Question

该法典允许将多份被拖入箱子的档案。存档器制作每个档案的书目,然后使用Ajax rq向服务器发送EACH文件:

$.each( e.dataTransfer.files, function(index, file){
    var fileReader = new FileReader();

    //..generate BLOB from file 


    fileReader.onloadend = (function(file) {                                
        return function(e) {                            
            uploadfilelist.append( &lt;li&gt;  + file.fileName +  &lt;/li&gt; );
            //send every single file to server
            var destfoldername=CURRENTPATH;
            var uploadfilename=file.fileName;

            var fd = new FormData();
            fd.append("param1", destfoldername);
            fd.append("param2", uploadfilename);
            fd.append("param3", blob);

            $.ajax({
                type:"POST",
                url:"url",
                data:fd,
                contentType: false,
                processData: false,             
                beforeSend:function (xhr){ 
                    //custom headers
                },  
                success: function(data, textStatus, jqXHR){

                },
                complete: function(jqXHR){    
                    alert("State after complete: " + jqXHR.statusText);                 
                }           
            });
        };
    })(file);
    fileReader.readAsBinaryString(file);
}

PROBLEM: the server internal crashes when receiving a next blob while not having processed the former one.

I found another Post discussing this: How to make all AJAX calls sequential? A "sequential" Request using async:false is not an option, it would block a whole lot of other things..

SOLUTION: ??? Call ajax forfile1, when call is done, call ajax for file2, ...call ajax for file-n

I would really like to use JQ Deferred (http://api.jquery.com/category/deferred-object/) e.g. as described here: http://api.jquery.com/jQuery.when/

$.when($.ajax(???how to refer to THIS ajax call).done(function(){
//call this ajax for the next file again or use $.ajax(code).then()?
});

我确实感到担忧,但我不知道如何正确。

Thanks for any suggestions! h.

Answer 1

这里有一个gin子:

http://plugins.jquery.com/project/ajaxqueue

它由John Resig本人撰写。

Answer 2

你们当然可以通过兑现下一个诺言,不给他们带来gin。

$.ajax(/*options*/)
.then(function(res) {
  // do something with result
  // call 2
  return $.ajax(/* options */);
})
.then(function(res) {
  // call 3
  return $.ajax(/* options */);
})  // and so on
.fail(function(err) {
  // catch any error from any ajax call
});

这几乎与把呼吁与可读性带来的附加利益和所有错误的追索结合起来。

友情链接