback, callbacks, content, data, dom, selector, selectors, _i, _len, _results;

  data = arguments[0], callbacks = 2 <= arguments.length ? __slice.call(arguments, 1) : [];
  selectors = data['selectors'];
  for (selector in selectors) {
    content = selectors[selector];
    if (!(content === '' || content === void 0)) {
      dom = $.toDom(selector);
      setText(dom, content, true);
    }
  }
  _results = [];
  for (_i = 0, _len = callbacks.length; _i < _len; _i++) {
    callback = callbacks[_i];
    if (isFunction(callback)) {
      _results.push(callback(data));
    } else {
      _results.push(callback);
    }
  }
  return _results;
};

/* jquery getJSON function simplified
*/


/* calls the callback function with retrieved data automatically
*/


getJSON = function() {
  var args, callback, data, url;

  url = arguments[0], data = arguments[1], callback = arguments[2], args = 4 <= arguments.length ? __slice.call(arguments, 3) : [];
  return $.getJSON(url, data, function(data, status) {
    if (isFunction(callback)) {
      return callback.apply(null, [data].concat(__slice.call(args)));
    }
  });
};