[英]AJAX to PHP Image Upload
我正在尝试将AJAX的多个图像上传到PHP。 这是我到目前为止所掌握的内容。
的HTML
<form>
<input class="images" type="file">
</form>
如果用户要上传多个图像,则可能会有多个输入字段。
Java脚本
var imageObject = GetAllFiles($('.images'));
function GetAllFiles(_selector)
{
var newObject = {};
for (var i = 0; i < $(_selector).length; i++)
{
var elem = $(_selector)[i];
newObject[i] = $(elem).files;
}
return newObject;
}
$(document).on('click', '#submit', function() {
var _data = JSON.stringify(imageObject);
$.post('upload.php', { action: 'ImageUpload', data: _data }, function (e){
alert(e);
});
)};
转换为JSON后通过AJAX发送数据
的PHP
if(isset($_POST['action']))
{
$action = $_POST['action'];
switch($action)
{
case 'ImageUpload' : ImageUpload($_POST['data']); break;
}
}
function ImageUpload($jsonData)
{
$images = json_decode($jsonData);
foreach($images as $image)
{
$directory = "../images/maschine/";
$target_file = $directory . basename($_FILES[$image])['name'];
if (move_uploaded_file($_FILES[$image]["tmp_name"], $target_file))
{
echo('Success');
} else
{
echo('Failure');
}
}
}
首先在文件输入中添加id,并记住表单中的enctype ='multipart / form-data'。
<input class="images" name="file[]" type="file" multiple>
之后在脚本中
<script>
$('YoursubmitbuttonclassorId').click(function() {
var filedata = document.getElementsByName("file"),
formdata = false;
if (window.FormData) {
formdata = new FormData();
}
var i = 0, len = filedata.files.length, img, reader, file;
for (; i < len; i++) {
file = filedata.files[i];
if (window.FileReader) {
reader = new FileReader();
reader.onloadend = function(e) {
showUploadedItem(e.target.result, file.fileName);
};
reader.readAsDataURL(file);
}
if (formdata) {
formdata.append("file", file);
}
}
if (formdata) {
$.ajax({
url: "/path to upload/",
type: "POST",
data: formdata,
processData: false,
contentType: false,
success: function(res) {
},
error: function(res) {
}
});
}
});
</script>
在php文件中
<?php
for($i=0; $i<count($_FILES['file']['name']); $i++){
$target_path = "uploads/";
$ext = explode('.', basename( $_FILES['file']['name'][$i]));
$target_path = $target_path . md5(uniqid()) . "." . $ext[count($ext)-1];
if(move_uploaded_file($_FILES['file']['tmp_name'][$i], $target_path)) {
echo "The file has been uploaded successfully <br />";
} else{
echo "There was an error uploading the file, please try again! <br
/>";
}
}
?>
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.