如何使用php mysql在弹出窗口中显示记录

Question

I want to show the record in the popup using PHP and MySQL. it is showing only the last record in the popup I want to show the selected record

    $s = mysql_query("Select * from student");

        while($sql = mysql_fetch_array($s))
        {
        echo'<div class="ammad">'. $sql["id"]."".$sql["Name"]."".$sql["Subject"].'</div>';
        echo '<input ammad="'.$sql["id"].'" type="submit" class="abc" id="abc"/>';

            $name=$sql["id"];

        }

                echo '  
<div id="dialog" title="Basic dialog">
  <input type="textbox" value="'.$name.'" />
</div>';

the dialog is my popup

<script type="text/javascript">

$(document).ready(function(){


$(".abc").click(function(){
$( "#dialog" ).dialog().close();    

    var b = $(this).attr("ammad");
    $( "#dialog" ).dialog();    
    });
});

</script>

It is showing only the last record in popup I want to show the selected record in the popup

Answer 1

Just you doesn't update the diaglog box.

<script type="text/javascript">

$(document).ready(function(){


$(".abc").click(function(){
$( "#dialog" ).dialog().close();    

    var b = $(this).attr("ammad");

    $("#dialog").html('<input type="textbox" value="' +b+ '/>');

    $( "#dialog" ).dialog();    
    });
});

</script>

Try like this and remove { id="abc" } inside loop because id selectors must be unique.

http://www.w3schools.com/cssref/sel_id.asp