Php 在mysql数据库中存储图像不会'；不行。现在怎么办？

我尝试了几天现在上传图像到我的数据库，然后显示它们。但是在我的脚本中似乎有一个错误，我找不到它：（所以现在我希望这个社区的人能在这种情况下帮助我-那真的太好了！非常感谢！我知道这很值得一读，对不起

以下是代码表单insert.php：

<?php
/*Enter your password on line 29*/
error_reporting(E_ALL);
ini_set("display_errors", 1);



if(!isset($_FILES['userfile']))
{
    echo '<p>Please select a file</p>';
}
else
{
    try {
    $msg= upload();  //this will upload your image
    echo $msg;  //Message showing success or failure.
    }
    catch(Exception $e) {
    echo $e->getMessage();
    echo 'Sorry, could not upload file';
    }
}

// the upload function

function upload() {
    $host="localhost";
$user="YYYY";
$pass="XXXX";
$db="ibm_sfreund";

    $maxsize = 10000000; //set to approx 10 MB

    //check associated error code
    if($_FILES['userfile']['error']==UPLOAD_ERR_OK) {

        //check whether file is uploaded with HTTP POST
        if(is_uploaded_file($_FILES['userfile']['tmp_name'])) {    

            //checks size of uploaded image on server side
            if( $_FILES['userfile']['size'] < $maxsize) {  

               //checks whether uploaded file is of image type
              //if(strpos(mime_content_type($_FILES['userfile']['tmp_name']),"image")===0) {
                 $finfo = finfo_open(FILEINFO_MIME_TYPE);
                if(strpos(finfo_file($finfo, $_FILES['userfile']['tmp_name']),"image")===0) {    

                    // prepare the image for insertion
                    $imgData =addslashes (file_get_contents($_FILES['userfile']['tmp_name']));

                    // put the image in the db...
                    // database connection
                    $con = mysql_connect("localhost", "...", "...");
                    mysql_select_db('...',$con);

                    // our sql query
                    $id_xd = $_POST["id"];
                    $sql = "UPDATE artikel SET image='{$imgData}', name='{$_FILES['userfile']['name']}', mime='{$_FILES['userfile']['type']}' WHERE id='$id_xd';";

                    // insert the image
                    mysql_query($sql) or die("Error in Query: " . mysql_error());
                    $msg='<p>Image successfully saved in database.</p>';
                }
                else
                    $msg="<p>Uploaded file is not an image.</p>";
            }
             else {
                // if the file is not less than the maximum allowed, print an error
                $msg='<div>File exceeds the Maximum File limit</div>
                <div>Maximum File limit is '.$maxsize.' bytes</div>
                <div>File '.$_FILES['userfile']['name'].' is '.$_FILES['userfile']['size'].
                ' bytes</div><hr />';
                }
        }
        else
            $msg="File not uploaded successfully.";

    }
    else {
        $msg= file_upload_error_message($_FILES['userfile']['error']);
    }
    return $msg;
}

// Function to return error message based on error code

function file_upload_error_message($error_code) {
    switch ($error_code) {
        case UPLOAD_ERR_INI_SIZE:
            return 'The uploaded file exceeds the upload_max_filesize directive in php.ini';
        case UPLOAD_ERR_FORM_SIZE:
            return 'The uploaded file exceeds the MAX_FILE_SIZE directive that was specified in the HTML form';
        case UPLOAD_ERR_PARTIAL:
            return 'The uploaded file was only partially uploaded';
        case UPLOAD_ERR_NO_FILE:
            return 'No file was uploaded';
        case UPLOAD_ERR_NO_TMP_DIR:
            return 'Missing a temporary folder';
        case UPLOAD_ERR_CANT_WRITE:
            return 'Failed to write file to disk';
        case UPLOAD_ERR_EXTENSION:
            return 'File upload stopped by extension';
        default:
            return 'Unknown upload error';
    }
}

已弃用：mysql_connect（）[function.mysql connect]：mysql扩展已弃用，将来将被删除：请在第7行的/users/ibm/www/beta/loadpic.php中使用mysqli或PDO

您应该重写代码以使用PDO或mysqli。要使其立即工作，请尝试使用：

 error_reporting(0);
 ini_set('display_errors', 0);

但这不是一个解决方案，而是要测试它是否工作。因为您可以向我们共享二进制数据，所以在开始时出现的错误会破坏您的脚本

标题：
此外，您还需要发送一个正确的标题，以便浏览器识别这是一个图像，而不是一个网页。在此处使用GIf后，您可以尝试：

header("Content-type: " . image_type_to_mime_type(IMAGETYPE_GIF));

在发送第一个输出之前。但在将图像保存到db之前，您需要先签出mime类型，然后再将其保存。然后，您可以使用二进制数据获取图像，并在显示时发回

addslashes不是很必要。请参考答案并调整您的代码。
执行
echo$img_final；
对于输出为
GIF87a^–ÌÌÌÌ…
是正常的。那么我该怎么做呢？如果您只将图像存储在文件服务器上，并将路径存储在数据库中，您的生活会轻松得多。@jeroen Stori在文件服务器上压缩图像并不总是最好的解决方案。@yergo可能并不总是，但如果您存储的图像最大为10Mb，并且在图像数据上使用
添加斜杠
之类的东西，则情况确实如此。现在它只显示字母等，而不显示警告。但是如何显示图像？@mm：您需要一个
内容类型
标题作为提示嗯..脚本中是否有任何其他错误可能会使所有内容崩溃？内容类型以前不工作或当前状态下不工作？请检查我的更新并测试它。好了，现在它说图像无法显示，因为它包含错误？
 error_reporting(0);
 ini_set('display_errors', 0);

header("Content-type: " . image_type_to_mime_type(IMAGETYPE_GIF));