当尝试使用AJAX上传图像而不直接提交表单并将FormData对象发送到服务器时,它返回空的$_FILES
数组。但是,如果我使用<input type="submit">
标记提交表单$_FILES
数组不为空,则会收到数据。
HTML
<form action="core/update.php" method="post" enctype="multipart/form-data" id="profile-photo" name="profile-photo-form">
<input type="file" id="photo-filename" name="avatar" class="edit-show panel photo-upload">
</form>
<button class="save-button" disabled="disabled">Save</button>
JS
$('#profile-photo').on('submit', function(e) {
e.preventDefault();
var form = $('#profile-photo')[0];
var formData = new FormData(form);
formData.append('avatar', $('#photo-filename')[0].files[0]);
$.ajax({
url: "core/update.php",
data: formData,
type: "POST",
contentType: false,
cache: false,
processData: false
});
console.log(formData);
});
$('.save-button').on('click', function() {
if ($('#photo-filename').val != '') {
$('#profile-photo').submit();
};
}
UPD
同样$('#profile-photo').serialize()
返回空字符串。
UPD 2
它可以与页面上的其他AJAX请求冲突吗?
答案 0 :(得分:3)
试试这个:
因为用户可能会上传多个文件
jQuery.each(jQuery('#photo-filename')[0].files, function(i, file) {
data.append('file-'+i, file);
});
而不是
formData.append('avatar', $('#photo-filename')[0].files[0]);
完整解决方案:
$('#profile-photo').on('submit', function(e) {
e.preventDefault();
var form = $('#profile-photo')[0];
var formData = new FormData(form);
jQuery.each(jQuery('#photo-filename')[0].files, function(i, file) {
formData.append('file-'+i, file);
});
$.ajax({
url: "core/update.php",
data: formData,
type: "POST",
contentType: false,
cache: false,
processData: false
});
console.log(formData);
});
答案 1 :(得分:0)
尝试以下方法,
$('#profile-photo').on('submit', function(e) {
e.preventDefault();
var formData = new FormData(this); // here I am editing
formData.append('avatar', $('#photo-filename')[0].files[0]);
$.ajax({
url: "core/update.php",
data: formData,
type: "POST",
contentType: false,
cache: false,
processData: false
});
console.log(formData);
});
答案 2 :(得分:0)
如果它只是您要发送的文件,那么您可以按照以下方式执行此操作,无需在此处将表单附加到formdata
:
$('#profile-photo').on('submit', function(e) {
e.preventDefault();
var formdata = new FormData();
var fileInput = $('#photo-filename');
//Use Either this
$.each($(fileInput).get(0).files, function (index, value) {
formdata.append('avatar', value);
});
//Or this
$.each($(fileInput).get(0).files, function (index, value) {
formdata.append(value.name, value);
});
//For single file use this
formData.append('avatar', $('#photo-filename')[0].files[0]);
$.ajax({
url: "core/update.php",
data: formData,
type: "POST",
contentType: false,
cache: false,
processData: false
});
console.log(formData);
});