Using HTML5 file uploads with AJAX and jQuery

Admittedly, there are similar questions lying around on Stack Overflow, but it seems none quite meet my requirements.

Here is what I'm looking to do:

  • Upload an entire form of data, one piece of which is a single file
  • Work with Codeigniter's file upload library
  • Up until here, all is well. The data gets in my database as I need it. But I'd also like to submit my form via an AJAX post:

  • Using the native HTML5 File API, not flash or an iframe solution
  • Preferably interfacing with the low-level .ajax() jQuery method
  • I think I could imagine how to do this by auto-uploading the file when the field's value changes using pure javascript, but I'd rather do it all in one fell swoop on for submit in jQuery. I'm thinking it's not possible to do via query strings as I need to pass the entire file object, but I'm a little lost on what to do at this point.

    Can this be achieved?


    It's not too hard. Firstly, take a look at FileReader Interface.

    So, when the form is submitted, catch the submission process and

    var file = document.getElementById('fileBox').files[0]; //Files[0] = 1st file
    var reader = new FileReader();
    reader.readAsText(file, 'UTF-8');
    reader.onload = shipOff;
    //reader.onloadstart = ...
    //reader.onprogress = ... <-- Allows you to update a progress bar.
    //reader.onabort = ...
    //reader.onerror = ...
    //reader.onloadend = ...
    
    
    function shipOff(event) {
        var result = event.target.result;
        var fileName = document.getElementById('fileBox').files[0].name; //Should be 'picture.jpg'
        $.post('/myscript.php', { data: result, name: fileName }, continueSubmission);
    }
    

    Then, on the server side (ie myscript.php):

    $data = $_POST['data'];
    $fileName = $_POST['name'];
    $serverFile = time().$fileName;
    $fp = fopen('/uploads/'.$serverFile,'w'); //Prepends timestamp to prevent overwriting
    fwrite($fp, $data);
    fclose($fp);
    $returnData = array( "serverFile" => $serverFile );
    echo json_encode($returnData);
    

    Or something like it. I may be mistaken (and if I am, please, correct me), but this should store the file as something like 1287916771myPicture.jpg in /uploads/ on your server, and respond with a JSON variable (to a continueSubmission() function) containing the fileName on the server.

    Check out fwrite() and jQuery.post() .

    On the above page it details how to use readAsBinaryString() , readAsDataUrl() , and readAsArrayBuffer() for your other needs (eg images, videos, etc).


    With jQuery (and without FormData API) you can use something like this:

    function readFile(file){
       var loader = new FileReader();
       var def = $.Deferred(), promise = def.promise();
    
       //--- provide classic deferred interface
       loader.onload = function (e) { def.resolve(e.target.result); };
       loader.onprogress = loader.onloadstart = function (e) { def.notify(e); };
       loader.onerror = loader.onabort = function (e) { def.reject(e); };
       promise.abort = function () { return loader.abort.apply(loader, arguments); };
    
       loader.readAsBinaryString(file);
    
       return promise;
    }
    
    function upload(url, data){
        var def = $.Deferred(), promise = def.promise();
        var mul = buildMultipart(data);
        var req = $.ajax({
            url: url,
            data: mul.data,
            processData: false,
            type: "post",
            async: true,
            contentType: "multipart/form-data; boundary="+mul.bound,
            xhr: function() {
                var xhr = jQuery.ajaxSettings.xhr();
                if (xhr.upload) {
    
                    xhr.upload.addEventListener('progress', function(event) {
                        var percent = 0;
                        var position = event.loaded || event.position; /*event.position is deprecated*/
                        var total = event.total;
                        if (event.lengthComputable) {
                            percent = Math.ceil(position / total * 100);
                            def.notify(percent);
                        }                    
                    }, false);
                }
                return xhr;
            }
        });
        req.done(function(){ def.resolve.apply(def, arguments); })
           .fail(function(){ def.reject.apply(def, arguments); });
    
        promise.abort = function(){ return req.abort.apply(req, arguments); }
    
        return promise;
    }
    
    var buildMultipart = function(data){
        var key, crunks = [], bound = false;
        while (!bound) {
            bound = $.md5 ? $.md5(new Date().valueOf()) : (new Date().valueOf());
            for (key in data) if (~data[key].indexOf(bound)) { bound = false; continue; }
        }
    
        for (var key = 0, l = data.length; key < l; key++){
            if (typeof(data[key].value) !== "string") {
                crunks.push("--"+bound+"rn"+
                    "Content-Disposition: form-data; name=""+data[key].name+""; filename=""+data[key].value[1]+""rn"+
                    "Content-Type: application/octet-streamrn"+
                    "Content-Transfer-Encoding: binaryrnrn"+
                    data[key].value[0]);
            }else{
                crunks.push("--"+bound+"rn"+
                    "Content-Disposition: form-data; name=""+data[key].name+""rnrn"+
                    data[key].value);
            }
        }
    
        return {
            bound: bound,
            data: crunks.join("rn")+"rn--"+bound+"--"
        };
    };
    
    //----------
    //---------- On submit form:
    var form = $("form");
    var $file = form.find("#file");
    readFile($file[0].files[0]).done(function(fileData){
       var formData = form.find(":input:not('#file')").serializeArray();
       formData.file = [fileData, $file[0].files[0].name];
       upload(form.attr("action"), formData).done(function(){ alert("successfully uploaded!"); });
    });
    

    With FormData API you just have to add all fields of your form to FormData object and send it via $.ajax({ url: url, data: formData, processData: false, contentType: false, type:"POST"})

    链接地址: http://www.djcxy.com/p/55774.html

    上一篇: 如何使用jQuery.ajax和FormData上传文件

    下一篇: 通过AJAX和jQuery使用HTML5文件上传