Question

我有这个代码，用户输入上传图像到我的页面然后它显示给他。那部分工作正常，但是当我想用他的图像制作缩略图时，它就不会显示出来。这是将图像转换为.jpg，然后是缩略图：

<?php
// connect to database
$link = mysqli_connect("localhost", "root", "iyadmajid", "moviesite");
if (!$link) {
    echo "Connection lost: " . mysqli_connect_error();
}

//make variables available
$image_caption = $_POST['image_caption'];
$image_username = $_POST['image_username'];
$file = $_FILES['image_filename'];
$image_tempname = $file['name'];
$today = date("Y-m-d"); 

// upload image and check for image type
$ImageDir = "/Applications/XAMPP/xamppfiles/htdocs/chapter7/images/";
$ImageThumb = $ImageDir . "thumbs/";
$ImageName = $ImageDir . $image_tempname;

if (move_uploaded_file($file['tmp_name'], $ImageName)) {
    // get info about the imag ebeing uploaded
    list($width, $height, $type, $attr) = getimagesize($ImageName);

    if ($type > 3) {
        // image isnt accepted
        echo "Sorry, but the file you uploaded was not a GIF, JPG, or 
PNG";
        echo "Please hit your browser's 'back' button and try again.";
    } else {
        // image is accepted
        $lastpic = uniqid('', true) ;
        $pic = explode(".", $lastpic);
        $lastpic = $pic[0];
    // insert info into images
    $insert = "INSERT INTO image
              (`image_caption`, `image_username`, `image_date`, 
 `image_code`)
              VALUES
              ('$image_caption', '$image_username', '$today', 
 '$lastpic')";
    $insertresults = mysqli_query($link, $insert);

    $picid = mysqli_insert_id($link);

    $newfilename = $ImageDir . $lastpic . ".jpg";

    if ($type == 2) {
        rename($ImageName, $newfilename);
    } else {
        if ($type == 1) {
        $image_old = imagecreatefromgif($ImageName);
    } elseif ($type == 3) {
        $image_old = imagecreatefrompng($ImageName);

    // 'convert' the image to jpg
    $image_jpg = imagecreatetruecolor($width, $height);
    imagecopyresampled($image_jpg, $image_old, 0, 0, 0, 0, $width, 
$height, $width, $height);
    imagejpeg($image_jpg, $newfilename);
    imagedestroy($image_old);
    imagedestroy($image_jpg);
    }

    $newthumbname = $ImageThumb . $image_code . ".jpg";

    //get dimensions for the thumbnail
    $thumb_width = $width * 0.10;
    $thumb_height = $height * 0.10;

    //create thumbnail
    $largeimage = imagecreatefromjpeg($newfilename);
    $thumb = imagecreatetruecolor($thumb_width, $thumb_height);
    imagecopyresampled($thumb, $largeimage, 0, 0, 0, 0, $thumb_width, 
$thumb_height, $width, $height);
    imagejpeg($thumb, $newthumbname);
    imagedestroy($largeimage);
    imagedestroy($thumb);
}
    $url = "Location:showimage.php?id=" . $picid;
    header($url);
}
}
?>

以下是应显示缩略图的代码：

<?php
// connect to database
$link = mysqli_connect("localhost", "root", "iyadmajid", "moviesite");
if (!$link) {
    die ("Connection lost: " . mysqli_connect_error($link));
}

$ImageDir = "../chapter7/images/";
$ImageThumb = $ImageDir . "thumbs/";
?>
<!DOCTYPE html>
<html>
<head>
<title>Welcome to our Photo Gallery</title>
</head>
<body>
<p align="center">Click on any image to see it full sized</p>
<table align="center">
    <tr>
        <td align="center">Image</td>
        <td align="center">Caption</td>
        <td align="center">Uploaded</td>
        <td align="center">Date Uploaded</td>
    </tr>

    <?php
        //get the thumbs
        $getpic = mysqli_query($link, "SELECT * FROM image");
        if (!$getpic) {
            die ("Invalid query: " . mysqli_error($link));
        }
        while ($row = mysqli_fetch_array($getpic)) {
            extract($row);
            echo "<tr>\n";
            echo "<td><a href='" . $ImageDir . $image_code . ".jpg>'";
            echo "<img src='" . $ImageThumb . $image_code . ".jpg' 
border='0'>";
            echo "<td>" . $image_caption . "</td>\n";
            echo "<td>" . $image_username . "</td>\n";
            echo "<td>" . $image_date . "</td>\n";
            echo "</tr>\n";
        }
    ?>
</table>
</body>
</html>

Here is what the page looks like. It should have the thumbnail in the 'image' column

非常感谢所有帮助。谢谢！

图像没有显示在php / html页面上

0 个答案: