我正在寻找这个问题的答案,并有我自己的问题。我在不同的地方找到了一些解决方案,并将它们放在我自己喜欢的答案中。
function exploreFolder(folderURL,options){
/* options: type explaination
**REQUIRED** callback: FUNCTION function to be called on each file. passed the complete filepath
then: FUNCTION function to be called after loading all files in folder. passed the number of files loaded
recursive: BOOLEAN specifies wether or not to travel deep into folders
ignore: REGEX file names matching this regular expression will not be operated on
accept: REGEX if this is present it overrides the `ignore` and only accepts files matching the regex
*/
$.ajax({
url: folderURL,
success: function(data){
var filesLoaded = 0,
fileName = '';
$(data).find("td > a").each(function(){
fileName = $(this).attr("href");
if(fileName === '/')
return; //to account for the (go up a level) link
if(/\/\//.test(folderURL + fileName))
return; //if the url has two consecutive slashes '//'
if(options.accept){
if(!options.accept.test(fileName))
//if accept is present and the href fails, dont callback
return;
}else if(options.ignore)
if(options.ignore.test(fileName))
//if ignore is present and the href passes, dont callback
return;
if(fileName.length > 1 && fileName.substr(fileName.length-1) === "/")
if(options.recursive)
//only recurse if we are told to
exploreFolder(folderURL + fileName, options);
else
return;
filesLoaded++;
options.callback(folderURL + fileName);
//pass the full URL into the callback function
});
if(options.then && filesLoaded > 0) options.then(filesLoaded);
}
});
}
然后,你可以这样调用:
var loadingConfig = {
callback: function(file) { console.log("Loaded file: " + file); },
then: function(numFiles) { console.log("Finished loading " + numFiles + " files"); },
recursive: true,
ignore: /^NOLOAD/,
};
exploreFolder('/someFolderURL/', loadingConfig);
这个例子将调用回调函数上的每个文件/文件夹中的指定文件夹除了对于那些与NOLOAD
开始。如果你想实际加载文件到页面中,那么你可以使用我开发的这个辅助函数。
function getFileExtension(fname){
if(fname)
return fname.substr((~-fname.lastIndexOf(".") >>> 0) + 2);
console.warn("No file name provided");
}
var loadFile = (function(filename){
var img = new Image();
return function(){
var fileref,
filename = arguments[0],
filetype = getFileExtension(filename).toLowerCase();
switch (filetype) {
case '':
return;
case 'js':
fileref=document.createElement('script');
fileref.setAttribute("type","text/javascript");
fileref.setAttribute("src", filename);
break;
case "css":
fileref=document.createElement("link");
fileref.setAttribute("rel", "stylesheet");
fileref.setAttribute("type", "text/css");
fileref.setAttribute("href", filename);
break;
case "jpg":
case "jpeg":
case 'png':
case 'gif':
img.src = filename;
break;
default:
console.warn("This file type is not supported: "+filetype);
return;
}
if (typeof fileref !== undefined){
$("head").append(fileref);
console.log('Loaded file: ' + filename);
}
}
})();
该函数接受JS | CSS | (通用图像)文件并加载它。它也会执行JS文件。 需要完整的调用来在你的脚本运行加载所有图片和*样式等脚本可能看起来像这样:
loadingConfig = {
callback: loadfile,
then: function(numFiles) { console.log("Finished loading " + numFiles + " files"); },
recursive: true,
ignore: /^NOLOAD/,
};
exploreFolder('/someFolderURL/', loadingConfig);
它工作得!
我认为这是执行OP所需的服务器端脚本的唯一方法。但我同意,最好避免。 – harpo 2010-11-13 22:29:53
我喜欢选项#3,我同意这不是一个好的选择。但是,我想先让它工作:)那么,我将如何使用'.getScript()'来获取*目录中的所有文件?你能提供一些代码吗? – Hristo 2010-11-13 22:36:01
Hristo,我只是说你可以为目录中的每个JavaScript文件调用一次getScrip()调用,然后每次将新文件添加到目录时添加一个新条目。 – 2010-11-13 22:55:58