問題描述

我收到以下錯誤

警告:mysqli_error() 需要 1 個參數，0 給定

問題在于這行代碼:

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

整個代碼是

session_start();require_once "scripts/connect_to_mysql2.php";//構建主導航菜單并在此處收集頁面數據$sqlCommand = "SELECT id, linklabel FROM pages ORDER BY pageorder ASC";$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='. $pid .'">'.$鏈接標簽.'</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("無法連接到mysql")；參考 $myConnection，為什么會出現此錯誤?

mysqli_error() 需要你將連接作為參數傳遞給數據庫.此處的文檔提供了一些有用的示例:

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

試著像這樣改變你的問題線，你應該處于良好的狀態:

$query = mysqli_query($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?

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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