我试图上传图片文件,我想显示该图片的信息。但是我的代码无法正常工作,我不确定是什么问题,我希望有人能指出我做错了什么。上传图片并显示信息
<!DOCTYPE html>
<html>
<head>
<title>Upload Image</title>
</head>
<body>
<form action="upload_image.php" action="post" enctype="multipart/form-data">
Select image: <input type="file" name="image">
<input type="submit" name="upload" value="Upload Now">
</form>
<?php
if(isset($_POST['upload'])){
echo $image_name = $_FILES['image']['name'];
echo $image_type = $_FILES['image']['type'];
echo $image_size = $_FILES['image']['size'];
echo $image_tmp_name = $_FILES['image']['tmp_name'];
if($image_name == ""){
echo "<script>alert('Please select an image!')</script>";
exit();
}
}
?>
</body>
</html>
PS:我还是新的PHP和文件上传,所以请去容易对me..Thank你提前
您使用的是哪个版本的PHP? –
@ChaibiAlaa php5 – blackhorse123
上传....,在哪里? –