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
    $query = new Query($conexao);
    $resultado = $query->dados();
    $imagem_banco = $resultado['Imagem_data'];
    $type = $resultado['Imagem_type'];
    if($imagem_banco != "") {
        HEADER("Content-type: $type");

and I call it from my page with <img src="lib_image.php?celebID='1'">.



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.
> 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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