当前位置： 文江博客 话题详情

显示图像 - 斑点 - 不正确

发布于 2024-12-08 16:27:11 字数 2893 浏览 1 评论 0 原文

我在某个地方犯了严重错误。我正在尝试在 php 中显示 MySQL 表中的图像。我将从头开始。

uploadform

<!DOCTYPE html>
    <head>
      <meta http-equiv="content-type" content="text/html; charset=UTF-8">
    </head>
    <body>
        <form action="add_file.php" method="post" enctype="multipart/form-data">
            <input type="file" name="uploaded_file"><br />
            <input type="submit" value="Upload file">
        </form>
        <p>
            <a href="list_files.php">See all files</a>
        </p>
    </body>
    </html>

add_file.php

<?php
// Check if a file has been uploaded
if(isset($_FILES['uploaded_file'])) {
    // Make sure the file was sent without errors
    if($_FILES['uploaded_file']['error'] == 0) {
        // Connect to the database
        $dbLink = new mysqli('localhost', 'root', '', 'mydb');
        if(mysqli_connect_errno()) {
            die("MySQL connection failed: ". mysqli_connect_error());
        }

        // Gather all required data
        $name = $dbLink->real_escape_string($_FILES['uploaded_file']['name']);
        $mime = $dbLink->real_escape_string($_FILES['uploaded_file']['type']);
        $data = $dbLink->real_escape_string(file_get_contents($_FILES  ['uploaded_file']['tmp_name']));
        $size = intval($_FILES['uploaded_file']['size']);

        // Create the SQL query
        $query = "
            INSERT INTO `file` (
                `name`, `mime`, `size`, `data`, `created`
            )
            VALUES (
                '{$name}', '{$mime}', {$size}, '{$data}', NOW()
            )";

        // Execute the query
        $result = $dbLink->query($query);

        // Check if it was successfull
        if($result) {
            echo 'Success! Your file was successfully added!';
        }
        else {
            echo 'Error! Failed to insert the file'
               . "<pre>{$dbLink->error}</pre>";
        }
    }
    else {
        echo 'An error accured while the file was being uploaded. '
           . 'Error code: '. intval($_FILES['uploaded_file']['error']);
    }

    // Close the mysql connection
    $dbLink->close();
}
else {
    echo 'Error! A file was not sent!';
}
?>

显示图像

<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("mydb", $con);
$id = 1;
if(!isset($id) || empty($id) || !is_int($id)){
     die("Please select your image!");
}else{  
$query = mysql_query("SELECT * FROM file WHERE id='".$id."'");
$row = mysql_fetch_array($query);
$content = $row['data'];  
header('Content-type: image/jpg');
     echo $content;
}  
?>

我的输出是这样的：����JFIFHH��C

原文

somewhere I'm going terribly wrong. I'm trying to display an image from a MySQL table in php. I'll start from the beginning.

uploadform

<!DOCTYPE html>
    <head>
      <meta http-equiv="content-type" content="text/html; charset=UTF-8">
    </head>
    <body>
        <form action="add_file.php" method="post" enctype="multipart/form-data">
            <input type="file" name="uploaded_file"><br />
            <input type="submit" value="Upload file">
        </form>
        <p>
            <a href="list_files.php">See all files</a>
        </p>
    </body>
    </html>

add_file.php

<?php
// Check if a file has been uploaded
if(isset($_FILES['uploaded_file'])) {
    // Make sure the file was sent without errors
    if($_FILES['uploaded_file']['error'] == 0) {
        // Connect to the database
        $dbLink = new mysqli('localhost', 'root', '', 'mydb');
        if(mysqli_connect_errno()) {
            die("MySQL connection failed: ". mysqli_connect_error());
        }

        // Gather all required data
        $name = $dbLink->real_escape_string($_FILES['uploaded_file']['name']);
        $mime = $dbLink->real_escape_string($_FILES['uploaded_file']['type']);
        $data = $dbLink->real_escape_string(file_get_contents($_FILES  ['uploaded_file']['tmp_name']));
        $size = intval($_FILES['uploaded_file']['size']);

        // Create the SQL query
        $query = "
            INSERT INTO `file` (
                `name`, `mime`, `size`, `data`, `created`
            )
            VALUES (
                '{$name}', '{$mime}', {$size}, '{$data}', NOW()
            )";

        // Execute the query
        $result = $dbLink->query($query);

        // Check if it was successfull
        if($result) {
            echo 'Success! Your file was successfully added!';
        }
        else {
            echo 'Error! Failed to insert the file'
               . "<pre>{$dbLink->error}</pre>";
        }
    }
    else {
        echo 'An error accured while the file was being uploaded. '
           . 'Error code: '. intval($_FILES['uploaded_file']['error']);
    }

    // Close the mysql connection
    $dbLink->close();
}
else {
    echo 'Error! A file was not sent!';
}
?>

Displaying the image

<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("mydb", $con);
$id = 1;
if(!isset($id) || empty($id) || !is_int($id)){
     die("Please select your image!");
}else{  
$query = mysql_query("SELECT * FROM file WHERE id='".$id."'");
$row = mysql_fetch_array($query);
$content = $row['data'];  
header('Content-type: image/jpg');
     echo $content;
}  
?>

My output is something like this: ����JFIFHH��C

如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论， 你可以免费 注册 一个本站的账号。

评论（2）

瑾兮 2024-12-15 16:27:11

您需要使用 imagecreatefromstring 和 imagejpeg

<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
    die('Could not connect: ' . mysql_error());
}

mysql_select_db("mydb", $con);
$id = 1;
if(!isset($id) || empty($id) || !is_int($id)){
     die("Please select your image!");
} else {  
    $query = mysql_query("SELECT * FROM file WHERE id='".$id."'");
    $row = mysql_fetch_array($query);

    $im = imagecreatefromstring($row['data']);

    if ($im !== false) {
        header('Content-Type: image/jpeg');
        imagejpeg($im);
        imagedestroy($im);
    }
}  
?>

You need to use imagecreatefromstring and imagejpeg

<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
    die('Could not connect: ' . mysql_error());
}

mysql_select_db("mydb", $con);
$id = 1;
if(!isset($id) || empty($id) || !is_int($id)){
     die("Please select your image!");
} else {  
    $query = mysql_query("SELECT * FROM file WHERE id='".$id."'");
    $row = mysql_fetch_array($query);

    $im = imagecreatefromstring($row['data']);

    if ($im !== false) {
        header('Content-Type: image/jpeg');
        imagejpeg($im);
        imagedestroy($im);
    }
}  
?>
回复 原文
轻拂→两袖风尘 2024-12-15 16:27:11

在您需要从数据库中编码 blob 的图像数据中尝试此操作

echo '<img src="data:image/jpeg;base64,' . base64_encode( $row['data'] ) . '" />';

try this in the image data you need encoded the blob from database

echo '<img src="data:image/jpeg;base64,' . base64_encode( $row['data'] ) . '" />';
回复 原文
~没有更多了~

关于作者

穿透光

暂无简介

0 文章
0 评论
25 人气
发私信

相关话题

更多

热门标签

操作系统 程序设计 IT运维 Linux系统管理 JavaScript 服务器应用 solaris C/C++ PHP Shell BSD Vue.js aix Oracle Python HTML 系统管理 HTML5 CSS 前端
更多

推荐作者

已经忘了多久

文章 0 评论 0

15867725375

文章 0 评论 0

LonelySnow

文章 0 评论 0

走过海棠暮

文章 0 评论 0

轻许诺言

文章 0 评论 0

信馬由缰

文章 0 评论 0

更多

友情链接

文江博客
    我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站，即表示您同意使用 Cookies 和您的相关数据。
    原文