警告:mysqli_query() 需要至少 2 個參數，1 個給定.什
Warning: mysqli_query() expects at least 2 parameters, 1 given. What?(警告:mysqli_query() 需要至少 2 個參數，1 個給定.什么?)
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本文介紹了警告:mysqli_query() 需要至少 2 個參數，1 個給定.什么?的處理方法，對大家解決問題具有一定的參考價值，需要的朋友們下面隨著小編來一起學習吧！
問題描述
我制作了一個 PHP 頁面，該頁面應該從數據庫中選擇兩個名稱并顯示它們.
它只是說:
<塊引用>
警告:mysqli_query() 需要至少 2 個參數，1 個在第 4 行的/home/tdoylex1/public_html/dorkhub/index.php 中給出
警告:mysqli_query() 需要至少 2 個參數，1 個在第 8 行的/home/tdoylex1/public_html/dorkhub/index.php 中給出
我的代碼是:
<?php mysqli_connect(localhost,tdoylex1_dork,dorkk,tdoylex1_dork);$name1 = mysqli_query("SELECT name1 FROM users按蘭德 () 排序限制 1");$name2 = mysqli_query("SELECT name FROM users按蘭德 () 排序限制 1");?><title>DorkHub.在線姓名評級網站.</title><link rel="樣式表";類型=文本/css"；href=style.css"><body bgcolor='EAEAEA'><中心><div id='標題'><h2>DorkHub.在線名稱評級網站.</h2><p><br><h3><?php echo $name1;?></h3><h4>反對</h4><h3><?php echo $name1;?></h3><br><br><h2 style='font-family:Arial, Helvetica, sans-serif;'>誰的聲音最笨?</h2><br><br><div id='投票'><h3 id='done' style='margin-right: 10px'>投票給第一個</h3><h3 id='done'>投票給最后一個</h3>
解決方案
問題是您沒有保存 mysqli 連接.將您的連接更改為:
$aVar = mysqli_connect('localhost','tdoylex1_dork','dorkk','tdoylex1_dork');然后將其包含在您的查詢中:
$query1 = mysqli_query($aVar, "SELECT name1 FROM users按蘭德 () 排序限制 1");$aName1 = mysqli_fetch_assoc($query1);$name1 = $aName1['name1'];另外不要忘記將您的連接變量作為字符串括起來，就像我上面所說的那樣.這就是導致錯誤的原因，但您使用的函數錯誤，mysqli_query 返回一個查詢對象，但要從中獲取數據，您需要使用類似 mysqli_fetch_assoc http://php.net/manual/en/mysqli-result.fetch-assoc.php 實際獲取數據一個變量，我上面有.
I made a PHP page that is supposed to select two names from a database and displays them.
It just says:
Warning: mysqli_query() expects at least 2 parameters, 1 given in /home/tdoylex1/public_html/dorkhub/index.php on line 4
Warning: mysqli_query() expects at least 2 parameters, 1 given in /home/tdoylex1/public_html/dorkhub/index.php on line 8
My code is:
<?php mysqli_connect(localhost,tdoylex1_dork,dorkk,tdoylex1_dork);
$name1 = mysqli_query("SELECT name1 FROM users
ORDER BY RAND()
LIMIT 1");
$name2 = mysqli_query("SELECT name FROM users
ORDER BY RAND()
LIMIT 1");
?>
<title>DorkHub. The online name-rating website.</title>
<link rel="stylesheet" type="text/css" href="style.css">
<body bgcolor='EAEAEA'>
<center>
<div id='TITLE'>
<h2>DorkHub. The online name-rating website.</h2>
</div>
<p>
<br>
<h3><?php echo $name1; ?></h3><h4> against </h4><h3><?php echo $name1; ?></h3>
<br><br>
<h2 style='font-family:Arial, Helvetica, sans-serif;'>Who's sounds the dorkiest?</h2>
<br><br>
<div id='vote'>
<h3 id='done' style='margin-right: 10px'>VOTE FOR FIRST</h3><h3 id='done'>VOTE FOR LAST</h3>
解決方案
The issue is that you're not saving the mysqli connection. Change your connect to:
$aVar = mysqli_connect('localhost','tdoylex1_dork','dorkk','tdoylex1_dork');
And then include it in your query:
$query1 = mysqli_query($aVar, "SELECT name1 FROM users
ORDER BY RAND()
LIMIT 1");
$aName1 = mysqli_fetch_assoc($query1);
$name1 = $aName1['name1'];
Also don't forget to enclose your connections variables as strings as I have above. This is what's causing the error but you're using the function wrong, mysqli_query returns a query object but to get the data out of this you need to use something like mysqli_fetch_assoc http://php.net/manual/en/mysqli-result.fetch-assoc.php to actually get the data out into a variable as I have above.
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