HOWTO: output a picture in the middle of a page

I want to generate a web page using PHP, and in the middle generate a
graphic I read from another server. So the graphic is in $picture. So I
can either write $picture to a temp file and then use an ordinary <img
src=filename> to get the browser to display it, or output it by itself
in a new page, because then I have a chance to output the correct header
so I don't just get garbage on the screen.

Evidently first approach is poor because I have to write a local file,
and second approach is not what I want anyway.

Is there a way to do this? That is, I have $picture that contains an
image, and I want to output it here and now in this HTML page I am in
the middle of creating, and have it show as a picture.

Thanks for any pointers.

"Tim Streater" <tim.streater@dante.org.uk> wrote in message
news:tim.streater-C7DF5E.17582804062004@individual.net...
> I want to generate a web page using PHP, and in the middle generate a
> graphic I read from another server. So the graphic is in $picture. So I
> can either write $picture to a temp file and then use an ordinary <img
> src=filename> to get the browser to display it, or output it by itself
> in a new page, because then I have a chance to output the correct header
> so I don't just get garbage on the screen.
>
> Evidently first approach is poor because I have to write a local file,
> and second approach is not what I want anyway.
>
> Is there a way to do this? That is, I have $picture that contains an
> image, and I want to output it here and now in this HTML page I am in
> the middle of creating, and have it show as a picture.
>
> Thanks for any pointers.

<?php // begin thispage.php
if ($_GET['action']="img") {
header("Content-type: image/jpeg");
// output image here
} else {
echo '<img src="thispage.php?action=img">';
}
?>

"kingofkolt" <jessepNOSPAM@comcast.net> wrote in message
news:952wc.50566$Ly.48896@attbi_s01...
> "Tim Streater" <tim.streater@dante.org.uk> wrote in message
> news:tim.streater-C7DF5E.17582804062004@individual.net...
> > I want to generate a web page using PHP, and in the middle generate a
> > graphic I read from another server. So the graphic is in $picture. So I
> > can either write $picture to a temp file and then use an ordinary <img
> > src=filename> to get the browser to display it, or output it by itself
> > in a new page, because then I have a chance to output the correct header
> > so I don't just get garbage on the screen.
> >
> > Evidently first approach is poor because I have to write a local file,
> > and second approach is not what I want anyway.
> >
> > Is there a way to do this? That is, I have $picture that contains an
> > image, and I want to output it here and now in this HTML page I am in
> > the middle of creating, and have it show as a picture.
> >
> > Thanks for any pointers.
>
> <?php // begin thispage.php
> if ($_GET['action']="img") {
> header("Content-type: image/jpeg");
> // output image here
> } else {
> echo '<img src="thispage.php?action=img">';
> }
> ?>
>

CORRECTION:

change line 2 to:

if ($_GET['action']=="image") { // I forgot the second '='

Won't this complain that the header has already been sent assuming that
thispage.php is sending HTML before the image? I have been trying to
find a solution around this too.

kingofkolt wrote:
> "kingofkolt" <jessepNOSPAM@comcast.net> wrote in message
> news:952wc.50566$Ly.48896@attbi_s01...
>
>>"Tim Streater" <tim.streater@dante.org.uk> wrote in message
>>news:tim.streater-C7DF5E.17582804062004@individual.net...
>>
>>>I want to generate a web page using PHP, and in the middle generate a
>>>graphic I read from another server. So the graphic is in $picture. So I
>>>can either write $picture to a temp file and then use an ordinary <img
>>>src=filename> to get the browser to display it, or output it by itself
>>>in a new page, because then I have a chance to output the correct header
>>>so I don't just get garbage on the screen.
>>>
>>>Evidently first approach is poor because I have to write a local file,
>>>and second approach is not what I want anyway.
>>>
>>>Is there a way to do this? That is, I have $picture that contains an
>>>image, and I want to output it here and now in this HTML page I am in
>>>the middle of creating, and have it show as a picture.
>>>
>>>Thanks for any pointers.
>>
>><?php // begin thispage.php
>>if ($_GET['action']="img") {
>> header("Content-type: image/jpeg");
>> // output image here
>>} else {
>> echo '<img src="thispage.php?action=img">';
>>}
>>?>
>>
>
> CORRECTION:
>
> change line 2 to:
>
> if ($_GET['action']=="image") { // I forgot the second '='
>
>

