Question

Mysql，Bootstrap，工具提示，我正在努力提高自己的能力，所以我想到了更好的学习方法，而不是跳入仪式。

我在这个网站和其他网站上找到了许多问题的答案，但随着我的项目变得越来越复杂，找到答案变得越来越困难。

我编写了这个脚本，它似乎在Firefox和IE中完美运行（我想要它的确切），但出于某些原因，在Dreamweaver的实时功能中，它显示了一些它不应该出现的Php。就好像有错误一样。但是Firefox和IE再次没有显示错误，事实上一切都在浏览器中完美运行。这是我正在使用的代码。

也许我的代码只是草率，因为你可以看到我进出PHP太多了。所以在Dreamweaver中有两个子弹点和代码片段：会员登录并激活。当我选择直播时，这就是我的Dreamweaver中显示的内容。

： =“2”）{echo $ hours; ？＆GT;小时，分钟分钟

我不确定为什么代码显示它不应该在实时版本中。

我可以从愿意帮助的人那里学到很多东西。谢谢。

<?php

include("db.php");

$con=mysql_connect($server, $db_user, $db_pwd) //connect to the database server
or die ("Could not connect to mysql because ".mysql_error());

mysql_select_db($db_name)  //select the database
or die ("Could not select to mysql because ".mysql_error());

#//////////////////////////////////////////////////////////////////////////

$time = time(); #Current time 
$previous = "86400"; #Time to check in seconds (24 hours)
$logout = $time-$previous; #Timeout=time - 24hours 


# /////////////////////////////////////////////////////////////////////////////

$time1 = time(); #Current time 
$previous1 = "600"; #Time to check in seconds  
$lagout1 = $time1-$previous1; #Timeout=time - 10 minutes 

#/////////////////////////loggedin AND lagged in //////////////////////////////

//prevent sql injection
//$username=mysql_real_escape_string($_POST["username"]);
//$password=mysql_real_escape_string($_POST["password"]);

$query = "SELECT * FROM ".$table_name." WHERE logout > \"$logout\" AND lagout >
    \"$lagout1\" ORDER BY username ASC";                                                 
$online = mysql_query($query); #Execute query 
$row_online = mysql_fetch_assoc($online); #Grab the users 

if (isset($row_online['username'])) { #If there is atleast one user online 
do { #Do this 

?>

<div id="effect2" >

<li><a href="#"><?php echo $row_online['username']; ?></a><span><?php echo             $row_online['username']; ?>:<br /><font color='yellow'>Member logged in and active.        </font></span></li>

</div>      

<? 

}    while($row_online = mysql_fetch_assoc($online)); #Until all records are displayed 

} 

?>

<?
#//////////////////Logged in BUT lagged out//////////////////////////////// 

$query = "SELECT * FROM ".$table_name." WHERE logout > \"$logout\" AND lagout <     \"$lagout1\" ORDER BY username ASC "; #Check and see who is online in the last 2 minutes 
$online = mysql_query($query); #Execute query 
$row_online = mysql_fetch_assoc($online); #Grab the users 

if (isset($row_online['username'])) { #If there is atleast one user online 
do { #Do this 

?>

<?
$T = time();
$lagT = $row_online['logout'];
$S = $T - $lagT;
$init = $S;
$hours = floor($init / 3600);
$minutes = floor(($init / 60) % 60);
$seconds = $init % 60;
?>

<div id="effect2" >

<li><a href="#"><font color='#333333'><?php echo $row_online['username']; ?></font>    </a><span><?php echo $row_online['username']; ?>:<br />

Idle for: <font color='red'>
<? if ($hours == "1") {

echo  $hours; ?> Hour, <? echo $minutes; ?> Minutes 
</font></span></li> 

<? }elseif ($hours >= "2") {

echo $hours; ?> Hours, <? echo $minutes; ?> Minutes
</font></span></li> 

<? } elseif ($hours <= "1") {

echo $minutes;  ?> Minutes
</font></span></li> 

<? } ?>
</div>

<?

} while($row_online = mysql_fetch_assoc($online)); #Until all records are displayed

} 

?>



<div style="position: absolute; 10px;  center: 14px; top: 286px; 90px; top: 294px;   left: 10px;">


 <?
#////////////////////////////////////  there are no more members online ///////////////
echo "_________________________<br /><center>Members who have logged <br />off do not show on this list.</center>"; #Inform user that no one is online 


?>
</div>
</div>

如何让我的代码在fireweaver中显示，就像firefox和Ie一样

0 个答案: