dynamic display of images

[misslois74]

im currently working on a page wherein i have to dynamically display an image in which the value is coming from the database, if the field contains null value meaning no image stored on it i set a default image which represent that there is no current image now my problem is if the field is null its not displaying the default image i have set but if there is an image stored its displaying that image:

here is my sample code:

<?php 
		include "connection_dbase_settings.php";
		$membr_id = $_GET['mem_id'];
		$pfile = $_POST['pic'];
		$quer = "Select * from tbl_member_profile where member_id = $membr_id";
		$result = mysql_query($quer);
		while($row = mysql_fetch_array($result))
		{
			$picfile = $row['pic_file'];
?>
        <div class="mid_in" id="mid_in">
        	<div class="mid_in_left">
            <div class="mid_in_left_box2">
            
                   <?php if(isset($picfile))
				         { ?>
                   			  <table>
                   		 	  <tr><td><img src="<?php echo $row['pic_file'] ?>" width="137" height="175"/></td></tr></table>
                           	  <!--<img src="images/avatar.gif" width="137" height="137" />-->
             	  <?php  }    
              	   		 
						 else
				   		 {  ?>
				   			  <table>
                   			  <tr><td><img src="../images/nopic_192.gif" width="137" height="175"/></td></tr> </table>
                     	       <!--<img src="images/avatar.gif" width="137" height="201" /><img src="members/alvin.jpg" width="137" height="201" />-->
                          	   
             	   <?php }  ?>

im not sure where the problem lies...
hope somebody could help me with these...
thanks in advance..

 

[johnwm]

You appear to be using both $_POST and $_GET at the same time.

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[Foamcow]

I might be wrong but pic_file will be set since you are saying...

$picfile = $row['pic_file'];

Try either checking whether $pic_file is null or $pic_file==""

Better still check the value of $row['pic_file'].

You might also want to try something like

if(fileexists("path/to/images/{$pic_file}"){
   echo("<img src=/"path/to/images/{$pic_file}/" />");
} else {
   echo("<img src=/"path/to/images/default.jpg/" />");
}


Finally, to reduce your code clutter. Start outputting the table, row and cell and [i]then[/i] do the image check and output the appropriate HTML. That way you'll only have a single block of HTML to work with. Might make maintenance easier.

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[southbeach]

I would edit your code to this:

<?php
include "connection_dbase_settings.php";
$membr_id = $_GET['mem_id'];
$pfile = $_POST['pic'];
$quer = "Select * from tbl_member_profile where member_id = $membr_id";
$result = mysql_query($quer);

while($row = mysql_fetch_array($result))
{
$picfile = $row['pic_file'];
if (!file_exists(realpath($row['pic_file'])) $picfile = '<img src="../images/nopic_192.gif" style="width: 137px; height: 175px; border: 0px;" alt="" />';

?>
<div class="mid_in" id="mid_in">
<div class="mid_in_left">
<div class="mid_in_left_box2">

<table>
<tr>
<td><img src="<?php echo $picfile; ?>" style="width: 137px; height: 175px; border: 0px;" /></td>
</tr>
</table>
<?PHP } ?> <!-- End Do/While Loop! -->

I think it makes things easier to read by simplifying the code. Also, following Foamcow advise, make sure targeted images does exist.

Hope this helps!


--
SouthBeach
The good thing about not knowing is the opportunity to learn - Yours truly, 2008.

 

[misslois74]

thanks southbeach now i was able to work it out already

 

[southbeach]

Already as in "prior to our postings" or already "thanks to our postings" ???

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--
SouthBeach
The good thing about not knowing is the opportunity to learn - Yours truly, 2008.

 

[misslois74]

yes i was able to work it out after going through all your postings so again thanks to all your postings!!!