显示图片上传

所以我有这个代码。我正在上传图片，并希望在用户上传图片后显示图片。请帮我用我的代码，这里只是一个PHP新手。 :(显示图片上传

connect.php 
<?php 
$servername = "localhost"; 
$user = "root"; 
$password = "12345"; 
$db = "demo"; 
$con = mysqli_connect($servername, $user, $password, $db); 
if (!$con) { 
    die('Could not connect: ' .mysql_error()); 
} 
?> 
try.php here 
<?php include 'connect.php'; ?> 

<?php 
if (isset($image)) { 
    $errors = array(); 
    $imageName = $_FILES['photoUpload']['name']; 
    $imageSize = $_FILES['photoUpload']['size']; 
    $imageType = $_FILES['photoUpload']['type']; 
    $imageTmp = $_FILES['photoUpload']['tmp_name']; 
    $imageExt = strtolower(end(explode('.', $imageName))); 
    $expensions = array("jpeg", "jpg", "png"); 

    if(in_array($imageExt, $expensions)===false) { 
     $errors[] = "Extensions not allowed. Only JPEG, JPG, PNG"; 
    } 
    if($imageSize > 2097152) { 
     $errors[] = "File size must not exceed 2MB"; 
    } else { print_r($errors); } 
} 
?> 

<div class="col-md-4"> 
<form action="upload.php" method="post" enctype="multipart/form-data"> 
    <input type="file" name="photoUpload"><br><br> 
    <input type="submit" name="submit" value="SUBMIT"/> 
    <ul> 
     <li>Sent File: <?php echo $imageName; ?></li> 
     <li>File Size: <?php echo $imageSize; ?></li> 
    </ul> 
</form> 
</div>

我知道我在这里缺少像使用echo和这样的显示它我也希望它显示在高度&宽度= 200像素;和边界半径：50％;

来源

2017-06-28 nesi

我相信你可以像这个例子一样打印图像：https://stackoverflow.com/questions/16284713/how-to-display-or-preview-a-uploaded-image-after-submitted-in-php- without-save-t –

[如何显示或预览上传的图像后提交的PHP没有保存到数据库？]（https://stackoverflow.com/questions/16284713/how-to-display-or-preview -a上传的图像，之后提交的，在-PHP-没有保存-T） –

是在保存它的名字在你的数据库？

如果是的话，试试这个下面

$id = $_SESSION['id'];  
$sql = "SELECT image FROM user WHERE `id` = $id"; 
$result = mysqli_query($conn, $sql);            

    while ($row = mysqli_fetch_array($result)) { 

     echo "<div id='img_div'>"; 
     //img tag fetching from folder 'images' - change to match the file it was uploaded to 
     echo "<img src='images/".$row['image']."' style='width:30%; height:30%;'>"; 
     echo "</div>"; 
    }

您需要更改查询，以适应你的代码，但这个想法应该让你朝着正确的方向前进。

来源

2017-06-28 02:48:53 sicksixstrings

显示图片上传

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