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Two Image Display Questions


Chezshire

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Hello Everyone - your friendly newb here

  As usual I'm trying to do things that are a touch beyond my grasp and so i'm reaching out to the experts. With a some guidance I've put together the following piece of code that is doing something I've wanted to about two weeks: It's displaying the character name and the name of actor associated with the character. It's also supplying image, but (and there is always a but isn't there?) the image is displaying as a broken link. I'm not sure if it's because i'm not explaining/executing the link correctly, or if my attempt to modify the code is screwing it up or what. But can some one please take a look and provide a touch of guidance - or a lot (i'm not choosey ;) Also the second thing is that it's only displaying the first and i'd like to display all the characters in the database.

 

So to clarify, I'm asking:

A. How can i get the image to display of the character/model named (I have an empty box right now)

B. How can i display all the characters, ideally alphabetically.

 

This is the code I'm working with:

<?php
//connect to the database
$sql = "SELECT id, modelname, codename FROM cerebra WHERE approved=''";
$result = mysql_query($sql) or trigger_error(mysql_error(),E_USER_ERROR); //execute the query. If it fails, throw an error message
$row = mysql_fetch_assoc($result);//put the result set in an array.
echo '<p>Character: '.$row['codename'];
echo ' = ';
echo 'Model: '.$row['modelname'];
echo '<br /><img src="/cerebra/images/<?php echo $id; ?>-.jpg" width="60" height="75" border="0"'.$row['image'].'" /</p>';
?>

 

 

And I would also like to give a shout of thanks out to Ben and to Singla who are both totally awesome people!

Chez

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echo <br /><img src="/cerebra/images/<?php echo $id; ?>-.jpg" width="60" height="75" border="0"'.$row['image'].'" /</p>';

 

you cannot echo inside of an echo

 

use this for starters

 

echo '<br /><img src="/cerebra/images/'.$id.'-.jpg" width="60" height="75" border="0"'.$row['image'].'" /</p>';

 

you also might want to double check the - before the .jpg to make sure thats not a typo

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