我已经创建了以下功能,该功能将在我网站的每个页面上运行。它运行得足够快,但是当它在每个页面上运行时,它似乎是一个很好的候选者,可以使用memoization进行缓存。在我对这个主题的研究中,memoization总是缓存用return
语句发出的结果。这是完全合理的,但由于我的功能没有返回任何东西,我想知道它是否仍然可以缓存?使用除了记忆之外的其他东西?
这是我的功能:
// A requirejs module
define(function() {
var loadMenu,
cssdisabled,
testcss,
currstyle;
/*
* Dynamically create a form that looks like this:
*
* <form action="/search.html" id="js-searchbox" class="form">
* <input type="text" name="q" id="tipue_search_input"
* placeholder="Search...">
* <input type="submit" id="tipue_search_button" value="Search">
* </form>
*/
loadMenu = function() {
var loadBox = document.getElementById("searchbox"),
footerBox = document.getElementById("search-about-column"),
frag = document.createDocumentFragment(),
form = document.createElement("form"),
searchTextBox = document.createElement("input"),
searchButton = document.createElement("input");
// set attributes for form
form.action = "/search.html";
form.setAttribute("role", "search");
form.id = "js-searchbox";
form.setAttribute("class", "form");
// set attributes for Search text box
searchTextBox.type = "text";
searchTextBox.name = "q";
searchTextBox.id = "tipue_search_input";
searchTextBox.placeholder = "Search...";
// set attributes for Submit button
searchButton.type = "submit";
searchButton.setAttribute("class", "btnSearch");
searchButton.value = "Search";
// Arrange elements
form.appendChild(searchTextBox);
form.appendChild(searchButton);
// Load arranged elements into document fragment
frag.appendChild(form);
// Load document fragment into #searchbox, which is already on the page
loadBox.appendChild(frag);
}
cssdisabled = false;
testcss = document.createElement('div');
testcss.style.position = 'absolute';
document.getElementsByTagName('body')[0].appendChild(testcss);
if (testcss.currentStyle) {
currstyle = testcss.currentStyle['position'];
}
else if (window.getComputedStyle) {
currstyle = document.defaultView.getComputedStyle(testcss, null).getPropertyValue('position');
}
cssdisabled = (currstyle === 'static') ? true : false;
document.getElementsByTagName('body')[0].removeChild(testcss);
if (cssdisabled === false) {
loadMenu();
} else {
return false;
}
});
答案 0 :(得分:0)
使用frag
作为备忘录:
frag.appendChild(form);
loadMenu.frag = frag;
然后更新作业:
loadMenu = loadMenu.frag || function() {...}
<强>参考强>