dsafq2131321 2013-08-31 04:46
浏览 61
已采纳

只有一个变量中的一个通过AJAX传递

I'm trying to pass 2 variables through AJAX. But right now only one of them gets passed and saved in my database.

$image_id gets passed.

$uid doesn't get passed.

<div class="interaction"><a href="#" class="like" id="<?php echo $image_id ?>"
xml-id="<?php echo $uid ?>"><?php echo number_format($image_likes) ?></a>
</div>    


<script type="text/javascript">
$(function() {
$(".like").click(function() 
{
var id = $(this).attr("id");
var uid = $(this).attr("uid");
var dataString = 'id='+ id +'&uid=' + uid ;

var parent = $(this);

$.ajax({
type: "POST",
url: "like.php",
data: dataString,
cache: false,

success: function(html)
{
parent.html(html);
}
});
return false;
});
});
</script>

like.php:

if(($_POST['id']) && ($_POST['uid'])) {
$image_id=$_POST['id'];
$user_id=$_POST['uid'];
*some mysql query*

What am I'm missing here?

ps: I tried the SQL query with some dummy values and it works fine.

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3条回答 默认 最新

  • dpd7122 2013-08-31 04:52
    关注

    You could change your anchor:

    <a href="#" class="like" id="<?php echo $image_id ?>" xml-id="<?php echo $user_id ?>">
    

    To:

    <a href="#" class="like" id="<?php echo $image_id ?>" data-uid="<?php echo $user_id ?>">
    

    And then grab the value for user_id like:

    var uid = $(this).data("uid");
    

    You can use the network tab of the web inspector in Chrome to see what parameters are actually getting passed to your server.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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