0 
<!DOCTYPE html>

<head>
    <title>Display Picture</title>
    <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
</head>
<body>
    <h1>A Picture from Our Collection</h1>
    <?php

    header("Content-type: image/jpeg");

    $id = $_GET["ident"];
    $pat = '/^[0-9]+$/';
    if(!preg_match($pat, $id)){
        exit; // broken image
    }
    $mysqli = new mysqli('localhost','root','9876543210','student13');
    $query = "select image from Picys where ident=$id";

    $result = $mysqli->query($query);

    $err = $mysqli->error;
    if(!empty($err)){
        exit;
    }

    $row = $result->fetch_array(MYSQLI_NUM);
    $bytes = $row[0];

    echo $bytes;
    ?>
    <hr />
    <h2>Tags</h2>
    <?php
    $stmt->close();
    $stmt = $mysqli->prepare("select Tagstr from PicyTags where Picid=?");
    $stmt->bind_param('s',$id);
    $stmt->execute();
    $stmt->bind_result($tag);
    $count = 0;
    while($stmt->fetch()){
        if($count==0){
            echo "<bold>Existing tags:</bold>";
        }
        $count++;
        echo "$tag<br />";
    }
    $mysqli->close();
    ?>
    <br />
    <h3>Add tag</h3>
    <p>You may add tags that characterize the contents of this picture.</p>
    <form action="./AddTag.php?ident=<?php echo $id ?>" method="POST">
        <fieldset>
            <legend>Your tag</legend>
                <input type="text" size="16" maxlength="16" />
        </fieldset>
        <input type="submit" value="Add Tag" />
    </form>
</body>

我无法显示我的数据库中的图片。我错在哪里？假设是正确的

并且从 ADD TAG<h3>没有什么可以显示出来。我的输出只是一个

<h1>A Picture from Our collection</h1>

在我的浏览器中！什么都没有出来

4

1 回答 1

2

您已经发送了带有 HTML 标记的响应，然后您尝试更改标题。在将任何内容输出到浏览器之前，应更改标头。

于 2013-06-08T17:35:21.137 回答