Question

我希望我的用户能够将图像上传到他们的帐户（我的MySQL数据库）。但是，当我尝试对其进行编码并上传时，似乎该文件从未上传且为空。我在PHP设置中检查了最大上传大小等。在此先感谢!!

$data = "";
if(isset($_FILES["up"])) {
   $data = file_get_contents($_FILES['up']['tmp_name']);
   $data = base64_encode($data);
   $data = $connection->real_escape_string($data);
} else {
   echo '<div style="position:absolute;height:100px;top:0px;left:0px;
                     border-top-right-radius:20px;border-top-left-radius:20px;
                     width:100%;background:white;z-index:100;"
          >
            <font style="color:#BB0000;font-size:2.2vw;">'.$_FILES['up']['error'].'</font>
        </div>';
   die('');
}

我的HTML是:(并且表单正确提交）

 <input type="file" accept=".jpg,.png,.jpeg" name="up" id="up"/>

Answer 1

建议：将图像存储在目录中。

为什么不问DB？

对数据库的读/写总是慢于文件系统
您的数据库备份将变得更加耗时

所以，这是我的解决方案。

步骤1：创建目录userPhotos

第2步：创建表单

<form action="upload.php" method="post" enctype="multipart/form-data">
 Select your profile picture:
 <input type="file" name="fileToUpload" id="fileToUpload">
 <input type="Upload" value="Upload Image" name="submit">
</form>

步骤3：创建一个名为upload.php的文件，用于处理文件上传。

<?php
$target_dir = "userPhotos/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$newfilename = ;//assign unique user ID
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
    echo "File is an image - " . $check["mime"] . ".";
    if (move_uploaded_file($_FILES["fileToUpload"][$newfilename.$imageFileType], $target_file)) {
      echo "The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded.";$uploadOk = 1;
} 
} else {
    echo "File is not an image.";
    $uploadOk = 0;
}
}
if($uploadOK==1){
 store the path of image in DB as "/userPhotos/".$newfilename
 echo "uploaded photo : <img src='userphotos/".$newfilename."'">
}
//to display the image fetch the path using user ID as put it in src of img tag. 
?>

如果有人有更好的解决方案，请告诉我。谢谢，祝你好运。

Answer 2

PHP代码

$data = "";
if(isset($_FILES["up"])) {
   $data = file_get_contents($_FILES['up']['tmp_name']);
   $data = base64_encode($data);
   $data = $connection->real_escape_string($data);
} else {
   echo '<div style="position:absolute;height:100px;top:0px;left:0px;
                     border-top-right-radius:20px;border-top-left-radius:20px;
                     width:100%;background:white;z-index:100;"
          >
            <font style="color:#BB0000;font-size:2.2vw;">'.$_FILES['up']['error'].'</font>
        </div>';
   die('');
}

HTML代码

<form method="POST" enctype="multipart/form-data">
<input type="file" accept=".jpg,.png,.jpeg" name="up" id="up"/>
</form>

为什么我的图像文件在PHP中上传？

2 个答案: