我收到以下错误
警告:mysqli_error()恰好需要1个参数,给定0
问题在于这行代码:
$query = mysqli_query($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);
包含的文件具有以下行
$myConnection = mysqli_connect("$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?
mysqli_error()需要您将连接作为参数传递给数据库。这里的文档提供了一些有用的示例:
http://php.net/manual/zh/mysqli.error.php
尝试像这样更改问题线,您应该处于良好状态:
$query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error($myConnection));