我收到以下错误
警告:mysqli_error() 需要 1 个参数,给定 0 个参数
问题出在这行代码上:
$query = mysqli_query($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。这里的文档有一些有用的示例:
http://php.net/manual/en/mysqli.error.php
尝试像这样改变你的问题线,你应该处于良好状态:
$query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error($myConnection));
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