当我运行下面的代码,它说为什么我不能在php中更改输入文件名?
“通知:未定义指数:uploadFile在C:\ XAMPP \ htdocs中\ ImageTest \ processImage.php 17行”
但是,当我更换使用fileToUpload的uploadFile的每个实例似乎都可以工作。为什么?
processImage.php
<!DOCTYPE html>
<html>
<head>
<title> hello world</title>
</head
<body>
<?php
echo ' hi';
$servername="localhost";
$username="root";
$password="";
$dbname="db_ImageTest";
$conn=new mysqli($servername, $username, $password, $dbname);
echo $_FILES["uploadFile"]["name"];
/*
echo $image;
$image_name=$_FILES['image']['name'];
$image_size=getimagesize($_FILES['image']['tmp_name']);
if($image_size==FALSE){
echo 'failed';
}
$query="INSERT INTO mytable(image, name) VALUES(' {$image}', '{$image_name}')
*/
?>
</body>
</html>
的index.php
<!DOCTYPE html>
<html>
<body>
<form action="processImage.php" method="post" enctype="multipart/form-data">
Select image to upload:
<input type="file" name="uploadFile" />
<input type="submit" value="Upload Image" name="submit"/>
</form>
</body>
</html>
当你加载html文件..表单中的文件没有上传,它试图访问它为什么会抛出警告,你可以解决'if($ _ POST){echo your code here};' –
Can你发布var_dump($ _ POST,$ _FILES)的结果; –
echo var_dump($ _ POST,$ _FILES); (1){[“submit”] => string(12)“Upload Image”} array(1){[fileToUpload“] => array(5){[”name“] => string(9) “pupil.png”[“type”] => string(9)“image/png”[“tmp_name”] => string(24)“C:\ xampp \ tmp \ phpEC09.tmp”[“error”] = > int(0)[“size”] => int(585)}} – user2350459