Question

我想点击上传文件后上传多个图片！下面没有输入更改事件的按钮是我的Jquery代码（我从这里下载代码http://net.tutsplus.com/tutorials/javascript-ajax/uploading-files-with-ajax/）

的index.html

<div id="main">
    <h1>Upload Your Images</h1>
    <form method="post" enctype="multipart/form-data"  action="upload.php">
    <input type="file" name="images" id="images" multiple />
    <button type="submit" id="btn">Upload Files!</button>
    </form>
</div>

upload.js

(function () {
    var input = document.getElementById("images"), 
        formdata = false;

    function showUploadedItem (source) {
        var list = document.getElementById("image-list"),
            li   = document.createElement("li"),
            img  = document.createElement("img");
        img.src = source;
        li.appendChild(img);
        list.appendChild(li);
    }   

    if (window.FormData) {
        formdata = new FormData();
    //  document.getElementById("btn").style.display = "none";
    }

    button.addEventListener("click", function (evt) {
        document.getElementById("response").innerHTML = "Uploading . . ."
        var i = 0, len = this.files.length, img, reader, file;

        for ( ; i < len; i++ ) {
            file = this.files[i];

            if (!!file.type.match(/image.*/)) {
                if ( window.FileReader ) {
                    reader = new FileReader();
                    reader.onloadend = function (e) { 
                        showUploadedItem(e.target.result, file.fileName);
                    };
                    reader.readAsDataURL(file);
                }
                if (formdata) {
                    formdata.append("images[]", file);
                }
            }   
        }

        if (formdata) {
            $.ajax({
                url: "upload.php",
                type: "POST",
                data: formdata,
                processData: false,
                contentType: false,
                success: function (res) {
                    document.getElementById("response").innerHTML = res; 
                }
            });
        }
    }, false);
}());

Answer 1

的index.html

<div id="main">
<h1>Upload Your Images</h1>
<form method="post" enctype="multipart/form-data"  action="upload.php">
<input type="file" name="images" id="images" multiple />
<button type="submit" id="btn">Upload Files!</button>
</form>
</div>

upload.js

(function () {
    var input = document.getElementById("images"), 
        formdata = false;

    function showUploadedItem (source) {
        var list = document.getElementById("image-list"),
            li   = document.createElement("li"),
            img  = document.createElement("img");
        img.src = source;
        li.appendChild(img);
        list.appendChild(li);
    }   

    if (window.FormData) {

            formdata = new FormData();
        formdata = new FormData();


        //document.getElementById("btn").style.display = "none";
    }

    $("#btn").click(function (evt) {


         evt.preventDefault();
        document.getElementById("response").innerHTML = "Uploading . . ."
        var i = 0, len = input.files.length, img, reader, file;

        for ( ; i < len; i++ ) {
            file = input.files[i];

            if (!!file.type.match(/image.*/)) {
                if ( window.FileReader ) {
                    reader = new FileReader();
                    reader.onloadend = function (e) { 
                        showUploadedItem(e.target.result, file.fileName);
                    };
                    reader.readAsDataURL(file);
                }
                if (formdata) {
                    formdata.append("images[]", file);
                }
            }   
        }

        if (formdata) {
            $.ajax({
                url: "upload.php",
                type: "POST",
                data: formdata,
                processData: false,
                contentType: false,
                success: function (res) {
                    document.getElementById("response").innerHTML = res; 
                }
            });
        }
    }); 
}());

单击上传文件后上传图片！按钮没有输入更改事件

1 个答案: