Question

我写了一段代码但是没有用。下面是一个代码，用于在用户输入上将图像输入数据库。但我的问题是，当我打电话给它打印在页面上时，没有什么是透露的，请在这里帮助我。

<?php
$conx=mysqli_connect('localhost','root','sultan','colour');
$sql = "SELECT theme from  colors WHERE username='shail' ";
$sqli=mysqli_query($conx,$sql);
$row=mysqli_fetch_row($sqli);
$theme=$row[0];


if(isset($_POST['submit']))
{
$theme=$_POST['theme'];
$sql = "UPDATE colors SET theme='$theme' WHERE username='shail' ";
$sqli=mysqli_query($conx,$sql);

}

?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<style>
body{ background-image:url(<?php echo $theme  ?>) ;}

</style>
</head>

<body>
<form method="post">
Input your picture<input type="file" name="theme" /><br />
<input type="submit" value="upload" name="submit" />

</form>
</body>
</html>

Answer 1

您需要终止该声明。

电流

<?php echo $theme  ?>

新

<?php echo $theme;  ?>

Answer 2

<?php
$conx=mysqli_connect('localhost','root','','colour');
$sql = "SELECT theme from  colors WHERE username='shail' ";
$sqli=mysqli_query($conx,$sql);
$row=mysqli_fetch_row($sqli);
$theme=$row[0];


if(isset($_POST['submit']))
{
        // new code
        $path = '/';
        $fileName = $_FILES['theme']["name"]; 
        $fileTmpLoc = $_FILES['theme']["tmp_name"];
        $temp = explode(".", $_FILES['theme']["name"]);
        $extension = end($temp);
        $temp = explode('.', $fileName );
        $ext = array_pop($temp );
        $name = implode('.', $temp );
        $FileNM = $name.time().".".$extension;
        $theme = $FileNM.$ext;
        move_uploaded_file($_FILES['theme']["tmp_name"], $theme);

        $sql = "UPDATE colors SET theme='$theme' WHERE username='shail' ";
        $sqli=mysqli_query($conx,$sql);
}

?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<style>
    body{ background-image:url(<?php echo $theme;  ?>) ;}
</style>
</head>

<body>
<form method="post" enctype="multipart/form-data">
Input your picture<input type="file" name="theme" /><br />
<input type="submit" value="upload" name="submit" />

</form>
</body>
</html>

Answer 3

<?php
$conx=mysqli_connect('localhost','root','','colour');
$sql = "SELECT theme from  colors WHERE username='shail' ";
$sqli=mysqli_query($conx,$sql);
$row=mysqli_fetch_row($sqli);
$theme=$row[0];


if(isset($_POST['submit']))
{
        // new code
        $error1="";
        $error2="";
        if ($_FILES["theme"]["type"] !== "image/jpeg"){
            echo $error1 = "File is not in JPG!";
        }else{
            $error1="";
        }

        if(($_FILES["theme"]["size"] > 1024)){
            echo $error2 = "File size not allowed more than 1 MB!";
        }else{
            $error2="";
        }

        if($error1 =="" && $error2==""){
            $fileName = $_FILES['theme']["name"]; 
            $fileTmpLoc = $_FILES['theme']["tmp_name"];
            $temp = explode(".", $_FILES['theme']["name"]);
            $extension = end($temp);
            $temp = explode('.', $fileName );
            $ext = array_pop($temp );
            $name = implode('.', $temp );
            $FileNM = $name.time().".".$extension;
            move_uploaded_file($_FILES['theme']["tmp_name"], "img/".$FileNM);

            $sql = "UPDATE colors SET theme='$FileNM' WHERE username='shail' ";
            $sqli=mysqli_query($conx,$sql);
        }

}

?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<style>
    body{ background-image:url(<?php echo "img/".$FileNM;  ?>) ;}
</style>
</head>

<body>
<form method="post" enctype="multipart/form-data">
Input your picture<input type="file" name="theme" /><br />
<input type="submit" value="upload" name="submit" />
</form>
</body>
</html>

我没有得到这个脚本的输出

3 个答案: