MFA*_*MFA 0 php mysql forms echo
我的代码似乎没有工作..单选按钮出现但旁边没有任何东西..似乎mysql_fetch_array由于某种原因不能正常工作,因为我已经玩了代码并反复测试它以找到代码似乎遇到的位置一个问题,并停止工作..有人可以请问有什么问题吗？欢呼ps.我是新手,最近几天才完成学习w3schools的php教程.
<body>
<?php
include 'dbyear2.php';
$qnumber = $_REQUEST['uqn']; // obtain question number from URL
$find = mysql_query("SELECT * FROM Renal WHERE UQN='$qnumber'");
while($retrieve=mysql_fetch_array($find));
{
$retrieve['question'] = $question;
$retrieve['MCQ_A'] = $a;
$retrieve['MCQ_B'] = $b;
$retrieve['MCQ_C'] = $c;
$retrieve['MCQ_D'] = $d;
$retrieve['MCQ_E'] = $e;
$retrieve['answer'] = $answer;
$retrieve['MCQ_correct'] = $correct;
}
?>
<form action='check.php' method='POST'>
<table>
<tr><td></td><td></td></tr>
<tr></tr>
<tr><td><input type='radio' name='group1' value='A' /></td><td> <?php echo $a; ?></td></tr>
<tr><td><input type='radio' name='group1' value='B' /></td><td> <?php echo $b; ?></td></tr>
<tr><td><input type='radio' name='group1' value='C' /></td><td> <?php echo $c; ?></td></tr>
<tr><td><input type='radio' name='group1' value='D' /></td><td> <?php echo $d; ?></td></tr>
<tr><td><input type='radio' name='group1' value='E' /></td><td> <?php echo $e; ?></td></tr>
<tr>
<?php
// sending the retrieved information from MYSQL via POST for use in check.php file
$qnumber;
$a;
$b;
$c;
$d;
$e;
$answer;
$correct;
?></tr>
<tr><td><input type="submit" value="Submit"></td></tr>
</table>
</form>
</body>
</html>
这部分是倒退的:
$retrieve['question'] = $question;
$retrieve['MCQ_A'] = $a;
$retrieve['MCQ_B'] = $b;
$retrieve['MCQ_C'] = $c;
$retrieve['MCQ_D'] = $d;
$retrieve['MCQ_E'] = $e;
$retrieve['answer'] = $answer;
$retrieve['MCQ_correct'] = $correct;
应该
$question = $retrieve['question' ;
$a = $retrieve['MCQ_A'];
$b = $retrieve['MCQ_B'];
$c = $retrieve['MCQ_C'];
$d = $retrieve['MCQ_D'];
$e = $retrieve['MCQ_E'];
$answer = $retrieve['answer'];
$correct $retrieve['MCQ_correct'];
请不要mysql_*在新代码中使用函数.它们不再维护,并且已被正式弃用.看到红色的盒子？了解准备好的语句,并使用PDO或MySQLi - 本文将帮助您确定哪些.如果您选择PDO,这是一个很好的教程.
你不应该使用w3schools.它不是可靠的信息来源,我们不想鼓励使用它.