Question

这可能很容易解决，但这段代码有什么问题？所有div的回声都没有到达终点，因此只有一部分被引用，从开头开始到结束但我不知道这有什么不对。

</header>
<?php $db_name = 'submissions';
$db_user = 'root';
$db_pass = '';
$db_host = 'localhost';
$link = mysql_connect("localhost", $db_user, $db_pass) or die(mysql_error());
mysql_select_db("submissions",$link) or die(mysql_error());
$query1 = mysql_query("SELECT actual_quote,poster,formtype FROM data WHERE formtype =    'Inspirational'",$link);   
$postid = 1;
$data = mysql_fetch_object(mysql_query("SELECT `like`, `unlike` FROM data WHERE     id='".$postid."'"));

while($info = mysql_fetch_array($query1)){
echo ' 
<div class="wrapper">
<div class="submissions">
<div class="logo-logo"><h2>Questions.</h2>
<div class="checkboxes">'.$info['formtype'].'
</div>

</div>
<div class="top-submit">
&#8220'. $info["actual_quote"] . '&#8221;
</div>
<div class="poster">- '. $info["poster"].'
<div class = "like">
<a href = "javascript:;" onclick = "doAction('<?php echo $postid; ?>', 'like');">Like    (<span id =' <?php echo $postid; ?>_likes'><?php echo $data->like;?></span>)</a>

</div>
<div class = "dislike">
<a href = "javascript:;" class = "btn btn-large" style = "float:right;   color:red;"   onclick = "doAction('<?php echo $postid;?>', 'unlike');">Dislike (<span id ='echo   $postid;_unlikes'><?php echo $data->unlike;?></span>)</a>

</div>
</div>
 <!-- use select to get the items to stay on the page-->

</div>
</div>
</div>
';
 }

 ?>

Answer 1

你有嵌入其他php代码的php代码。

这不起作用：

<?php
<?php
echo 'hello';
?>
?>

报价搞乱了

1 个答案: