如何在php中以变量的形式获取上传文件的相对路径(repost)
如何在php中以变量的形式获取上传文件的相对路径(repost)
提问于 2017-01-06 19:30:47
如何获取上传文件的相对路径？例如，如果我上传test.png，我会得到/ upload /test.png。这是我的HTML：
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Jquery Ajax File Upload</title>
</head>
<body>
<div class="col-sm-6">
<div class="form-group ">
<label>Profile image</label>
<div class="input-group">
<span class="input-group-addon"><i class="fa fa-image"></i></span>
<input type="text" class="form-control" name="profile_image" autocomplete="off" value="" placeholder="" >
<label class="btn btn-default btn-file input-group-addon">
Browse <input type="file" name="image" style="display: none;" onchange="myFunction()" id="image" >
</label>
<div class="result"></div>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script>
$('#image').change(function(e){
var file = this.files[0];
var form = new FormData();
form.append('image', file);
$.ajax({
url : "http://192.168.1.147/upload.php",
type: "POST",
cache: false,
contentType: false,
processData: false,
data : form,
success: function(response){
$('.result').html(response.html)
}
});
});
</script>
</div>
</div>
</div>
</body>
</html>PHP：
<?php
$file = $_FILES['image'];
/* Allowed file extension */
$allowedExtensions = ["gif", "jpeg", "jpg", "png", "svg"];
$fileExtension = explode(".", $file["name"]);
/* Contains file extension */
$extension = end($fileExtension);
/* Allowed Image types */
$types = ['image/gif', 'image/png', 'image/x-png', 'image/pjpeg', 'image/jpg', 'image/jpeg','image/svg+xml'];
if(in_array(strtolower($file['type']), $types)
// Checking for valid image type
&& in_array(strtolower($extension), $allowedExtensions)
// Checking for valid file extension
&& !$file["error"] > 0)
// Checking for errors if any
{
if(move_uploaded_file($file["tmp_name"], 'uploads/'.$file['name'])){
header('Content-Type: application/json');
echo json_encode(['html' => /*return uploded file path and name*/ ]);
echo $_FILES['upload']['name'];
}else{
header('Content-Type: application/json');
echo json_encode(['html' => 'Unable to move image. Is folder writable?']);
}
}else{
header('Content-Type: application/json');
echo json_encode(['html' => 'Please upload only png, jpg images']);
}
?>代码起作用了，那就是上传文件，但我不知道如何取回路径。路径可能会更改，因为它用于用户配置文件图像，稍后我会将上传路径更改为/$username。如果你只知道如何获得名字，请张贴。提前谢谢。
回答 1
Stack Overflow用户
回答已采纳
发布于 2017-01-06 20:19:18
假设是test1.png，$file['name']中有文件的名称
这是程序的一部分，用于将文件保存在特定位置。在本例中，是相对于当前工作目录的文件夹uploads。
if(move_uploaded_file($file["tmp_name"], 'uploads/'.$file['name'])){
header('Content-Type: application/json');
echo json_encode(['html' =>
/* you might want to output something like */
getcwd().'/uploads/'.$file['name']
// getcwd() current working directory
// it may be that you need a relative path based on
// where your application lives to generate a url, for example
// 'http://your_app_base_url/'.'/uploads/'.$file['name']
]);
echo $_FILES['upload']['name']; // this will be empty unless there is a file form field with the name 'upload'
// compare with $_FILE['image'] above
}else{
header('Content-Type: application/json');
echo json_encode(['html' => 'Unable to move image. Is folder writable?']);
}如果您想在以后移动该文件，您可以有一个名为profile_pics的目录，并将其移动到该目录中，然后重命名为用户名。
例如：
$oldfilename = $file['name'];
//create a name for the file based on username and uploaded file extension (.jpg/.png)
$newfilename = $username.'_profile.'.pathinfo($oldfilename,PATHINFO_EXTENSION);
//this will rename /move the file from uploads to profile_pics (directory must exist already)
rename('uploads/'.$oldfilename,'profile_pics/'.$newfilename);页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/41504769
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