This is really bugging me.

civicboy

I am having problems with this code.

<table>
<tr>
<td colspan="2"></td>
</tr>
<?php
    $chandle = mysql_connect("servername", 'username', 'pswd');
    $boolval = mysql_select_db('dbname', $chandle);
	$cuid_txt = $_GET['CUID'];
    $query = "SELECT * FROM photos where CUID ='$cuid_txt'";
    $result = mysql_query($query)
    	or die("Query failed");
?>
<tr>
<td>
<img src="images/CARS/<?php print( $row[2] ); ?>">
</td>
<td><?php print( $row[3] ); ?></td>
</tr>

</table>

Ok here is the problem. I have 3 rows in this database (CUID, PHOTO_FILE, DESCRIPTION). I am trying to load the pictures in the script. The only thing it will do though is show a red x where the picture should be. If you have any info let me know. thanks.

mattotto77

are you storing the images in the database or the url to the picture if you have the dreaded red cross it looks the pathe to the image is not correct

civicboy

i am storing jsut the image name in the database (image.jpg). i have another script that is uploading the imapge to the right directory. I have ftp'd into the server to make the the upload script is inserting the picture in the right directory and it is doing that. but it wont show the pictures when i try to view them in this script.

Actually when i check the properties on the red x. it is showing the correct directory the picture should be in but it does not show the actual picture name (image.jpg)

devinemke

you are not defining $row. try this:

<?
$chandle = mysql_connect('servername', 'username', 'pswd');
$boolval = mysql_select_db('dbname', $chandle);
$result = mysql_query("SELECT * FROM photos where CUID = '" . $_GET['CUID'] . "'") or die('Query failed');
if (mysql_num_rows($result))
{
	$row = mysql_fetch_array($result);

echo '
<table>
<tr>
<td><img src="images/CARS/' . $row[2] . '"></td>
<td>' . $row[3] . '</td>
</tr>
</table>
';
}
else {echo 'no records matching CUID #' . $_GET['CUID'];}
?>

civicboy

cool. that is working now but the description for the pic (row 3) is not showing. thanks for the help..

EDIT... got it working thanks alot.. i had to switch $row[2] to 1 and [3] to 2.

civicboy

ok now i have another problem. Say i have more then one picture to show with the same CUID how will i be able to see all the pictures. I got it showing the first picture the database sees but it wont show the other pictures in the database with the same CUID

devinemke

<?
$chandle = mysql_connect('servername', 'username', 'pswd'); 
$boolval = mysql_select_db('dbname', $chandle); 
$result = mysql_query("SELECT * FROM photos where CUID = '" . $_GET['CUID'] . "'") or die('Query failed'); 
if (mysql_num_rows($result)) 
{ 
    echo '<table>';

	while ($row = mysql_fetch_array($result))
	{
		echo '
		<tr>
		<td><img src="images/CARS/' . $row[2] . '"></td>
		<td>' . $row[3] . '</td>
		</tr>
		';
	}
	echo '</table>';
} 
else {echo 'no records matching CUID #' . $_GET['CUID'];}
?>

civicboy

works great. thanks alot.