我看不到:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
任何人都可以提供帮助,因为我的代码看起来很准确:
<?php
//retreive questions from database and put into question box
$query = "SELECT `QuestionId`, `Question`, `Opt1`, `Opt2`, `Opt3`, `Opt4`,`Answer` FROM `pf_questions`";
$question = mysql_query($query);
while($row = mysql_fetch_array($question)){
$id = $row['QuestionId'];
$question = $row['Question'];
$opt1 = $row['Opt1'];
$opt2 = $row['Opt2'];
$opt3 = $row['Opt3'];
$opt4 = $row['Opt4'];
$answer = $row["Answer"];
?>
<div id="ContainerQuestion">
<span class="Question">Question <?php echo $id; ?>. <?php echo $question; ?></span>
<p><input type=radio name='q<?php echo $id; ?>' value="<?php echo $opt1; ?>"> <?php echo $opt1; ?> </p>
<p><input type=radio name='q<?php echo $id; ?>' value="<?php echo $opt2; ?>"> <?php echo $opt2; ?> </p>
<p><input type=radio name='q<?php echo $id; ?>' value="<?php echo $opt3; ?>"> <?php echo $opt3; ?> </p>
<p><input type=radio name='q<?php echo $id; ?>' value="<?php echo $opt4; ?>"> <?php echo $opt4; ?> </p>
</div>
<?php
}
已经尝试过 mysql_error() 并且没有任何输出,所以我假设我的查询是正确的?
非常感谢