【发布时间】:2016-08-31 01:21:51
【问题描述】:
我有一个使用拖放上传文件的功能,它调用 php 函数来检查文件并上传它。但是我希望在成功上传时显示一些消息,但它没有出现。这是代码
<div id="upload" class="border"> <i class="fa fa-upload"></i> Drag file here</div>
<div id="result" style="display: none;">
<?php
if (isset($_FILES["csv"]["name"])) {
$uploadOk = 1;
if (!file_exists("./uploads/")) { mkdir("./uploads/", 0700, true); }
$temp = explode(".", $_FILES["csv"]["name"]);
$newfilename = round(microtime(true)).".".end($temp);
$target_dir = "./uploads/";
$target_file = $target_dir . $newfilename;
$csvFileType = pathinfo($target_file, PATHINFO_EXTENSION);
$iscsv = array('application/vnd.ms-excel','text/plain','text/csv','text/tsv');
if (file_exists($target_file)) { $uploadOk = 0; }
if ($_FILES["csv"]["size"] > 200000) { $uploadOk = 0; }
if ($csvFileType != "csv") { $uploadOk = 0; }
if(!in_array($_FILES['csv']['type'], $iscsv)) { $uploadOk = 0; }
if ($uploadOk == 1) {
$tmp_files = glob($target_dir."/*");
if (move_uploaded_file($_FILES["csv"]["tmp_name"], $target_file)) {
echo "Uploaded";
}
}
}
?>
</div>
和上传文件的JS函数:
<script type="text/javascript">
function sendFile(file) {
var uri = "index.php";
var xhr = new XMLHttpRequest();
var fd = new FormData();
xhr.open("POST", uri, true);
xhr.onreadystatechange = function() {
if (xhr.readyState == 4 && xhr.status == 200) {
$('#result').slideDown('slow');
}
};
fd.append('csv', file);
xhr.send(fd);
}
window.onload = function() {
var dropzone = document.getElementById("upload");
dropzone.ondragover = dropzone.ondragenter = function(event) {
event.stopPropagation();
event.preventDefault();
}
dropzone.ondrop = function(event) {
event.stopPropagation();
event.preventDefault();
var filesArray = event.dataTransfer.files;
for (var i=0; i<filesArray.length; i++) {
sendFile(filesArray[i]);
}
}
}
</script>
它成功上传文件,但没有显示我的消息“已上传”
为什么?
【问题讨论】:
-
删除文件时网络请求说什么?它是否按预期返回 200 响应?检查浏览器中开发人员工具中的网络选项卡(如果您已经知道这一点,我深表歉意。不完全了解您的技能水平)
-
文件大小是否大于
200000?
标签: javascript php file-upload upload