publish JPEG or HTML from a database

I have a jpeg file and a HTML page that are stored in two separate
blob fields of a databses, namely interbase, what is the best method
of pulling each of these items out so that they can be displayed in a
web browser.

UnixUser wrote:[color=blue]
> I have a jpeg file and a HTML page that are stored in two separate
> blob fields of a databses, namely interbase, what is the best method
> of pulling each of these items out so that they can be displayed in a
> web browser.[/color]

The method that has worked for me for dispalying images is:

show_image.php
--------------
<?php
header('Content-type: image/jpeg');
$query='SELECT field_name FROM table_name WHERE id = '.$_GET['id'];
// add your database code to perform query and store
// data in $data variable
echo $data;
?>

Then in the HTML, I use:
<img src="show_image.php?id=24" alt="">

You can do the same with HTML, but your Content-type should be
'text/html' instead.

--
Justin Koivisto - spam@koivi.com
PHP POSTERS: Please use comp.lang.php for PHP related questions,
alt.php* groups are not recommended.

Image files are commonly displayed from a MySQL database by linking to
another php page . . .

<image 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;
?>

I'm not sure about the HTML in a BLOB field but I assume it would be roughly
the same . . .

-Jamie



On 11/14/03 1:34 PM, in article
a1c0482e.0311141034.65f6f6a2@posting.google.com, "UnixUser"
<rafel.coyle@pfshouston.com> wrote:
[color=blue]
> I have a jpeg file and a HTML page that are stored in two separate
> blob fields of a databses, namely interbase, what is the best method
> of pulling each of these items out so that they can be displayed in a
> web browser.[/color]



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Jamie Davison wrote:[color=blue]
> <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;
> ?>[/color]

This would display garbage on the screen... You need to send header
information for the Content-type (see my previous post).

--
Justin Koivisto - spam@koivi.com
PHP POSTERS: Please use comp.lang.php for PHP related questions,
alt.php* groups are not recommended.

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 . . .

JD

On 11/14/03 2:18 PM, in article pU9tb.912$Uz.26563@news7.onvoy.net, "Justin
Koivisto" <spam@koivi.com> wrote:
[color=blue]
> Jamie Davison wrote:[color=green]
>> <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;
>> ?>[/color]
>
> This would display garbage on the screen... You need to send header
> information for the Content-type (see my previous post).[/color]



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Jamie Davison wrote:
[color=blue]
> On 11/14/03 2:18 PM, in article pU9tb.912$Uz.26563@news7.onvoy.net, "Justin
> Koivisto" <spam@koivi.com> wrote:
>
>[color=green]
>>Jamie Davison wrote:
>>[color=darkred]
>>><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;
>>>?>[/color]
>>
>>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=blue]
> 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.

--
Justin Koivisto - spam@koivi.com
PHP POSTERS: Please use comp.lang.php for PHP related questions,
alt.php* groups are not recommended.

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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Jamie Davison wrote:[color=blue]
> 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.[/color]

Check my reply to the OP

--
Justin Koivisto - spam@koivi.com
PHP POSTERS: Please use comp.lang.php for PHP related questions,
alt.php* groups are not recommended.