> Won't this complain that the header has already been sent assuming that
> thispage.php is sending HTML before the image? I have been trying to
> find a solution around this too.

[snip]
> >>>
> >>>Thanks for any pointers.
> >>
> >><?php // begin thispage.php
> >>if ($_GET['action']="img") {
> >> header("Content-type: image/jpeg");
> >> // output image here
> >>} else {
> >> echo '<img src="thispage.php?action=img">';
> >>}
> >>?>
> >>
> >
> > CORRECTION:
> >
> > change line 2 to:
> >
> > if ($_GET['action']=="image") { // I forgot the second '='
> >
> >
>
Try to use separate script for image output like this:
image.html
<html><head><title>Image output</title></head><body><img
src=image.php?imagename=imagename></body></html>

image.php
<?
header("Content-type: image/gif");
readfile($_GET['imagename']);
?>

> Try to use separate script for image output like this:
> image.html
> <html><head><title>Image output</title></head><body><img
> src=image.php?imagename=imagename></body></html>
>
> image.php
> <?
> header("Content-type: image/gif");
> readfile($_GET['imagename']);
> ?>
>
>

Yes, that is one way of doing it and is what I am doing now. However,
there are limitations to it.

First, myphpfile is a separate script and one has to pass the bits in
$picture into it which is inefficient and perhaps not much better than
creating a file.

The second more serious drawback is when you want to display multiple
pictures. The following will not work and will display the second
picture twice.

<img src="myphpfile.phtml">

$picture=<second picture>

<img src="myphpfile.phtml">

Of course you can get by with something like the following after
modifying the myphpfile script

<img src="myphpfile.phtml">

$picture2=<second picture>

<img src="myphpfile.phtml?num=2">

but this makes everything even more inefficient as you now have to pass
more data from one script to another.

It would be neat to find a way to display the image directly from the
original script but I am not sure it can be done in php.

In article <E4adnTb887G7g1rdRVn-jw@comcast.com>,
Jin Mazumdar <mazumdar_REMOVE_@REMOVE.comcast.net> wrote:

> > Try to use separate script for image output like this:
> > image.html
> > <html><head><title>Image output</title></head><body><img
> > src=image.php?imagename=imagename></body></html>
> >
> > image.php
> > <?
> > header("Content-type: image/gif");
> > readfile($_GET['imagename']);
> > ?>
> >
> >
>
> Yes, that is one way of doing it and is what I am doing now. However,
> there are limitations to it.
>
> First, myphpfile is a separate script and one has to pass the bits in
> $picture into it which is inefficient and perhaps not much better than
> creating a file.
>
> The second more serious drawback is when you want to display multiple
> pictures. The following will not work and will display the second
> picture twice.
>
> <img src="myphpfile.phtml">
>
> $picture=<second picture>
>
> <img src="myphpfile.phtml">
>
> Of course you can get by with something like the following after
> modifying the myphpfile script
>
> <img src="myphpfile.phtml">
>
> $picture2=<second picture>
>
> <img src="myphpfile.phtml?num=2">
>
> but this makes everything even more inefficient as you now have to pass
> more data from one script to another.
>
> It would be neat to find a way to display the image directly from the
> original script but I am not sure it can be done in php.

In the end I did something like this too. As I was obtaining the picture
in the second script, and just in effect passing a pointer to this
script so it knew which picture to get, there was no extra overhead
because that was work needing to be done anyway.

Thanks for all feedback.

--tim