以下代码仅抓取PHP while循环中最后一个结果的值。任何/所有的帮助非常感谢。谢谢!jquery只返回php while循环的最后结果
PHP/HTML/JS:
<section id="info">
<?php
$user = $session->username;
$q = sprintf("SELECT * FROM mail WHERE UserTo = '%s' ORDER BY SentDate DESC",
mysql_real_escape_string($user));
$getMail = mysql_query($q, $link) or die(mysql_error());
if(mysql_num_rows($getMail) == 0) {
echo "<p>you have no mail</p>";
}
else {
?>
<form id="inbox" class="mail">
<fieldset>
<ul>
<li style="border: 2px solid purple; width: 100%;">
<span style="width: 8%; margin-left: 13%;">Status</span>
<span style="width: 15%;">From</span>
<span style="width: 45%;">Subject</span>
<span style="width: 16%;">Time</span>
</li>
<?php
while($mail = mysql_fetch_object($getMail)) {
$status = $mail->status;
$mailId = $mail->mail_id;
$from = $mail->UserFrom;
$subject = $mail->Subject;
$received = $mail->SentDate;
$theMessage = $mail->Message;
?>
<li class="outerDiv" style="border: 2px dotted purple;">
<button style="display: inline;" class="viewButton">View</button>
<button style="display: inline;">Delete</button>
<?php
echo "<span>" . $mail_id . "</span>";
echo "<span style="display: inline-block; width: 8%; border: 1px solid red;'>" . $status . "</span>";
echo "<span style='display: inline-block; width: 15%; border: 1px solid red;'>" . $from . "</span>";
echo "<span style='display: inline-block; width: 45%; border: 1px solid red;'>" . $subject . "</span>";
echo "<span style='display: inline-block; font-size: x-small; width: 17%; border: 1px solid red;'>" . $received . "</span>";
?>
</li>
<?php }
} ?>
</ul>
</fieldset>
</form>
</section>
<section id="details">
<div class="theMessage" style="display: none;"><?php echo $theMessage; ?></div>
<script type="text/javascript">
$(document).ready(function() {
$(".outerDiv").click(function(e) {
if($(e.target).is(".viewButton")) {
$(".theMessage").fadeIn(1000);
}
});
return false;
});
</script>
</section>
我不认为这个问题与HTML或jQuery有任何关系。请仅放置相关代码,以便其他人能够轻松理解并回答您的问题。 – Starx
等等究竟是什么问题? – kaveman
你的代码是正确的,据我所见,只是可以肯定的是,在使用'while'循环之前,尝试查看输出mysql_num_rows($ getMail)'它必须是一个.... – Starx