图片未使用Ajax发布(无需提交表单)
问题描述：
我的表单是-
<form id="fileupload" method="post" enctype="multipart/form-data">
<input type="file" id="headerimage" spellcheck="true" class="typography" name="headerimage">
</form>
我的Ajax代码是-
var fileData = new FormData($('#fileupload'));
fileData.append('imagefile', $('#headerimage')[0].files);
$.ajax({
type : 'post',
data : fileData,
url : 'UploadImage.php',
dataType: 'json',
processData: false,
success : function(data)
{
alert("done");
},
});
Php代码-
<?php
# Data Base Connection
require_once('conn/dbConn.php');
var_dump($_REQUEST);
if (!empty($_FILES)) {
var_dump($_FILES);
}
请帮助.在php页面上，我没有获取文件数据.
Please Help. On the php page i am not getting file data.
答
JS代码-
var form = document.getElementById('fileupload');
var fileInput = document.getElementById('headerimage');
var file = fileInput.files[0];
var formData = new FormData();
var filename = '';
formData.append('file', file);
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
if (xhr.readyState == 4) {
filaname = xhr.responseText;
}
}
// Add any event handlers here...
xhr.open('POST', form.getAttribute('action'), true);
xhr.send(formData);
PHP代码-
<?php
# Data Base Connection
require_once('conn/dbConn.php');
if (!empty($_FILES)) {
$file = $_FILES;
if ($file['file']['error'] == 0) {
$name = explode('.', $file['file']['name']);
$newName = "header.".$name[count($name)-1];
if (move_uploaded_file($file['file']['tmp_name'], "../assets/Themes/Default/".$newName)) {
echo $newName;
exit;
}
} else {
echo "";
exit;
}
} else {
echo "";
exit;
}