Javascript 文件上载在twitter引导中不起作用
我正在使用twitter引导文件上传,我想将图像存储在Database上,还想移动到一个文件夹中,我花了很多时间,无法获取文件名值。请参阅输入名字并在提交表单后选择一个图像,现在我获得了名字值,但我无法获得照片上载值Javascript 文件上载在twitter引导中不起作用,javascript,php,ajax,Javascript,Php,Ajax,我正在使用twitter引导文件上传,我想将图像存储在Database上,还想移动到一个文件夹中,我花了很多时间,无法获取文件名值。请参阅输入名字并在提交表单后选择一个图像,现在我获得了名字值,但我无法获得照片上载值 <form class="form-horizontal form-bordered" method="POST" id="newUserForm" enctype="multipart/form-data"> <d
<form class="form-horizontal form-bordered" method="POST" id="newUserForm" enctype="multipart/form-data">
<div class="form-group">
<label class="col-md-3 control-label">First Name<span class="star_mark"> *</span></label>
<div class="col-sm-6">
<input type="text" class="form-control" id="fname" name="fname" value="" aria-required="true" required="" data-msg-required="Please enter your firstname" placeholder="Enter your firstname">
</div>
</div>
<div class="form-group">
<label class="col-md-3 control-label">Photo Upload<span class="star_mark"> *</span></label>
<div class="col-md-6">
<div class="fileupload fileupload-new" data-provides="fileupload">
<div class="input-append">
<div class="uneditable-input">
<i class="fa fa-file fileupload-exists"></i>
<span class="fileupload-preview"></span>
</div>
<span class="btn btn-default btn-file">
<span class="fileupload-exists">Change</span>
<span class="fileupload-new">Select file</span>
<input type="file" id="file" name="file">
</span>
<a href="#" class="btn btn-default fileupload-exists" data-dismiss="fileupload">Remove</a>
</div>
</div>
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-3 col-sm-6">
<button class="btn btn-info btn-block" type="submit" id="user-submit">Submit</button>
</div>
</div>
</form>
<script type="text/javascript">
$(document).ready(function(){
$('#user-submit').click(function(event){
event.preventDefault();
if($('#newUserForm').valid()){
//console.log('success');
$.ajax({
type:'POST',
url :"php/register-newuser.php",
data: $('form#newUserForm').serialize(),
success: function(data) {
//var res=jQuery.parseJSON(data);// convert the json
console.log(data);
},
error:function(exception){
alert('Exeption:'+exception);
}
});
}
else{
console.log('please select all fields');
}
});
});
</script>
例如,要使用AJAX进行上传,可以使用jQUery插件。引导与上传无关。我不明白,如果可能的话,我的代码怎么办
$fstname=$_POST['fname'];// i got ans here
$filename = basename($_FILES['file']['name']);
$newuser = array("fstName" => $fstname,'photoname' => $filename);
echo json_encode($newuser);
print_r($_FILES)//means nothing will happen, i didn't get any value like filename,size,extension....