All image information gets stored in the database except for the last 2

[sandhya_sep14]

Hi Everyone,

 

I have a program that generates 200 unique images keeping the first image static in each run.The images keep scrolling on to the screen pause for 3 seconds and scroll off I'm able to generated all 200 unique images without repetition, everything is working well except for the lase two images the last two  images are scrolling on to the screen but are not been displayed in the database, Moreover The last image is a duplicate of 197th image.I don't know what is happening.....

 

Here is MY code..........

 

 

 

<?php

session_start();

$sid = $_SESSION['id'];

$_SESSION['imageDispCnt'] = 0;

$myQuery = "SELECT * from image";

$conn = mysql_connect("localhost","User","Passwd");

mysql_select_db("database_Name",$conn);

$result = mysql_query($myQuery);

$img =Array();

$id =Array();

$i =0;

$imagepath = 'http://localhost/images/';

while ($row = mysql_fetch_array($result))

  {

  $img[$i] =  $imagepath.$row['img_name'];

  $id[$i] =  $row['imageid'];

  $i = $i + 1;

      }

?>

</head>

<script language="JavaScript1.2">

var scrollerwidth='800px';

var scrollerheight='600px';

var scrollerbgcolor='white';

var pausebetweenimages=3200;

var s;

var sec;

var d;

var j;

var imgid;

var milisec = 0;

var seconds = 0;

var flag = 1;

var ses_id = '<?php echo $sid;?>';

var count = 0;

var i = 0;

var imgname;

var imgid;

var k =0;

var slideimages=new Array();

var img_id = new Array();

var index;

 

<?php

  $l =0;   

$count = array();

$j = rand(0,199);

while($l < 200)

{

while(in_array($j, $count))

        {

        $j = rand(0,199);

        }

        $count[$l] =  $j;

        $l++;

       

    }?>

    <?php

  $k = 0;

  for($k = 0;$k<count($count);$k++){

  ?>

  index = <?php echo $k;?>;

  <?php

  $indx = $count[$k];?>

  if(index == 0){

  slideimages[0] = '<img src="http://localhost/images/hbag044.jpg" name="r_img" id="0"/>';

  img_id[0] = '<input type="hidden" value="0" id="imgId" />';

  }

 

  else if(index > 0)

  {

slideimages[<?php echo $k?>] = '<img src="<?php echo $img[$indx]?>" name="r_img" id="<?php echo $id[$indx]?>"/>';

    img_id[<?php echo $k?>] = '<input type="hidden" value="<?php echo $id[$indx]?>" id="imgId" />';

  }

<?php

}

?>

 

 

Can Any one plese help me

 

Appreciate your help...

 

 

Thanks

[denno020]

Well if the last two images display in your scroller but don't appear in the database, then I would suggest making sure there isn't another page in the database view in phpMyAdmin (assuming you're using this). Having 200 records on the one page seems like alot, so it might be broken up into a couple of pages.

 

Also, you have a typo near the bottom with index.

$indx = $count[$k];?>

should be

$index = $count[$k];?>

I do believe....

 

As for the 200th picture, this is infact the 201st picture, as indexing starts at 0. So maybe you're having a problem there somewhere?

 

Denno

 

P.S. Use the code tags or php tags in the future so it's easier for us to read your code.