我创建了一个PHP页面,它应该从数据库中选择两个名称并显示它们.
它只是说:
<?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>
我的代码是:
<?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>
Chi*_*tte 26
问题是你没有保存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
ORDER BY RAND()
LIMIT 1");
$aName1 = mysqli_fetch_assoc($query1);
$name1 = $aName1['name1'];
另外,不要忘记将连接变量作为字符串包含在上面.这就是导致错误的原因,但是你使用的函数是错误的,mysqli_query返回一个查询对象但是为了获取数据,你需要使用像mysqli_fetch_assoc这样的用户http://php.net/manual/en/mysqli-result .fetch-assoc.php实际上将数据输出到变量中,如上所述.
小智 5
除mysqli_query2个参数外，第一个变量是mysqli_connect等效变量，第二个是您提供的查询
$name1 = mysqli_connect(localhost,tdoylex1_dork,dorkk,tdoylex1_dork);
$name2 = mysqli_query($name1,"SELECT name FROM users ORDER BY RAND() LIMIT 1");