Question

在两页中，我遇到了问题。第一个问题是在我提交之前我收到此提醒：

（注意：未定义的索引：在第15行的C：\ xampp \ htdocs \ art-legend \ 12 \ admin \ add.php中提交）

，此页面中的第15行是

(if ($_POST['submit'] && !empty($_FILES)){)

第二个问题是在我放置任何图像之前，我有一个图像在顶部，但我仍然没有放任何图像，我有这张图片

第三个错误是当我上传图片时我看不到它，而且我的屏幕就像上一个屏幕一样。

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=windows-1256" />
<title>Untitled Document</title>
<title>Untitled Document</title>
</head>

<body>
<?
if ($_POST['submit'] && !empty($_FILES)){

    $formok = TRUE;

    $title = $_POST['title'];
    $thread = $_POST['elm1'];
    $date =  date("d/m/y h;i:s");

   $path = $_FILES['upload']['tmp_name'];
   $name = $_FILES['upload']['name'];
   $size = $_FILES['upload']['size'];
   $type = $_FILES['upload']['type']; 
   $error = $_FILES['upload']['error'];

    if (!is_uploaded_file($path)){
        $formok = FALSE; 
        echo "there is no file uploaded try again";

        }

        if (!in_array($type,array('image/png','image/jpg','image/jpeg'))){
            $formok = FALSE ;
            echo "the file is not image try again ";
            }

        if (filesize($path)>800000){
            $formok = FALSE ;
            echo "the file is bigger ";
            }

        if ($formok){
            if ($connect = mysqli_connect('localhost','root','adminpass','flip')){
                $content = file_get_contents($path);
                $safetitle = mysqli_real_escape_string($connect,$title);
                $safethread = mysqli_real_escape_string($connect,$thread);
                $safeimage = mysqli_real_escape_string($connect,$thread);


                $sqltitle ="insert into title(title) values ('$safetitle')";
                $sqlthread = "insert into threads (topic,date) values ('$safethread','$date')";
                $sqlimge ="insert into images (name,size,type,content) values ('$name','$size','$type','$safeimage')";

                $querytitle = mysqli_query($connect,$sqltitle);
                $querythread = mysqli_query($connect,$sqlthread);
                $queryimage = mysqli_query($connect,$sqlimge);


                if ($querytitle && $querythread && $queryimage){

                    $imageid = mysqli_insert_id($connect);
                    }

                    mysqli_close($connect);

                }
                else {
                    echo "There is error in database connect";
                    }

                    echo "insert done all ";
            }
    }
?>

<img src="image.php?id=<?=$imageid; ?>" />
<hr />

<form action="<?php echo $_SERVER["PHP_SELF"];?>" method="post" enctype="multipart/form-data">
<p>
title :
<input type="text" name="title" />
</p>
<div>
    <textarea id="elm1" name="elm1" rows="15" cols="80" style="width: 80%">

    </textarea>
</div>
<input type="hidden" name="MAX_FILE_SIZE" value="800000" />
<p>
<input type="file" name="upload" />
</p>
<p>
<input type="submit" name="submit" value="process" />
</p>
</form>

</body>
</html>

-

<?

$imageid = $_GET['id'];

if ( mysqli_connect('localhost','root','adminpass','flip')){

    $escape = mysqli_real_escape_string($connect,$content);
    $sql = "select type,content from images where id =$imageid";
    $query = mysqli_query($connect, $sql);


    $fetch = mysqli_fetch_array($query,MYSQLI_ASSOC);
    mysqli_free_result($query);
    mysqli_close($connect);

    }

        else {
            echo"ther is no connection 2";
            }


if (!empty($fetch)){

    header("Content-type:{$fetch['type']}");
    echo $fetch['content'];

    }

?>

Answer 1

使用isset

isset - 确定变量是否设置且不是NULL

<?
if (isset($_POST['submit']) && !empty($_FILES)){

为什么我有这些错误？

1 个答案: