Where you you include this "content-type" header? I have used the very
method below many times and never had to sent a content header simply
because PHP and HTML expect an image from the <img src> tag.
-JD
On 11/14/03 3:41 PM, in article j6btb.915$Uz.26871@news7.onvoy.net, "Justin
Koivisto" <spam@koivi.com> wrote:
[color=blue]
> Jamie Davison wrote:
>[color=green]
>> On 11/14/03 2:18 PM, in article pU9tb.912$Uz.26563@news7.onvoy.net, "Justin
>> Koivisto" <spam@koivi.com> wrote:
>>
>>[color=darkred]
>>> Jamie Davison wrote:
>>>
>>>> <img src="display_image.php?imageid=<?php echo $yourimageid; ?>">
>>>>
>>>> Your display image page (display_image.php in this case)
>>>>
>>>> <?
>>>> require("config.inc");
>>>> $sql = "SELECT src FROM images WHERE id=\"$id\"";
>>>> $result = mysql_query($sql,$connection) or die("Couldn't execute get sector
>>>> types query");
>>>> while ($row = mysql_fetch_array($result)) {
>>>> $src = $row['src'];
>>>> }
>>>> echo $src;
>>>> ?>
>>>
>>> This would display garbage on the screen... You need to send header
>>> information for the Content-type (see my previous post).[/color]
>>[/color]
>
> **Fixed top-posting **
>[color=green]
>> I might have been a bit unclear. You must link "to" the display_image.php
>> page from the calling php page with
>> <img src="display_image.php?imageid=<?php echo $yourimageid; ?>">
>>
>> I have used this same format with greatb success . . .[/color]
>
> See my post that came in just before yours....
>
> In any case, you still need to send the content-type headers for the
> browsers to display the image properly, or you will get a blank image.
> Maybe *some* browsers will automatically figure it out, but when I first
> did this, if the header wasn't sent, the image didn't display.[/color]
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