我最近一直在尝试PHP,到目前为止一直很好,直到我碰到了一堵砖墙.这是我的一小段代码.它允许我上传一个文件,但我想要的是能够上传多个文件.
<html>
<head>
<Meta charset="utf-8" />
<title>Ajax upload form</title>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript">
function sendfile(){
var fd = new FormData();
for (var i = 0, len = document.getElementById('myfile').files.length; i < len; i++) {
fd.append("myfile", document.getElementById('myfile').files[i]);
}
$.ajax({
url: 'uploadfile.PHP',
data: fd,
processData: false,
contentType: false,
type: 'POST',
success: function(data){
alert(data);
}
});
}
</script>
</head>
<body>
<form action="uploadfile.PHP" method="post" enctype="multipart/form-data" id="form-id">
<p><input id="myfile" type="file" name="myfile" multiple=multiple/>
<input type="button" name="upload" id="upload" value="Upload" onclick="sendfile()" id="upload-button-id" /></p>
</form>
</body>
</html>
<?PHP
$target = "uploadfolder/";
//for($i=0; $i <count($_FILES['myfile']['name']); $i++){
if(move_uploaded_file($_FILES['myfile']['tmp_name'], $target.$_FILES['myfile']['name'])) {
echo 'Successfully copied';
}else{
echo 'Sorry, Could not copy';
}
}//
?>
任何帮助将受到高度赞赏.
解决方法:
的index.html
<html>
<head>
<title>Load files</title>
<script src="jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$('#myfiles').on("change", function() {
var myfiles = document.getElementById("myfiles");
var files = myfiles.files;
var data = new FormData();
for (i = 0; i < files.length; i++) {
data.append('file' + i, files[i]);
}
$.ajax({
url: 'load.PHP',
type: 'POST',
contentType: false,
data: data,
processData: false,
cache: false
}).done(function(msg) {
$("#loadedfiles").append(msg);
});
});
});
</script>
</head>
<body>
<div id="upload">
<div class="fileContainer">
<input id="myfiles" type="file" name="myfiles[]" multiple="multiple" />
</div>
</div>
<div id="loadedfiles">
</div>
</body>
</html>
load.PHP
<?PHP
$path="myfiles/";//server path
foreach ($_FILES as $key) {
if($key['error'] == UPLOAD_ERR_OK ){
$name = $key['name'];
$temp = $key['tmp_name'];
$size= ($key['size'] / 1000)."Kb";
move_uploaded_file($temp, $path . $name);
echo "
<div>
<h12><strong>File Name: $name</strong></h2><br />
<h12><strong>Size: $size</strong></h2><br />
<hr>
</div>
";
}else{
echo $key['error'];
}
}
?>
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