警告:mysqli_query() 需要至少 2 个参数，1 个给定.什么?
Warning: mysqli_query() expects at least 2 parameters, 1 given. What?(警告:mysqli_query() 需要至少 2 个参数，1 个给定.什么?)
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问题描述
我制作了一个 PHP 页面，该页面应该从数据库中选择两个名称并显示它们.
它只是说:
<块引用>
警告:mysqli_query() 需要至少 2 个参数，1 个在第 4 行的/home/tdoylex1/public_html/dorkhub/index.php 中给出
警告:mysqli_query() 需要至少 2 个参数，1 个在第 8 行的/home/tdoylex1/public_html/dorkhub/index.php 中给出
我的代码是:
<?php mysqli_connect(localhost,tdoylex1_dork,dorkk,tdoylex1_dork);$name1 = mysqli_query("SELECT name1 FROM users按兰德 () 排序限制 1");$name2 = mysqli_query("SELECT name FROM users按兰德 () 排序限制 1");?><title>DorkHub.在线姓名评级网站.</title><link rel="样式表";类型=文本/css"；href=style.css"><body bgcolor='EAEAEA'><中心><div id='标题'><h2>DorkHub.在线名称评级网站.</h2>
<p><br><h3><?php echo $name1;?></h3><h4>反对</h4><h3><?php echo $name1;?></h3><br><br><h2 style='font-family:Arial, Helvetica, sans-serif;'>谁的声音最笨?</h2><br><br><div id='投票'><h3 id='done' style='margin-right: 10px'>投票给第一个</h3><h3 id='done'>投票给最后一个</h3>
解决方案
问题是您没有保存 mysqli 连接.将您的连接更改为:
$aVar = mysqli_connect('localhost','tdoylex1_dork','dorkk','tdoylex1_dork');
然后将其包含在您的查询中:
$query1 = mysqli_query($aVar, "SELECT name1 FROM users按兰德 () 排序限制 1");$aName1 = mysqli_fetch_assoc($query1);$name1 = $aName1['name1'];
另外不要忘记将您的连接变量作为字符串括起来，就像我上面所说的那样.这就是导致错误的原因，但您使用的函数错误，mysqli_query 返回一个查询对象，但要从中获取数据，您需要使用类似 mysqli_fetch_assoc http://php.net/manual/en/mysqli-result.fetch-assoc.php 实际获取数据一个变量，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>
解决方案
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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