Question

这是我的HTML：

<!DOCTYPE html>
<html>
    <body>
        <form action="uploaded_img.php" method="post" enctype="multipart/form- 
         data">
            Select image to upload:
            <input type="file" name="fileToUpload" id="fileToUpload">
            <input type="submit" value="Upload Image" name="submit">
        </form>

    </body>
</html>

PHP代码：

<?php 
 if(isset($_POST['submit'])){
    $name = $_FILES['file']['name'];  
    $temp_name = $_FILES['file']['tmp_name'];  
    if(isset($name)){
        if(!empty($name)){      
            $location = 'uploads/';      
            if(move_uploaded_file($temp_name, $location.$name)){
                echo 'File uploaded successfully';
            }
        }       
    }  else {
        echo 'You should select a file to upload !!';
    }
}
?>

我的代码给了我这些错误：

注意：未定义索引：文件位于 C：\ xampp \ htdocs \ phptutorial \ uploaded_img.php 在第3行注意：未定义索引：C：\ xampp \ htdocs \ phptutorial \ uploaded_img.php中的文件在第4行上，您应该选择要上传的文件！

我想将上传图像保存在c：xamp \ htdocs \ phptutorials \ uploads

Answer 1

您应该在文件输入元素上使用name后面的值，例如fileToUpload而不是file，

$name = $_FILES['fileToUpload']['name'];  
$temp_name = $_FILES['fileToUpload']['tmp_name'];

Answer 2

在代码中添加文件的正确名称。像这样

<?php 
 if(isset($_POST['submit'])){
    $name = $_FILES['fileToUpload']['name'];  
    $temp_name = $_FILES['fileToUpload']['tmp_name'];  
    if(isset($name)){
        if(!empty($name)){      
            $location = 'uploads/';      
            if(move_uploaded_file($temp_name, $location.$name)){
                echo 'File uploaded successfully';
            }
        }       
    }  else {
        echo 'You should select a file to upload !!';
    }
}
?>

Answer 3

尝试一下

rand()

如何使用php上传文件夹中的图像

3 个答案: