PHP：为什么不获取我将插入数据库的图像数据？

Question

I've the next problem: The code to validate the name, address and email works, but to validate the image does not work and do not know why that happens. As always prints "Select an image"

I did a test and set the line to print the temporary name of the image with "print $photo" but nothing prints!

form.php

<form id="formregister" name="formregister" action="register.php" method="post">
    <input id="name" name="name" type="text"/>
    <input id="address" name="address"  type="text"/>
    <input id="email" name="email" type="email" />
    <input id="photouser" name="imguser" type="file"/>
    <input id="bregister" name="bregister" type="submit" value="Registrer"/>
</form>

register.php

..............
$name = $_POST['name'];
$address = $_POST['address'];
$email = $_POST['email'];
$photo = $_FILES['imguser']['tmp_name'];
$errors = array();

    if(!isset($photo)) {
        $errors[] = "Select an image";
    } else {
        $fototemp = addslashes(file_get_contents($_FILES['imguser']['tmp_name']));
        $fotoname = addslashes($_FILES['imguser']['name']);
        $fotosize = getimagesize($_FILES['imguser']['tmp_name']);
        if ($fotosize == false) {
            $errors[] = "Invalid format";
        }
    } 
//Code to validate the other fields (name, address, email)
....................

Answer 1

You are missing the proper enctype for file uploads on your form. It should be enctype='multipart/form-data'

<form id="formregister" name="formregister" action="register.php" method="post" enctype='multipart/form-data'>

Rather than accessing the secondary tmp_name key directly, test if it is set first to avoid undefined index notices:

if (isset($_FILES['imguser']) {
  $photo = $_FILES['imguser']['tmp_name'];
}