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Re: displaying image

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On 5/1/2013 9:31 AM, richard wrote:
> http://mroldies.net/showtable.php?year=1960
> Click on first link only.
>
> I am using a 2d array and want to display an image based upon the number
> key of that array.
>
> <?php echo "<img src='$list[1]['cover']'>"; ?>
>
> This line gives a place holder in firefox and in IE.
> It should at least show this image.
> http://mroldies.net/covers/1960-001.jpg
>

I think the problem you are having is you are using a 'complex' var in a 
simple parse method (double quotes).

Once the engine hit the first closing array index bracket ']', the var 
is done being parsed.

Try concatenate with '.'

<?php echo "<img src='" . $list[1]['cover'] . "'>"; ?>

or my favorite, use the complex type parsing with curlies.

<?php echo "<img src='{$list[1]['cover']}'>"; ?>

Scotty

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displaying image richard <noreply@example.com> - 2013-05-01 12:31 -0400
  Re: displaying image richard <noreply@example.com> - 2013-05-01 13:33 -0400
  Re: displaying image Scott Johnson <noonehome@chalupasworld.com> - 2013-05-01 12:19 -0700

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