[英]php variable for javascript ajax
我有一個表,該表從經常更新的數據庫中查詢,以及在哪里
<span id="totalvotes1"></span>
在哪里
<span id="totalvotes2"></span>
我需要能夠識別行中的那些
success: function(data) { $('#totalvotes1').text(data); } });
在我的ajax中,查詢的每個對應行...現在設置的方式,我的ajax只會將信息顯示回
<span id="totalvotes1"></span>
在最后一行中查詢...。
<?php
$sql = mysql_query("SELECT * FROM blogData ORDER BY id DESC");
$sql2=mysql_query("SELECT * FROM messages WHERE mod(mes_id,2) = 0 ORDER BY mes_id DESC");
$sql3=mysql_query("SELECT * FROM messages WHERE mod(mes_id,2) = 1 ORDER BY mes_id DESC");
$count_variable = 0;
while(($row = mysql_fetch_array($sql))AND($row2 = mysql_fetch_array($sql2))AND($row3 = mysql_fetch_array($sql3)) ){
$id = $row['id'];
$title = $row['title'];
$content = $row['content'];
$category = $row['category'];
$podcast = $row['podcast'];
$datetime = $row['datetime'];
$message1=$row2['msg'];
$mes_id1=$row2['mes_id'];
$totalvotes1=$row2['totalvotes'];
$message2=$row3['msg'];
$mes_id2=$row3['mes_id'];
$totalvotes2=$row3['totalvotes'];
?>
<table class="content">
<tr>
<td>
<div id="main">
<div id="left">
<span class='up'><a href="" class="vote" name="up" data-options="key1=<?php echo $mes_id1;?>&key2=<?php echo $mes_id2;?>"><img src="up.png" alt="Down" /></a></span><br />
<span id="totalvotes1"><?php echo $totalvotes1; ?></span><br />
</div>
<div id="message">
<?php echo $message1; ?>
</div>
<div class="clearfix"></div>
</div>
<div id="main">
<div id="right">
<br />
<span id="totalvotes2"><?php echo $totalvotes2; ?></span><br />
<span class='down'><a href="" class="vote" name="down" data-options="key1=<?php echo $mes_id1;?>&key2=<?php echo $mes_id2;?>"><img src="down.png" alt="Down" /></a></span>
</div>
<div id="message">
<?php echo $message2; ?>
</div>
<div class="clearfix"></div>
</div>
</td>
</tr>
</table>
<?php
}
?>
這是我的general.js文件
$(".vote").click(function()
{
var id = $(this).attr("id");
var name = $(this).attr("name");
var eData = $(this).attr("data-options");
var dataString = 'id='+ id + '&' + eData ;
var parent = $(this);
if(name=='up')
{
$(this).fadeIn(200).html('');
$.ajax({
type: "POST",
url: "up.php",
data: dataString,
cache: false,
success: function(data) { $('#totalvotes1').text(data); }
});
}
else
{
$(this).fadeIn(200).html('');
$.ajax({
type: "POST",
url: "down.php",
data: dataString,
cache: false,
success: function(data) { $('#totalvotes2').text(data); }
});
}
});
});
});
可能是您的最新問題的早期版本, JSON編碼,ajax並以html顯示
在您對此的最新問題的回答中,我給出了詳細的示例,說明如何在PHP和jQuery之間使用JSON數據。
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