警告：mysqli_error()只需要1个参数，给定错误为0

Question

我得到以下错误

警告：mysqli_error()只需要1个参数，给定0

问题出在代码的这一行：

$query = mysqli_query($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); 

包含的文件有以下一行

$myConnection = mysqli_connect("$db_host","$db_username","$db_pass","$db_name") or die ("could not connect to mysql"); with reference to $myConnection, why do I get this error?

Answer 1

mysqli_error函数需要$myconnection作为参数，这就是您得到警告的原因

Answer 2

mysqli_error()需要您将连接作为参数传递给数据库。这里的文档有一些有用的示例：

http://php.net/manual/en/mysqli.error.php

试着像这样修改你的问题行，你应该会处于良好的状态：

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