首页 \ 问答 \ 如何获取上传文件的相对路径作为变量在PHP（转贴）(How to get relative path of uploaded file as variable in php (repost))

如何获取上传文件的相对路径作为变量在PHP（转贴）(How to get relative path of uploaded file as variable in php (repost))

 我怎样才能获得我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，即上传文件，但我不知道如何恢复路径。 该路径可能会因用户配置文件图像而改变，稍后我会将上传路径更改为/ $ username。 如果您知道如何获取名称，请将其发布。 提前致谢。 

How can I get the relative path to my uploaded file? For example if I upload test.png I would get /upload/test.png. Here is my 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']);
}

?>
 
The code works, that is upload the file but I don't know how to get the path back. The path may change because its for a user profile image and later I will change the upload path to one that is /$username. If you know how get the name only please post that anyway. Thanks in advance.

原文：https://stackoverflow.com/questions/41504769

更新时间：2022-03-31 20:03

最满意答案

 如果您没有要返回的默认值，则可以返回Option[Int] ，并将其与getOrElse结合使用：  
def a(n: Int): Option[Int] = {
  if (n < 100) {
    Some(n * 2)
  } else {
    None
  }
}

a(10).getOrElse("Something else")
 
 
 另一种可能性是使用部分功能 ，因为在您的情况下，您的第一个功能并不涵盖所有情况，并且您希望有一个后备：  
val a: PartialFunction[Int, Int] = {
  case n if n < 100 =>
    n * 2
}

val b: PartialFunction[Int, String] = {
  case _ =>
    "Something else"
}
 
 然后你可以使用applyOrElse ：  
// If function a is not defined for the input, then call function b
val result = a.applyOrElse(10, b)
 
 或将两个部分函数合并到另一个函数中，并调用它：  
// Combine a and b
val cf = a.orElse(b)
// Call
val result = cf(10)

If you don't have a default value to return, then you could return an Option[Int] instead, and combine this with getOrElse: 
def a(n: Int): Option[Int] = {
  if (n < 100) {
    Some(n * 2)
  } else {
    None
  }
}

a(10).getOrElse("Something else")
 
 
Another possibility is to use partial functions, because in your case your first function does not cover all cases, and you want to have a fallback: 
val a: PartialFunction[Int, Int] = {
  case n if n < 100 =>
    n * 2
}

val b: PartialFunction[Int, String] = {
  case _ =>
    "Something else"
}
 
Then you can use applyOrElse: 
// If function a is not defined for the input, then call function b
val result = a.applyOrElse(10, b)
 
or combine both partial functions into another function, and call that one: 
// Combine a and b
val cf = a.orElse(b)
// Call
val result = cf(10)

如何获取上传文件的相对路径作为变量在PHP（转贴）(How to get relative path of uploaded file as variable in php (repost))

最满意答案

相关问答

下列中不属于面向对象的编程语言的是?[2022-05-30]

在Scala返回(Return in Scala)[2023-08-16]

如何在scala中返回值类型为String的Map(How to return a Map with Value Type as String in scala)[2022-05-22]

如何从scala中的if块返回值(how to return value from a if block in scala)[2023-07-01]

如何从scala中的if-else块而不是anyVal返回Int？(How to return Int from an if-else block instead of anyVal in scala?)[2023-11-08]

在Scala中同时打印和返回值的习惯性方式[复制链接](Idiomatic way of print and return value at the same time in Scala [duplicate])[2023-04-17]

你可以在Scala中返回一个可赋值的左值吗？(Can you return an assignable lvalue in Scala?)[2022-02-24]

在Scala中返回finally块的值[复制](Return value of finally block in Scala [duplicate])[2023-03-09]

scala隐式返回值问题(scala implicit return value issue)[2023-10-15]

如何在scala中检查返回值类型(how to check return value type in scala)[2023-06-08]

相关文章

最新问答