Question

我有一个网站，用户上传他们的个人资料图片，我已将图片存储在一个文件夹中，他们的链接反映在数据库中如何在我必须展示照片的地方显示该链接？

上传照片的代码

    define ("MAX_SIZE","1000"); 
    function getExtension($str)
    {
         $i = strrpos($str,".");
         if (!$i) { return ""; }
         $l = strlen($str) - $i;
         $ext = substr($str,$i+1,$l);
         return $ext;
    }

    $errors=0;
    $image=$_FILES['image']['name'];

    if ($image) 
    {

$filename = stripslashes($_FILES['image']['name']);
$extension = getExtension($filename);
$extension = strtolower($extension);
if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") 
    && ($extension != "gif")&& ($extension != "JPG") && ($extension != "JPEG") 
    && ($extension != "PNG") && ($extension != "GIF")) 
{
    echo '<h3>Unknown extension!</h3>';
    $errors=1;
}
else
{
    $size=filesize($_FILES['image']['tmp_name']);

    if ($size > MAX_SIZE*1024)
    {
        echo '<h4>You have exceeded the size limit!</h4>';
        $errors=1;
    }

    $image_name=time().'.'.$extension;
    $newname="photo/".$image_name;

    $copied = copy($_FILES['image']['tmp_name'], $newname);
    if (!$copied) 
    {
        echo '<h3>Copy unsuccessfull!</h3>';
        $errors=1;
    }
    else echo '<h3>uploaded successfull!</h3>';

    //mysql_query("insert into tutor (photo) values('".$newname."')");
    $myphoto= ".$newname.";
}

插入代码

    $mysqli = new mysqli("localhost", "***", "***", "***");
    if ($mysqli->connect_errno) {
        echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " .                         $mysqli->connect_error;
    }

    /* Prepared statement, stage 1: prepare */
    if (!($stmt = $mysqli->prepare("INSERT INTO `table` (`photo` ) VALUES (  ?)"))) {
        echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
    }

    /* Prepared statement, stage 2: bind and execute */

    if (!$stmt->bind_param('s',$myphoto )) {
        echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error;
    }

    if (!$stmt->execute()) {
        echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
    }

Answer 1

您需要以某种方式将用户链接到图片，例如向您的users表添加外键，然后将当前用户ID与图片一起插入。

在您的上传/插入页面中，您最终会得到与

类似的内容

// Assuming you're saving your logged in user's data in sessions
$current_user_id = $_SESSION['user_id'];
// Assuming you already have a $mysqli variable with an open connection
$mysqli->exec("INSERT INTO picture_table SET user = $current_user_id, picture = '$picture_file';");

然后，在您的个人资料/显示页面上，您必须使用类似于

的内容从数据库中获取它

// Assuming you already have a $mysqli variable with an open connection
$q = $mysqli->query("SELECT picture WHERE user = $current_user_id");
$row = $q->fetch_object();

printf('<img src="http://yoursite.com/yourpath/%s"  />', $row->picture);

使用PHP显示用户上传的个人资料照片

1 个答案: