Mysqli_query返回false而Mysqli_error返回NULL？

$query = "INSERT INTO appointmentinformation (ClientID, Length, Date, Month, Year, Time, Price) VALUES ('$clientid', '$length', '$date', '$month', '$year', '$time', '$price')";
    $result = mysqli_query($connection, $query);
    if ($result) {
        header("Location:UserCP.php");
    } else {
        echo ("Could not insert data : " . mysqli_error($result) . " " . mysqli_errno());
    }

mysqli_query返回的结果为null.这会将您发送到代码的else分支.然后你编写了mysqli_error($ result),它等于mysqli_error(null).

我读过的文档说是用查询的"链接"初始化变量.你这样做:

 $cOnnection= mysqli_connect("$servername", "$username", "$password", "$dbname") or die("Could not connect to the database");

您现在想将其编码为mysqli_error($ connection)和mysqli_errno($ connection).

另外一个建议.在mysqli_connect语句后面添加此代码.

if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }