1

我已经阅读了几乎所有与此相关的主题。我不知道是什么原因造成的。

这是显示图像的代码

<html>
<body>
<?php 

                    mysql_connect('localhost','root','') or die("Unable to Connect: ".mysql_error());
                    mysql_select_db("punjabi") or die("Unable to Select Database: ".mysql_error());

                    $sql = "SELECT imagename, mimetype, imagedata
                            FROM images WHERE ID = 1";

                    $result = @mysql_query($sql);
                    if(!$result)
                    {
                        die(mysql_error());
                    }

                    $file = mysql_fetch_array($result);

                    $imagename = $file['imagename'];
                    $mimetype = $file['mimetype'];
                    $imagedata = $file['imagedata'];

                    header("content-type: $mimetype");

                    echo($imagedata);

?>

这是插入图像的代码

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Add Post</title>
<style type="text/css">
    *
    {
        margin:0px;
        padding:0px;
    }
</style>

</head>

<body bgcolor="#333333">

<?php if(isset($_POST['submit'])):
        {


            if (!is_uploaded_file($_FILES['uploadfile']['tmp_name']))
            {
                die("$uploadfile is not an uploaded file!");
            }

            $uploadfile = $_FILES['uploadfile']['tmp_name'];
            $uploadname = $_FILES['uploadfile']['name'];
            $uploadtype = $_FILES['uploadfile']['type'];
            $uploaddesc = $_POST['desc'];


            $tempfile = fopen($uploadfile,'rb');
            $filedata = fread($tempfile,filesize($uploadfile));
            $filedata = addslashes($filedata);
            $sql = "INSERT INTO images SET
            imagename = '$uploadname',
            mimetype = '$uploadtype',
            description = '$uploaddesc',
            imagedata = '$filedata'";
            $ok = @mysql_query($sql);
            if (!$ok) die("Database error storing file: " .mysql_error());

            $sql = "Select id from images where imagename = '$uploadname'";
            $result = @mysql_query($sql);
            if(!$result) die(mysql_error());
            if(mysql_num_rows($result)!=1) die(mysql_error());

            $row = mysql_fetch_array($result);
            $imageid = $row['id'];

            $title = $_POST['title'];
            $content = $_POST['content'];

            $sql = "INSERT into posts SET
                    title = '$title',
                    content = '$content',
                    imageid = '$imageid'";
            if(!@mysql_query($sql))
            {
                die(mysql_error());
            }


            header("Location:http://localhost/punjabi/adminhome.php");  

        }


?>

<?php else: ?>

    <div id="post">
        <form action="<?php echo($_SERVER['PHP_SELF']) ?>" method="post" enctype="multipart/form-data" >
            Title: <input type="text" name="title" /><br /><br /><br />
            Content:<br /> <textarea cols=100 rows="40" wrap="hard" name="content" /></textarea><br /><br />
            Image: <input type="file" name="uploadfile" /><br /><br />
            Image Desc: <input type="text" name="desc" /><br /><br />
            <input type="submit" value="Submit" name="submit" />
        </form>
    </div>

<?php endif; ?>
</body>
</html>

这是表结构：

CREATE TABLE filestore (
-> ID INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
-> FileName VARCHAR(255) NOT NULL,
-> MimeType VARCHAR(50) NOT NULL,
-> Description VARCHAR(255) NOT NULL,
-> FileData MEDIUMBLOB
->);

帮助将不胜感激！

4

1 回答 1

4

问题是，当您这样做时，标头已经发送：

<html>
<body>
<?php

所以你不能再为图像设置标题：

header("content-type: $mimetype");

只需在 php 标记解决该问题之前摆脱 html（图像不是 html / 文本）即可。

于 2012-04-30T13:48:11.667 回答