即时通讯熟悉PHP,但在js严重的学习者,我已经尝试了3种方法来实现我的目标,我已经看到了下面使用的方法,似乎是最容易实现的,所以我可以实现我的目标,但由于某种原因,它不起作用。我只包含了不需要的特定函数的必要位,这一切都在基于db中条目的while循环内工作,并与{$ page_trackid}分开以区分每个函数并链接到它。显示隐藏JS div css
我明白'这个'风格的功能可以使用,但正如我所说我是如此新的js我只是试图达到必要之前,我扩展到变得更加复杂。
这里有什么我在做完全错误的东西吗?
干杯
<head>
<style>
.hidden {display:none;}
.visible {display:block;}
.subtext_img {
width: 100px; height: 100px; padding-top: 10px; padding-right: 10px; float: right;
}
.subtext {
padding-left: 10px;
}
.arrow_box { padding-left: 100px; position: absolute; z-index: 100; background: #88b7d5; border: 4px solid #c2e1f5; width: 580px; height: 120px;} .arrow_box:after, .arrow_box:before { left: 100%; border: solid transparent; content: " "; height: 0; width: 0; position: absolute; pointer-events: none; } .arrow_box:after { border-color: rgba(136, 183, 213, 0); border-left-color: #88b7d5; border-width: 10px; top: 30%; margin-top: -10px; } .arrow_box:before { border-color: rgba(194, 225, 245, 0); border-left-color: #c2e1f5; border-width: 16px; top: 30%; margin-top: -16px; }
</style>
</head>
<body>
<?
$data = mysql_query("SELECT * FROM user_Uploaded_Tracks WHERE username = '$user' ORDER BY datetime DESC")
or die(mysql_error());
$page_trackid = '1'; //reset page track id
// get a whole bunch of data, track info & user info based on some cross referencing via id
while ($info = mysql_fetch_array($data)) {
$db_trackid = $info['id'];
$username = $info['username'];
//data 2 track info based on db track id
$data2 = mysql_query("SELECT * FROM user_Uploaded_Tracks WHERE id = '$db_trackid' LIMIT 1")
or die(mysql_error());
$info2 = mysql_fetch_array($data2);
//data 3 is user profile data based on db track id
$data3 = mysql_query("SELECT * FROM users WHERE username = '$username' LIMIT 1")
or die(mysql_error());
$info3 = mysql_fetch_array($data3);
//data 4 profile image based on user id, track uploader*
$data4 = mysql_query("SELECT * FROM user_profile_pic WHERE username = '$username' ORDER BY datetime DESC LIMIT 1")
or die(mysql_error());
$info4 = mysql_fetch_array($data4);
echo "
<script>
function showbox(userinfo_{$page_trackid}){
document.getElementById(userinfo_{$page_trackid}).style.visibility='visible';
}
function hidestuff(userinfo_{$page_trackid}){
document.getElementById(userinfo_{$page_trackid}).style.visibility='hidden';
}
</script>
<div id='userinfo_{$page_trackid}' class='arrow_box' style='display: none;'> <img src='" . $info4['Image'] . "' class='subtext_img'>
<h2 class='subtext'><a href='http://XXXX/XXX/" . $info2['username'] . "'>" . $info2['username'] . "</a></h2>
<p class='subtext'>" . $info3['user_title'] . "</p>
<p class='subtext'><a href='" . $info3['website_link'] . "' target='_blank'>" . $info3['website_link'] . "</a>
</p>
</div>";
echo "
<div style='position: absolute; z-index: 1; width: 20px; height: 20px; padding-top: 50px; padding-left: 699px;'>
<a href='javascript:showbox_{$page_trackid}()'><img style='height: 20px;' alt='Track stats' src='http://XXXXXX/play1/skin/user-profile2.png' style=''></a>
</div>";
}
?>
</body>
呃!不好的缩进,跳过问题。 – elclanrs 2013-04-09 23:01:21
您是否考虑过使用连接或两个从数据库中获取数据? – PeeHaa 2013-04-09 23:08:11
“不起作用”不是一个合适的描述。请描述你正在做什么和实际发生的事情,包括错误消息。由于这是一个客户端JavaScript问题,请张贴在客户端收到的HTML,而不是生成它的PHP。 – RobG 2013-04-09 23:13:41