Question

我正在尝试将$userid插入我的数据库，但不是插入用户ID而是插入0.当我echo $user['id'];时，它返回用户ID。

if (!empty($_FILES)) {
    // Validate the file type
    $userid = trim(sanitize($user['id'])); //GOD DAMMIT HELP HERE
    $fileTypes = array('jpg', 'jpeg', 'gif', 'png', 'rar', 'zip', 'exe'); // File extensions
    $fileParts = pathinfo($_FILES['Filedata']['name']);
    $ext = trim(sanitize($fileParts['extension']));
    $file_name = trim(sanitize($_FILES['Filedata']['name']));
    $file_size = trim(sanitize(format_bytes($_FILES['Filedata']['size'])));

    if (in_array($fileParts['extension'],$fileTypes)) {
        mysql_query("INSERT INTO files (userid, name, ext, size) VALUES ('$userid', '$file_name', '$ext', '$file_size')"); //Upload the file and add to MySQL
        $fileid = mysql_insert_id();
        move_uploaded_file($_FILES['Filedata']['tmp_name'], $_SERVER['DOCUMENT_ROOT']."/uploads/$fileid." .$fileParts['extension']);
        echo '1';
    } else {
        echo 'Invalid file type.';
    }
}
?>

$ user ['id']函数

  //Use $user['value'] to retrieve data
  if ($_SESSION['username'] != null) {
    $u_query = mysql_query("SELECT * FROM users WHERE `username` = '$_SESSION[username]' LIMIT 1");
    $user = mysql_fetch_array($u_query);
  }

$ user的var_dump

array(20) { [0]=> string(2) "14" ["id"]=> string(2) "14" [1]=> string(4) "test" ["username"]=> string(4) "test" [2]=> string(128) "f94f06eac339c5c59685d008318b88cf74d047742cca202271a49ac90c44e0ca4e0953c57b5dcf589b3deeedd04a3d0807ba134c3e0cf0e371066e69ae92d22e" ["password"]=> string(128) "f94f06eac339c5c59685d008318b88cf74d047742cca202271a49ac90c44e0ca4e0953c57b5dcf589b3deeedd04a3d0807ba134c3e0cf0e371066e69ae92d22e" [3]=> string(14) "test@gmail.com" ["email"]=> string(14) "test@gmail.com" [4]=> string(14) "test@gmail.com" ["ppemail"]=> string(14) "test@gmail.com" [5]=> string(0) "" ["ip"]=> string(0) "" [6]=> string(1) "0" ["banned"]=> string(1) "0" [7]=> string(0) "" ["reason"]=> string(0) "" [8]=> string(1) "1" ["active"]=> string(1) "1" [9]=> string(8) "28127712" ["code"]=> string(8) "28127712" }

Answer 1

也许你的字段是数字的，因此你会给出一个字符串（你在'$ userid'的查询参数中使用你的userid周围的引号）mysql假设为0 ...

Answer 2

$userid = trim(sanitize($user['id']));
echo '$userid=' . $userid;
var_dump($user);

这会输出所需的值吗？另外，在您的第一个代码之前定义了$user吗？如果它返回0，则可能在您使用时未设置。

MySQL插入0而不是变量值

2 个答案: