PHP和MySQL提交错误消息问题

Question

When I submit a new name and not a new avatar I get the following avatar error message Please upload a .gif, .jpeg, .jpg or .png image!. I want to be able to send a new name only without having to upload a new avatar each time I submit the form without getting the avatar error message Please upload a .gif, .jpeg, .jpg or .png image! can someone help me fix this problem?

Here is the php code.

if (isset($_POST['submitted'])) {

    $mysqli = mysqli_connect("localhost", "root", "", "sitename");
    $dbc = mysqli_query($mysqli,"SELECT users.*
                                 FROM users 
                                 WHERE user_id=3");

    $first_name = mysqli_real_escape_string($mysqli, htmlentities(strip_tags($_POST['first_name'])));

$user_id = '3';

if(isset($_FILES["avatar"]["name"]) && $_FILES['avatar']['size'] <= 5242880) {

    if($_FILES["avatar"]["type"] == "image/gif" || $_FILES["avatar"]["type"] == "image/jpeg" || $_FILES["avatar"]["type"] == "image/jpg" || $_FILES["avatar"]["type"] == "image/png" || $_FILES["avatar"]["type"] == "image/pjpeg") {

        if (file_exists("../members/" . $user_id . "/images/" . $_FILES["avatar"]["name"])) {
          echo '<p class="error">' . mysqli_real_escape_string($mysqli, htmlentities(strip_tags(basename($_FILES["avatar"]["name"])))) . ' already exists! ';
        } else if($_FILES["avatar"]["name"] == TRUE) {
          move_uploaded_file($_FILES["avatar"]["tmp_name"],
          "../members/" . $user_id . "/images/" . mysqli_real_escape_string($mysqli, htmlentities(strip_tags(basename($_FILES["avatar"]["name"])))));
          $avatar = mysqli_real_escape_string($mysqli, htmlentities(strip_tags(basename($_FILES["avatar"]["name"]))));
        }

    } else if($_FILES["avatar"]["type"] != "image/gif" || $_FILES["avatar"]["type"] != "image/jpeg" || $_FILES["avatar"]["type"] != "image/jpg" || $_FILES["avatar"]["type"] != "image/png" || $_FILES["avatar"]["type"] != "image/pjpeg") {
        echo '<p class="error">Please upload a .gif, .jpeg, .jpg or .png image!</p>';
    }

} else if($_FILES['avatar']['size'] >= 5242880) {
    echo '<p class="error">Please upload a smaller pic!</p>';
} else if($_FILES["avatar"]["name"] == NULL) {
    $avatar = NULL;
}


if(isset($_FILES["avatar"]["name"]) && $_FILES['avatar']['size'] <= 5242880) {

        if (mysqli_num_rows($dbc) == 0) {
                $mysqli = mysqli_connect("localhost", "root", "", "sitename");
                $dbc = mysqli_query($mysqli,"INSERT INTO users (user_id, first_name, avatar) 
                                             VALUES ('$user_id', '$first_name', '$avatar')");
        }

        if ($dbc == TRUE) {
                $dbc = mysqli_query($mysqli,"UPDATE users 
                                             SET first_name = '$first_name', avatar = '$avatar' 
                                             WHERE user_id = '$user_id'");

                echo '<p class="changes-saved">Your changes have been saved!</p>';

        }

        if (!$dbc) {
                print mysqli_error($mysqli);
                return;
        }

    }

}

Answer 1

You simply need to add a check to see if they specified a new image for their avatar (or left it blank). In that case, without knowing how your form is set up, you could simply check that the size is greater than zero (isset()/empty() can return unexpected results in this case):

if($_FILES['avatar']['size'] > 0) { ... }

Edit In response to your comment, you'd change the following line (including fixing a basic logic error):

} else if($_FILES["avatar"]["type"] != "image/gif" || $_FILES["avatar"]["type"] != "image/jpeg" || $_FILES["avatar"]["type"] != "image/jpg" || $_FILES["avatar"]["type"] != "image/png" || $_FILES["avatar"]["type"] != "image/pjpeg") {

to:

} else if($_FILES["avatar"]["size"] <= 0 && $_FILES["avatar"]["type"] != "image/gif" && $_FILES["avatar"]["type"] != "image/jpeg" && $_FILES["avatar"]["type"] != "image/jpg" && $_FILES["avatar"]["type"] != "image/png" && $_FILES["avatar"]["type"] != "image/pjpeg") {