下面是我...
函数CHECKDB(code,用户ID)
{
$(#notice_div)HTML(中..')。
$阿贾克斯({
键入:POST,
网址:< PHP代码bloginfo('template_url');> /profile/check_$c$c.php?,
数据:{code:code,用户ID:用户ID},
//数据:'code ='+ code +'和;用户id ='+用户ID,
数据类型:HTML,
成功:函数(结果){
如果(结果== 0)
{
$('#成功)HTML(code +'已经赎回!);
//警报(成功); //测试的目的
}
否则,如果(结果== 2)
{
$('#ERR)HTML(code +'已经存在,并已兑现....');
//警报('故障'); //测试目的
}否则,如果(结果== 1){
$('#ERR)HTML(code +'赎回code犯规存在');
}
$(#notice_div)的HTML('')。
// $('#originalquery)隐藏()。
//$('#mainquery').fadeIn('slow');
//$("#originalquery").replaceWith($('#mainquery',$(HTML)));
//警报(结果);
}
})
}
这是在同一个页面,因为这对:
< PHP
$ SQL =SELECT * FROM wp_scloyalty其中userid ='$ USER_ID';
$结果= mysql_query($ SQL);
?>
<表ID =便便的风格=保证金:10px的0px 0px 0px; WIDTH =100%的边界=0的cellpadding =0CELLSPACING =0>
&其中; TR>
< TD><强>产品< / STRONG>< / TD>
< TD><强> code< / STRONG>< / TD>
< TD><强>值小于; / STRONG>< / TD>
< / TR>
<?PHP而($行= mysql_fetch_array($结果)){>
&其中; TR>
< TD>&LT ;?回声$行['产品']; ?>< / TD>
< TD>&LT ;?回声$行['code']; ?>< / TD>
< TD>&LT ;?回声$行['值']; ?>< / TD>
< / TR>
&LT ;? ?}>
< /表>
<形式方法=后级=sc_ajaxxxID =sc_add_voucherxNAME =sc_ajax行动=的onsubmit =CHECKDB(document.sc_ajax.sc_voucher_ code.value,<?PHP的echo $ USER_ID;?>);返回false;>
<按钮ID =呈交code型=提交>提交< /按钮>
< /形式GT;
这是HTML ...
在Ajax请求被提交,我想表自动显示新添加的信息。我知道为什么它不工作,但不知道如何得到它刷新。我想刷新DIV心不是真的有可能,因为我想它是..即时通讯想着做一个新的查询,并成功函数把它吗?
感谢:)
修改
表查询:
< PHP
$ SQL =SELECT * FROM wp_scloyalty其中userid ='$ USER_ID';
$结果= mysql_query($ SQL);
?>
<表ID =便便的风格=保证金:10px的0px 0px 0px; WIDTH =100%的边界=0的cellpadding =0CELLSPACING =0>
&其中; TR>
< TD><强>产品< / STRONG>< / TD>
< TD><强> code< / STRONG>< / TD>
< TD><强>值小于; / STRONG>< / TD>
< / TR>
<?PHP而($行= mysql_fetch_array($结果)){>
&其中; TR>
< TD>&LT ;?回声$行['产品']; ?>< / TD>
< TD>&LT ;?回声$行['code']; ?>< / TD>
< TD>&LT ;?回声$行['值']; ?>< / TD>
< / TR>
&LT ;? ?}>
< DIV ID =新的code>< / DIV>
< /表>
解决方案
所以基本上要根据 $ USER_ID
从数据库中提取的记录,并由此产生生成的表的需求要显示在页面请求使用Ajax?对?
显示新添加的信息
在哪里是它被添加呢?
Here is what I have...
function checkDB(code, userid)
{
$("#notice_div").html('Loading..');
$.ajax({
type: "POST",
url: "<?php bloginfo('template_url'); ?>/profile/check_code.php",
data: { code: code, userid: userid},
//data: 'code='+code+'&userid='+userid,
datatype: "html",
success: function(result){
if(result == 0)
{
$('#success').html( code + ' has been redeemed!');
// alert('success');//testing purposes
}
else if(result == 2)
{
$('#err').html( code + ' already exists and has already been redeemed....');
//alert('fail');//testing purposes
}else if(result == 1){
$('#err').html( code + ' redeem code doesnt exist');
}
$("#notice_div").html('');
//$('#originalquery').hide();
//$('#mainquery').fadeIn('slow');
//$("#originalquery").replaceWith($('#mainquery', $(html)));
//alert(result);
}
})
}
This is on the same page as this:
<?php
$sql="SELECT * FROM wp_scloyalty WHERE userid = '$user_id'";
$result=mysql_query($sql);
?>
<table id="poo" style="margin:10px 0px 0px 0px;" width="100%" border="0" cellpadding="0" cellspacing="0">
<tr>
<td><strong>Product</strong></td>
<td><strong>Code</strong></td>
<td><strong>Value</strong></td>
</tr>
<?php while($rows=mysql_fetch_array($result)){ ?>
<tr>
<td><? echo $rows['product']; ?></td>
<td><? echo $rows['code']; ?></td>
<td><? echo $rows['value']; ?></td>
</tr>
<? } ?>
</table>
<form method="post" class="sc_ajaxxx" id="sc_add_voucherx" name="sc_ajax" action="" onsubmit="checkDB(document.sc_ajax.sc_voucher_code.value, <?php echo $user_id; ?>); return false;">
<button id="submit-code" type="submit" >Submit</button>
</form>
This is the HTML...
When the ajax request is submitted, I want the table to automatically show the newly added information. I know why its not working but not sure how to get it to refresh. I assume refreshing the div isnt really possible as I would like it to be.. Im thinking about making a new query and bringing it in on success function?
Thanks :)
EDIT
The table query:
<?php
$sql="SELECT * FROM wp_scloyalty WHERE userid = '$user_id'";
$result=mysql_query($sql);
?>
<table id="poo" style="margin:10px 0px 0px 0px;" width="100%" border="0" cellpadding="0" cellspacing="0">
<tr>
<td><strong>Product</strong></td>
<td><strong>Code</strong></td>
<td><strong>Value</strong></td>
</tr>
<?php while($rows=mysql_fetch_array($result)){ ?>
<tr>
<td><? echo $rows['product']; ?></td>
<td><? echo $rows['code']; ?></td>
<td><? echo $rows['value']; ?></td>
</tr>
<? } ?>
<div id="newCode"></div>
</table>
解决方案
so basically you want to pull the record from database based on $user_id
and the resulting generated table needs to be shown on requesting page with ajax? Right?
show the newly added information
where's it being added anyway?