Question

我的意思是我这里有这段代码，该文件上传带有图像的文件，在托管它后在我的本地主机上运行良好，我意识到我没有得到图像预览，但是它在数据库中的发布非常好

我已经将其托管在一个实时网站上，正如我所说的，它在我的本地主机上运行得很好

define('BASEURL', $_SERVER['DOCUMENT_ROOT'].'/images/');

if (!empty($_FILES)){
$photo = $_FILES['description_image'];
$name = $photo['name'];
$nameArray = explode('.',$name);
$fileName = $nameArray[0];
$fileExt = $nameArray[1];
$mime = explode('/',$photo['type']);
$mimeType = $mime[0];
$mimeExt = $mime[1];
$tmpLoc = $photo['tmp_name'];
$fileSize = $photo['size'];
$allowed = array('png','jpg','jpeg','gif','mp4','mp3','wma','MP4');
$uploadName = md5(microtime()).'.'.$fileExt;
$uploadPath = BASEURL.'/images/uploads/'.$uploadName;
$dbpath = '/holyfamilycatholicchurch- 
foso.com/httpdocs/images/uploads/'.$uploadName;

/*if($mimeType != 'image'){
    $errors[] = 'The file must be an image.';
}*/
/*if(!in_array($fileExt, $allowed)){
    $errors[] = 'The File extension must be a png, jpg, jpeg or gif, 
mp4, mp3, wma, MP4';
}*/
if($fileSize > 1500000000){
    $errors[] = 'The file size must be under 15MB';
}
if($fileExt != $mimeExt && ($mimeExt == 'jpeg' && $fileExt != 
'jpg')){
    $errors[] = 'File extension does not match the file.';
}
}

if(!empty($errors)){
echo display_errors($errors);   
}else{
//upload files into db
move_uploaded_file($tmpLoc,$uploadPath);
$insert_post = "INSERT INTO posts 
(`title`,`post_cat`,`description`,`description_image`,`main_content`,`date_posted`) VALUES ('$title','$post_cat','$description','$dbpath','$main_content',now())";


$insert_pst = mysqli_query($con, $insert_post);
if($insert_pst){
    echo "<script>alert('Post has been inserted')</script>";
    echo "<script>window.open('new_post.php?insert_product','_self') 
</script>";
}
}
}

我希望图像输出

Answer 1

尝试替换您的第一个变量：

define('BASEURL', $_SERVER['DOCUMENT_ROOT'].'/images/');

带有标准变量声明：

$BASEURL = $_SERVER['DOCUMENT_ROOT'] . "/images/" ;

请确保每次致电$BASEURL时都以$开头。我注意到您在没有BASEURL的情况下致电$。

在我的实时网站中查看我的图像帖子时遇到麻烦

1 个答案: