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我想上傳文件後上傳多個圖像!下面的按鈕不輸入變化事件是我的jQuery代碼(我下載代碼從這裏http://net.tutsplus.com/tutorials/javascript-ajax/uploading-files-with-ajax/)點擊上傳文件後上傳圖片!按鈕沒有輸入改變事件
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);
}());
感謝您花時間回答您自己的問題。你能不能添加一些解釋,以便我幫助別人。 – Win