Question

我对jquery比较新，不能为我的生活得到这个以返回正确的youtube id它只是总是返回表中的第一个。

PHP。我从数据库中加载了youtube id，然后将其显示在表格中。

require_once 'library/Zend/Loader.php';
Zend_Loader::loadClass('Zend_Gdata_YouTube');
echo '<br/>';
echo '<table width="95%" cellpadding="0" cellspacing="0" align="center" border="0">'; 

while ($row = mysqli_fetch_assoc($videoquery))
{
$vidid = ($row['id']);
$vidtitle = ($row['title']);
$vidip = ($row['ip']);
$page2 = ($row['page']);
$q = ($row['youtubeid']);

$yt = new Zend_Gdata_YouTube();
$yt->setMajorProtocolVersion(2);
$videoEntry = $yt->getVideoEntry($q);
$thumbnail = '<img src="' . $videoEntry->mediaGroup->thumbnail[0]->url . '" width="100" height="92"';

echo '<tr bgcolor="#333">
 <td width="100px" rowspan="2" nowrap="nowrap" style="border:2px solid #999999">'.$thumbnail.'</td>
 <td colspan="2" align="left">' .$vidtitle. '
 </td></tr>
 <tr bgcolor="#3B3B3B"><td nowrap="nowrap">
 <form method="post">
 <input name="parse" type="hidden" value="playvid" /><input name="youtubeid" id="youtubeid" type="hidden" value="'.$q.'" />;
 <input type="button" name="play" id="play" value="Play Video" />
 </form></td>
 <td align="right">
 <form method="post" action="">
 <input type="hidden" name="page2" value="'.$_SERVER['REQUEST_URI'].'">
 <input name="parse" type="hidden" value="deletepost2" />
 <input name="postid2" type="hidden" value="'.$vidid.'"/>';
if ($topuserid == $_GET['id']){
 echo '<input type="submit" name="deletepost2" id="deletepost2" value="Delete Video" /';>
}  
echo  '</form></td></tr>';  
}  
echo   '</table>';  
}

Jquery代码

<script type="text/javascript">
$(document).ready(function(){
$('#play').live('click', function() {
$('#video').empty()
var vidid = $('#youtubeid').val(); 
alert(vidid);
var playvid = '<object width="300" height="193">'+
'<param name="movie" value="http://www.youtube.com/v/'+vidid+'&amp;hl=en_US&amp;fs=1"></param>'+
'<param name="allowFullScreen" value="true"></param>'+
'<param name="allowscriptaccess" value="always"></param>'+
'<embed src="http://www.youtube.com/v/'+vidid+'&amp;hl=en_US&amp;fs=1" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="300" height="193"></embed>'+
'</object>';
$('#video').append(playvid);    });
});
</script>

Answer 1

内森有正确的想法，但在他的代码中有一些错误。试试这个：

<script type="text/javascript">
$(document).ready(function(){
  $('#play').live('click', function() {
    var tr = $(this).closest('tr');
    tr.find('#video').empty();
    var vidid = tr.find('#youtubeid').val(); 
    alert(vidid);
  });
});
</script>

Answer 2

你需要给你的选择器一些上下文。在您的html中不执行任何更改，您可以将您的js更改为

$(document).ready(function(){
var $tr = $(this).closest('tr')
$('#play').live('click', function() {
tr.find('#video').empty()
var vidid = tr.find('#youtubeid').val(); 
alert(vidid);
var playvid = '<object width="300" height="193">'+
'<param name="movie" value="http://www.youtube.com/v/'+vidid+'&amp;hl=en_US&amp;fs=1"></param>'+
'<param name="allowFullScreen" value="true"></param>'+
'<param name="allowscriptaccess" value="always"></param>'+
'<embed src="http://www.youtube.com/v/'+vidid+'&amp;hl=en_US&amp;fs=1" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="300" height="193"></embed>'+
'</object>';
tr.find('#video').append(playvid);    });
});

如何为动态表中按下的特定按钮选择正确的ID

2 个答案: