警告:mysqli_error() 需要 1 个参数，0 给定错误 [英] Warning: mysqli_error() expects exactly 1 parameter, 0 given error

问题描述

我收到以下错误

<块引用>

警告:mysqli_error() 需要 1 个参数，0 给定

问题在于这行代码:

$query = mysqli_query($myConnection, $sqlCommand) 要么死 (mysqli_error());

整个代码是

session_start();require_once "scripts/connect_to_mysql2.php";//构建主导航菜单并在此处收集页面数据$sqlCommand = "SELECT id, linklabel FROM pages ORDER BY pageorder ASC";$query = mysqli_query($myConnection, $sqlCommand) 要么死 (mysqli_error());$menuDisplay = '';而 ($row = mysqli_fetch_array($query)) {$pid = $row["id"];$linklabel = $row["linklabel"];$menuDisplay .= '<a href="index.php?pid='. $pid .'">'.$链接标签.'</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") 要么死("无法连接到mysql")；参考 $myConnection，为什么会出现此错误?

解决方案

mysqli_error() 需要你将连接作为参数传递给数据库.此处的文档提供了一些有用的示例:

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

试着像这样改变你的问题线，你应该处于良好的状态:

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

I get the following error

Warning: mysqli_error() expects exactly 1 parameter, 0 given

The problem is with this line of the code:

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

The whole code is

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); 

The included file has the following line

$myConnection = mysqli_connect("$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?

解决方案

mysqli_error() needs you to pass the connection to the database as a parameter. Documentation here has some helpful examples:

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

Try altering your problem line like so and you should be in good shape:

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

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