Php 上传图像时如何获取图像的完整路径
我试图上传图像,但它并没有采取图像的填充路径。它只接受扩展名为的文件名Php 上传图像时如何获取图像的完整路径,php,jquery,Php,Jquery,我试图上传图像,但它并没有采取图像的填充路径。它只接受扩展名为的文件名 <!DOCTYPE html> <html> <head> <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script> <script type="text/javascript"> $(
<!DOCTYPE html>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("#sok").on('click',function(e){
//alert("Ok");
e.preventDefault();
var imf = $("#FoodieProfileImage").val();
alert(imf);
$.ajax({
url: 'Image.php',
type: 'POST',
data: imf,
async: false,
cache: false,
contentType: false,
processData: false,
success: function () {
//alert(returndata);
// $("#target").html(data);
},
error: function(){
}
});
});
});
</script>
</head>
<div class="modal-body">
<form id="editUserForm" method="post" action="Image.php" enctype="multipart/form-data">
<div class="form-group">
<label>Profile Image</label>
<input type="file" accept="image/*" name="FoodieProfileImage" id="FoodieProfileImage"/>
</div>
</form>
<div id="target"></div>
</div>
<div class="modal-footer">
<input type="button" id="sok" value="ok"/>
</div>
</html>
$(文档).ready(函数(){
$(“#sok”)。在('click',函数(e){
//警报(“正常”);
e、 预防默认值();
var imf=$(“#FoodieProfileImage”).val();
警报(货币基金组织);
$.ajax({
url:'Image.php',
键入:“POST”,
数据:国际货币基金组织,
async:false,
cache:false,
contentType:false,
processData:false,
成功:函数(){
//警报(返回数据);
//$(“#目标”).html(数据);
},
错误:函数(){
}
});
});
});
轮廓图像
php脚本如下所示:
<?php
if(!empty($_FILES["FoodieProfileImage"]["name"]) && $_FILES["FoodieProfileImage"]["size"] > 0) {
//Set upload path
$uploadPath = "images/";
echo $uploadPath;//print_r();
exit();
//Check for is directoty exist ot not, if not then create new directoty
if (!is_dir($uploadPath)) {
mkdir($uploadPath);
chmod($uploadPath, 0755);
}
if(isset($_FILES["FoodieProfileImage"]['name'])) {
$fileName = $_FILES["FoodieProfileImage"]["name"];
//$fileName = realpath($fileName);
//print_r(S_FILES);exit();
//$newUserDetail['profileImage'] = $fileName;
//pathinfo - Is a function that seperates file name, extension, basename
$path_parts = pathinfo($fileName);
$fileName = str_replace(" " , "_", strtolower($path_parts['filename'])); // Replaces all spaces with hyphens.
$fileName = preg_replace('/[^A-Za-z0-9\-]/', '', $fileName); // Removes special chars.
$fileName .= 'Foodie'.$userId ;
//Create new file name
move_uploaded_file($_FILES['FoodieProfileImage']["tmp_name"], $uploadPath . $fileName . '.' . $path_parts['extension']);
//Get Path for Uploaded File
$newUserDetail['profileImage'] = $uploadPath . $fileName. '.' . $path_parts['extension'];
//print_r($newUserDetail['profileImage']);exit();
//Get page image detail by page id
}
}
?>
服务器本身只能显示它所获得的数据,即使它知道文件路径,也无法访问它(因为您通过HTTP请求接收数据,而不是通过任何拉动方式)
可能可以解决您的任务的是通过JavaScript使用API,并实现一些逻辑,在上载图像的同时在客户端创建缩略图。为什么任意Web服务器都应该知道客户端文件系统的详细信息?这对于完成任务不是必需的,因此客户端将只传输文件内容本身。确定。。我想在上传图像时显示图像预览,所以我选择图像的完整路径并尝试设置为#目标。