Home/ Questions/Q 8198569
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-06-07T05:58:44+00:00

My code $sql=SELECT * FROM `room_lists`; $sqlquery = mysql_query($sql); while($result = mysql_fetch_array($sqlquery)){ $toDay =

  • 0

My code

$sql="SELECT * FROM `room_lists`";
$sqlquery = mysql_query($sql);
while($result = mysql_fetch_array($sqlquery)){

$toDay = strtotime($result['timeStop']); 
//echo $toDay;

?>

<head>
<meta http-equiv="Content-Type" content="text/html; charset=tis620" />
<script type="text/javascript" src="core/jquery-latest.js"></script>
<title>test</title>
<script type="text/javascript">

function countDown(times){
    var toDay=Math.round(new Date().getTime()/1000); 
    var difTime=times-toDay;
    var day=0,hours=0,minutes=0,seconds=0;
    if(difTime>0){
    day=Math.floor(difTime/84600);
    hours=Math.floor((difTime/3600))%24;
    minutes=Math.floor(difTime/60)%60;
    seconds=Math.floor(difTime)%60; 
    countDown_onLoad();
    }
    else{
        alert('asd');   //time out
    }
    $('.show').html(day+' Day '+hours+' Hour '+minutes+' Min '+seconds+' Sec ');
}
function countDown_onLoad(){
    setTimeout("countDown(<?=$toDay ?>);",1000);
}
$(document).ready(function() {
countDown_onLoad();
});

</script>

</head>

<body>
<div class="show"></div>

</body>


</html>
<? } ?>

Hello, I am a newbie using javascript. I have created countdown timer to use like auction website with js function that call from database(timestamp type) with php query, if the result is 1 row, it works fine but if more than 1 the result are all the same as last row in database. How can I fix this?

This is my output

0 Day 1 Hour 41 Min 25 Sec
0 Day 1 Hour 41 Min 25 Sec
Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    You appear to be outputting an entire HTML document (less the opening <html> tag and <!doctype>) for every counter.

    The core problem is that you are using $(".show").html(...) to output the counter, which is essentially telling it to put that content in every .show element. The net result is that they all show the same thing: whatever the last counter is at.

    Your approach is almost unsalvageable, but here is my best attempt:

    <!DOCTYPE html>
<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=tis620" />  
    <script type="text/javascript" src="core/jquery-latest.js"></script>  
    <title>test</title>  
    <script type="text/javascript">
      function countDown(){  
          var toDay=Math.round(new Date().getTime()/1000);   
          $(".show").each(function() {
              var elm = $(this);
              var difTime=this.timestamp-toDay;  
              var day=0,hours=0,minutes=0,seconds=0;  
              if(difTime>0){  
                  day=Math.floor(difTime/84600);  
                  hours=Math.floor((difTime/3600))%24;  
                  minutes=Math.floor(difTime/60)%60;  
                  seconds=Math.floor(difTime)%60;   
              }  
              else{  
                  alert('asd');   //time out  
                  elm.removeClass("show");
              }  
              elm.html(day+' Day '+hours+' Hour '+minutes+' Min '+seconds+' Sec ');  
          });
      }  
      function countDown_onLoad(){  
          $(".show").each(function() {
              this.timestamp = parseInt(this.firstChild.nodeValue,10);
          });
          setInterval(countDown,1000);
      }  
      $(document).ready(function() {  
          countDown_onLoad();  
      });  
    </script>  
</head>  
<body>
    <?php
      $sql="SELECT * FROM `room_lists`";   
      $sqlquery = mysql_query($sql);   
      while($result = mysql_fetch_array($sqlquery)){   
          $toDay = strtotime($result['timeStop']);  
          echo "<div class=\"show\">".$toDay."</div>";
      }
    ?>
</body>
</html>

    Please note that I do not use jQuery. I am only using it here because you were already using it. With that in mind I may have incorrectly used .each(). Please let me know if you encounter any errors and I will try to fix them.

    • 0