How can I echo an image from other location than the mysql using this kind
of script?? since I'm using php to output the image from database, there's a
way to change it to a ftp link for example if image doesn't exists??
this is my script:
<?
$sql = "SELECT Imagem_data,Imagem_type FROM imagens WHERE
CelebID='$celebID'";
$query = new Query($conexao);
$query->executa($sql);
$resultado = $query->dados();
mysql_close();
$imagem_banco = $resultado['Imagem_data'];
$type = $resultado['Imagem_type'];
if($imagem_banco != "") {
HEADER("Content-type: $type");
echo($imagem_banco);
}
}
?>
and I call it from my page with <img src="lib_image.php?celebID='1'">.
Thank's
Rodrigo
on 1/9/02 9:16 PM, Bogdan Stancescu at [EMAIL PROTECTED] wrote:
> My way around this is using two fallback stages: for one you test
> mysql_num_rows() after performing the query. If that's 0, use a default image
> placed in the database beforehand. If that also returns a null
> mysql_num_rows()
> then echo an image (probably the same default) from a known location in the
> filesystem.
>
> HTH
>
> Bogdan
>
> Rodrigo Peres wrote:
>
>> Hi list,
>>
>> I have some images stored in a blob in mysql. I've made this code to output
>> them to html, but the problem is if there's no image at a given id the page
>> tooks a long time to load and display the "broken image", there's a way to
>> avoid this, I mean, there's a way to print a message or another image if
>> there's none in mysql ???
>
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