警告:mysqli_query() 需要至少 2 个参数，1 个给定.什么?

2021-12-25 00:00:00 php mysqli

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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