Jam*_*ant 0 html php mysql phpmyadmin
这是我的html表单的代码:
<!DOCTYPE html>
<head>
<title> Question </title>
<style type = "text/css">
body {
font-family:cursive;
}
a:link {
text-decoration:none;
background-color:#D0D0D0;
color:#000000;
width:100px;
display:block;
text-align:center;
padding:4px;
}
a.visited {
text-decoration:none;
background-color:#D0D0D0;
color:#000000;
width:100px;
display:block;
text-align:center;
padding:4px;
}
a.active {
text-decoration:none;
background-color:#D0D0D0;
color:#000000;
width:100px;
display:block;
text-align:center;
padding:4px;
}
a:hover {
background-color:#686868;
color:#FFFFFF;
}
#title {
text-align:center;
}
</style>
</head>
<body>
<?php
session_start();
?>
<h1 id="title"> Question 1 </h1>
<br/>
<form action="q15.php" method="POST" >
<fieldset>
<legend>Who wrote the music we all recognise from the Paralympics?</legend>
<p>
<input
type="checkbox"
value="your friend"
name="answer"
/>Your Friend
</p>
<p>
<input
type="checkbox"
value="public friend"
name="answer"
/>Public Friend
</p>
<p>
<input
type="checkbox"
value="your enemy"
name="answer"
/>Your Enemy
</p>
<p>
<input
type="checkbox"
value="public enemy"
name="answer"
/>Public Enemy
</p>
<p>
<input
type="submit"
value="Submit"
/>
</p>
</fieldset>
</form>
</body>
</html>
这是我的页面的代码,它将处理数据并更新一个空白空格的数据库,以便稍后填写(如现在)
<body>
<h1 id="title"> Quiz </h1>
<?php
session_start();
$connection = mysql_connect("mysql15.000webhost.com", "a4987634_quiz", "********")
or die (mysql_error());
mysql_select_db("a4987634_quiz", $connection)
or die (mysql_error());
$fname = $_SESSION['fname'];
$lname = $_SESSION['lname'];
$id = $_SESSION['ID'];
$answer = $_POST['answer'];
$id = mysql_query("SELECT ID FROM users WHERE fname=$fname LIMIT 1");
if(isset($_POST['answer']) &&
$_POST['answer'] == 'public enemy')
{
?>
<h3 id = "correct"> Correct </h3>
<?php
$sqlcorrect = "UPDATE users SET q1 = correct WHERE ID = $ID LIMIT 1";
mysql_query($sqlcorrect);
(mysql_error());
}
else {
?>
<h3 id = "incorrect"> Incorrect </h3>
<?php
$sqlwrong = "UPDATE users SET q1 = 'wrong' WHERE ID = $ID LIMIT 1";
mysql_query($sqlwrong);
(mysql_error());
}
?>
</body>
</html>
我可以完美地连接到数据库,它知道什么时候你的问题正确或不正确但我的问题是当你尝试更新数据库时它不会这样做.有没有人有任何解决方案？也没有错误消息.它没有意义!
您的查询有几个问题:
当id提供权利时,这应该有效:
$sqlwrong = "UPDATE `users` SET `q1` = 'some text' WHERE `ID` = $id";