这是我正在制作的一个投票系统,这段代码向用户显示了他们可以从中挑选的问题列表:
<div class="main_questions">
<p class="style1 style2"><strong>Select Your Question</strong></p>
<p class="style1">
<form action="vote_list.php" method="post" name="form1" class="style1">
<?php
$sql = "SELECT DISTINCT (question_tba) FROM question ORDER BY answer_id DESC";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result)){
?>
<p>
<?php echo $row['question_tba']; ?>
<input type="radio" name="questions" value="<?php echo $row['question_tba']; ?>">
</p>
<?php
}
?>
<input type="submit" name="submit" value="Vote">
</p>
</div>然后,此代码应将选择的问题发布到此页面,以允许他们投票:
<?php
include('core/initialise.inc.php');
$q = $_POST['question_tba'];
if (isset($_GET['vote'], $_GET['id'])){
add_vote($_GET['id'], $_GET['vote'], $q);
echo "<META HTTP-EQUIV=\"REFRESH\" CONTENT=\"0; URL=vote_logged.php\">";
}
?>
</div>
<div class="main_questions">
<p class="style1 style2"><strong>Place Your Vote!</strong></p><?php
foreach (get_answers($q) as $id => $answer){
?>
<p>
<?php echo $answer; ?>
<a href="?vote=up&id=<?php echo $id; ?>">Vote</a>
</p>
<?php
}
?>
</div>然后,当他们点击一个答案进行投票时,此页面上的函数应使用以下函数将所选答案的投票增加1:
<?php
function get_answers($q){
$q = $q;
$sql = "SELECT answer_id, answer FROM question WHERE question_tba = '$q'";
$result = mysql_query($sql);
echo "$result";
$answers = array();
while (($row = mysql_fetch_assoc($result)) or die(mysql_error()))
{
$answers[$row['answer_id']] = $row['answer'];
}
return $answers;
}
function add_vote($answer_id, $vote, $q2){
$q2 = $q2;
$answer_id = (int)$answer_id;
$vote = '+';
$sql = "UPDATE question SET answer_votes = answer_votes $vote 1 WHERE question_tba = $q2 AND answer_id = $answer_id";
mysql_query($sql);
}
?>然而,我的问题是,当我点击我想要投票的问题时,它只显示了资源id #6,而不是显示我可以选择投票的答案。有人能告诉我我的代码有什么问题吗?
发布于 2012-03-21 09:58:59
好了,我把这事解决了。事实证明,在发布单选按钮值时,您必须使用name="“而不是value="”非常简单的解决方案。
发布于 2012-03-08 21:49:23
$result是由mysql_query()调用返回的资源,而不是实际的行对象/数组。与在代码的其他区域使用mysql_fetch_assoc()提取数据的方式相同,如果您想要显示数据,则需要在echo之前执行此操作。
https://stackoverflow.com/questions/9618604
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