如何在弹出窗口中显示php mysql选择值
我想将mysql选择值(UID)传递到css弹出窗口。这是我尝试过的代码。但它显示为空文本框。单击“删除”按钮时会显示此弹出按钮。我想将相关用户UID传递到弹出窗口如何在弹出窗口中显示php mysql选择值,php,html,mysql,css,ajax,Php,Html,Mysql,Css,Ajax,我想将mysql选择值(UID)传递到css弹出窗口。这是我尝试过的代码。但它显示为空文本框。单击“删除”按钮时会显示此弹出按钮。我想将相关用户UID传递到弹出窗口 <table class="table table-striped table-bordered responsive"> <thead> <tr>
<table class="table table-striped table-bordered responsive">
<thead>
<tr>
<th>User ID</th>
<th>Facebook ID</th>
<th>Name</th>
<th>E-mail</th>
<th></th>
</tr>
</thead>
<tbody>
<tr>
<?php
$query = 'SELECT * FROM Users ';
$result = mysql_query($query);
while($row = mysql_fetch_assoc($result))
{
echo '<tr>';
echo '<td>'.$row['UID'].'</td>';
echo '<td class="center">'.$row['Fuid'].'</td>';
echo '<td class="center">'.$row['Ffname'].'</td>';
echo '<td class="center">'.$row['Femail'].'</td>';
echo '<td>'.'<a href="#" class="btn btn-info btn-setting" >'.'Delete'.'</a>'.'</td>';
echo '</tr>';
}
?>
</tr>
</tbody>
</table>
<form action="db_sql/db_delete_supplier.php" method="post" name="frm1" id="frm1" >
<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel"
aria-hidden="true">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal">×</button>
<h3>Delete Supplier</h3>
</div>
<div class="modal-body">
<p>Are you sure to delete this supplier from system ...? </p>
</div>
<div class="modal-footer">
<input type="text" name="UID" value="<?php echo $row['UID'];?>" />
<!--<a href="user.php" class="btn btn-default" data-dismiss="modal" name="no">Close</a>
<a href="db_sql/db_delete_supplier.php" class="btn btn-primary" data-dismiss="modal" name="yes">Delete</a>-->
<input type="submit" class="btn btn-default" name="no" value="Close" />
<input type="submit" class="btn btn-primary" name="yes" value="Delete"/>
</div>
</div>
</div>
</div>
</form>
用户ID
Facebook ID
名称
电子邮件
×
删除供应商
您确定要从系统中删除此供应商吗
将此代码放入While循环,这样将打开带有cat-id查询字符串的弹出窗口
现在,您可以从该ID获取弹出窗口中的所有记录
这是你能做的可能的解决方案
注意:您可以传递任何唯一的ID,以便您可以在弹出页面上执行select查询,它将从数据库中获取数据如何呈现popup.html
?请提供您的文件structure@CodeGodie抱歉,它不是popup.html,它是popup.php和popup代码在同一页面中的html代码。您是否在浏览器中看到500错误?如果,是的,它说什么?使用Chrome Inspector(或类似工具)@TJ-无任何错误。请稍候,带有$row['UID']
的弹出代码不在循环内-正确吗?它如何访问该变量?
<?php echo "<a href=\"popup.php?cat_id=".$row["cat_id"]." \">Edit</a>"; ?>