Im使用此代码从数据库中调用名称，并尝试为每个名称创建一个新的Jquery对话框和按钮，其中每个按钮都在first_name下面，姓氏将显示在弹出对话框中。我接近了吗？或者这可能是我尝试的方式吗？
<!doctype html>
<html lang="en">
<head>
<link rel="stylesheet" href="http://code.jquery.com/ui/1.10.3/themes/smoothness/jquery-ui.css" />
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script src="http://code.jquery.com/ui/1.10.3/jquery-ui.js"></script>
</head>
<body>
<?php
$server = 'localhost';
$user = 'root';
$pass = 'root';
$db = 'OCVC';
$connection = mysql_connect($server, $user, $pass)
or die ("Could not connect to server ... 'n" . mysql_error ());
mysql_select_db($db)
or die ("Could not connect to database ... 'n" . mysql_error ());
$result = mysql_query("SELECT * FROM members")
or die(mysql_error());
while($row = mysql_fetch_array( $result )) {
$id = $row['id'];
$name = $row['first_name'];
echo $row['first_name'];
echo '<div id="'.$row['first_name'].'" title="Basic dialog">';
echo $row['last_name'];
echo '</div>';
echo '<button id="'.$row['id'].'">Open Dialog</button>';
}
?>
<script>
var name = "<?php echo $name; ?>";
var id = "<?php echo $id; ?>";
$(function() {
$( "#" + name ).dialog({
autoOpen: false,
show: {
effect: "",
duration: 0
},
hide: {
effect: "",
duration: 0
}
});
$( "#" + id ).click(function() {
$( "#" + name ).dialog( "open" );
});
});
</script>
</body>
</html>
echo '<div class="popups" id="'.$row['first_name'].'" title="Basic dialog">';
echo $row['last_name'];
echo '</div>';
echo '<button class="showLastname" data-for="'.$row['first_name'].'">Show Last Name</button>';
以及在jquery 中
$('.popups').hide();
$(document).on('click','.showLastname',function(){
var id=$(this).data('for');
$('#'+id).dialog({
show: {
effect: "",
duration: 0
},
hide: {
effect: "",
duration: 0
}
});
});