Newbie Help With Image Display

[artisticre]

Here is my database
CREATE TABLE card_images (
card_imageid tinyint(4) NOT NULL auto_increment,
image1 varchar(50) NOT NULL default '',
image2 varchar(50) NOT NULL default '',
image3 varchar(50) NOT NULL default '',
image4 varchar(50) NOT NULL default '',
image5 varchar(50) NOT NULL default '',
PRIMARY KEY (card_imageid)
) TYPE=MyISAM;

I have greetingcard packs that we sell. When I enter in the packs, some have 5 images, some have 4, some have 3. This will create blanks in some entries. Not a problem. Except when I want to display the packs, it creates broken links. I think I need to create an array but I am new and not sure what to do. I want to pull only the images and not the blanks.

Someone please help

[ultraslacker]

I would change the design -

card
card_id

card_image
card_image_id
card_id
image

[Danni]

What about if you put the images for the paticular product all in one large image?

Then you could do this:

CREATE TABLE card_images (
id varchar(50),
url varchar(255) NOT NULL,
);

#
#Dumping data for table 'card_images'
#

INSERT INTO card_images VALUES ('youridname', 'imagesurl');

🙂

[mtombs]

I agree with the others here, you need to change your design. What if one day you release a pack with 6 different images? You're screwed then, aren't you?
I suggest a structure like:

table card_packs

pack_id int auto_increment

table cards

card_id int autoincrement
pack_id int
image

Then you can have as many card images as you like for a pack (including none). Then when you get the data out, you do something like

$sql = "select * from cards where pack_id = x";
$qry = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_object($qry)){
     // do whatever it is you do...
     echo "<img src=" . $row->image . "/>";
}

[artisticre]

OK I will try this but how would I insert the data into the table?

[artisticre]

OK here is what I have and I am getting a parse error on line 13. And another question, how do I insert the data in with this new design. Please bear with me, I am trying to learn as I go.

thanks again

<?

$server = "**";
$userid = "";
$pass = "";
$database = "**";

$con = mysql_connect("$server","$userid","$pass") or die ("Huh? What Server");
$db = mysql_select_db("$database",$con) or die("I said WHAT database");

$sql = "select * from cards where pack_id = $id";

$qry = mysql_query($sql) or die(mysql_error));

while ($row = mysql_fetch_object($qry)) {

echo "<img src=". $row->image . "/">; 

}
?>

[AshtrayWaterloo]

There seems to be an extra close-bracket in there...

$qry = mysql_query($sql) or die(mysql_error));

Should probably be:

$qry = mysql_query($sql) or die(mysql_error);

Ash

[artisticre]

OK now I have a parse error in line 17

$con = mysql_connect("$server","$userid","$pass") or die ("Huh? What Server");
$db = mysql_select_db("$database",$con) or die("I said WHAT database");

$sql = "select * from cards where pack_id = $id";

$qry = mysql_query($sql) or die(mysql_error);

while ($row = mysql_fetch_object($qry)) {

[B] echo "<img src=". ../greetingcards/$row->image . "/">; [/B] 

}
?>

[ultraslacker]

To import data into the new schema, either use a script or simply a series of sql statements. You will be using INSERT INTO...SELECT.

[artisticre]

OK everything is working but I still don't understand how to insert with this. I know how to do it if I had a field in the database for each image, but I am not sure how to do it with only one field with several images.

[mtombs]

First of all, your error should look like this:

echo "<img src='/images/greetingcards/". $row->image . "'/>";

Inserting, then. More than one solution to this, but heres one. Call this file addimage.php or change the action of the form. This script assumes that register_globals is on. Note the hidden input.

<?php 

// connect to the database...

// a bit of code to insert new images

if ($image!=NULL){
     $sql = "insert into cards (pack_id, image) values ($id,$image)";
    mysql_query($sql) or die(mysql_error());
}

// now show a list of the images in the pack
$sql = "select * from cards where pack_id = $id";
$qry = mysql_query($sql) or die(mysql_error());
echo "<table><tr>";
echo "<th>id</th><th>url</th>";
echo "</tr>";
while ($row=mysql_fetch_object($qry)){
   echo "<tr>";
   echo "<td>" . $row->card_id . "</td>";
   echo "<td>" . $row->image . "</td>";
   echo "</tr>";
}
echo "</table>";

// now a form to add new ones
?>
<form method="POST" action="addimage.php">
<input name="image"/>
<input type="submit" value="submit" />
<input type="hidden" name="id" value="<?php echo $id ?>"/>
</form>

[artisticre]

This doesn't makes sense. First how does it insert more than one image into one field and also how does the pack_id in the greetingcardpacks get populated

[mtombs]

Some confusion here, I think. I'm not trying to insert lots of images into one field. I have one record per image (not field, record). Lots of images, lots of records. One card pack = 1 record in 'card_packs' table, lots of images = lots of records in 'cards' table. Maybe you should do a little general reading about relational databases.
As you have two tables you need to do two inserts. You will probably need two scripts, one for creating card packs, one for adding the images.