我试图创建一个脚本来将文件上传到我的服务器。
真正困扰我的是,虽然文件未上传,但仍会执行成功通话。
HTML
<form action="processupload.php" method="post" enctype="multipart/form-data" id="MyUploadForm">
<label>Castellà</label><input name="mapa_es" id="mapa_es" type="file" />
<div id="progressbox_es" style="display:none;">
<div id="progressbar_es">
<div id="statustxt_es">0%
</div>
</div>
</div>
</form>
JS
$(document).ready(function() {
$('#mapa_es').change(function() {
var formData = (this.files[0]);
$.ajax({
xhr: function()
{
var xhr = new window.XMLHttpRequest();
xhr.upload.addEventListener("progress", function(evt) {
if (evt.lengthComputable) {
var percentComplete = Math.floor((evt.loaded / evt.total) * 100);
console.log(percentComplete + '%');
//Progress bar
$('#progressbox_es').show();
$('#progressbar_es').width(percentComplete + '%')
$('#progressbar_es').css('background-color', '#6699FF');
$('#statustxt_es').html(percentComplete + '%');
if (percentComplete > 50)
{
$('#statustxt_es').css('color', '#000');
}
}
}, false);
xhr.addEventListener("progress", function(evt) {
if (evt.lengthComputable) {
var percentComplete = Math.floor((evt.loaded / evt.total) * 100);
//Progress bar
$('#progressbox_es').show();
$('#progressbar_es').width(percentComplete + '%');
$('#progressbar_es').css('background-color', '#6699FF');
$('#statustxt_es').html(percentComplete + '%');
if (percentComplete > 50)
{
$('#statustxt_es').css('color', '#000');
}
console.log(percentComplete + '%');
}
}, false);
return xhr;
},
url: 'processupload.php',
type: 'POST',
data: formData,
async: true,
beforeSend: function() {
},
success: function(msg) {
console.log(msg);
},
cache: false,
contentType: false,
processData: false
});
return false;
});
});
和processupload.php中的PHP代码
<?php
if (isset($_FILES["mapa_es"]) && $_FILES["mapa_es"]["error"] == UPLOAD_ERR_OK) {
$UploadDirectory = 'mapes/';
if ($_FILES["mapa_es"]["size"] > 5242880) {
echo "File size is too big!";
}
$File_Name = strtolower($_FILES['mapa_es']['name']);
$File_Ext = substr($File_Name, strrpos($File_Name, '.'));
$Random_Number = rand(0, 9999999999);
$NewFileName = $Random_Number . $File_Ext;
if (move_uploaded_file($_FILES['mapa_es']['tmp_name'], $UploadDirectory . $NewFileName)) {
echo 'Success! File Uploaded.';
} else {
echo 'error uploading File!';
}
} else {
echo 'Something wrong with upload!';
}
?>
我将感谢你能给我的任何帮助。感谢。
编辑:我遵循了gtryonp建议,我遇到了更多问题。 当我尝试var_dump($ _ FILES)时,我得到的只是一个空字符串。我还尝试使用$()提交表单。提交而不是$()。更改并且它有效,我认为这可能是因为&#34; enctype =&#34; multipart / form-data&# 34;在表单标签上。 有没有办法实现它而无需提交整个表格?
答案 0 :(得分:1)
您的processupload.php以die(消息)结尾,将作为正常程序结束,因此将触发success事件并且console.log(&#39; FILE UPLOADED&#39;)将成为即使在错误的情况下,你的反应总是如此使用 echo消息更改所有 die(消息),例如:
if (move_uploaded_file($_FILES['mapa_es']['tmp_name'], $UploadDirectory . $NewFileName)) {
echo 'Success! File Uploaded to '.$UploadDirectory . $NewFileName;
} else {
echo 'error uploading File!';
}
...
并更改成功功能以回应可能的屏幕答案。类似的东西:
success: function(msg) {
console.log(msg);
},
HAGD