小编Aas*_*zam的帖子

我正在按照教程,但本教程的作者不回答问题 - 但这是我的查询

我收到以下错误警告:mysqli_error()预计正好1个参数,0给出,问题是这行代码 -

$query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error()); 

整个代码是

session_start();

require_once "scripts/connect_to_mysql2.php";

//Build Main Navigation menu and gather page data here

$sqlCommand = "SELECT id, linklabel FROM pages ORDER BY pageorder ASC";

$query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error()); 

$menuDisplay = '';
while ($row = mysqli_fetch_array($query)) { 
    $pid = $row["id"];
    $linklabel = $row["linklabel"];

    $menuDisplay .= '<a href="index.php?pid=' . $pid . '">' . $linklabel . '</a><br />';

} 
mysqli_free_result($query); 

包含的文件包含以下行

$myConnection = mysqli_connect("$db_host","$db_username","$db_pass","$db_name") or die ("could not …