Question

我写这个时有问题：

<body>
<form action="new 1.php" method="POST" enctype="multi/form-data">
File:
<input type="file" name="image"><input type="submit" value="upload">

</form>
<?php


$file = $_FILES['image']['tmp_name'];

$dbhost     = "localhost";
$dbname     = "testtest";
$dbuser     = "root";
$dbpass     = "";


$conn = new PDO("mysql:host=$dbhost;dbname=$dbname",$dbuser,$dbpass);


?>

它写给我：未定义的索引：C：\ Program Files（x86）\ EasyPHP-DevServer-14.1VC11 \ data \ localweb \ projects \ guysite2 \ new 1.php中的图像在第11行

我尝试做：if（isset（$ _ FILES [＆＃39; image＆＃39;]））但是，有人可以帮助我吗？

Answer 1

替换

<form action="new 1.php" method="POST" enctype="multi/form-data">

到

<form action="new 1.php" method="POST" enctype="multipart/form-data">

<强> UPD：这段代码对我有用：

<?php
    if (isset($_FILES['image'])) {
        $file = $_FILES['image']['tmp_name'];
        echo $file . '<br>';
    }
?>

<form action="new 1.php" method="POST" enctype="multipart/form-data">
    File: <input type="file" name="image">
    <input type="submit" value="upload">
</form>

php图像数据库出错

1 个答案: