Thanks Shawn and Dollah
-Sashi
Shawn McKenzie wrote:
Sashikanth Gurram wrote:
haliphax wrote:
On Mon, Mar 9, 2009 at 10:52 AM, Sashikanth Gurram <sashi34u@xxxxxx>
wrote:
haliphax wrote:
On Mon, Mar 9, 2009 at 10:24 AM, Sashikanth Gurram <sashi34u@xxxxxx>
wrote:
Nathan Nobbe wrote:
On Mon, Mar 9, 2009 at 7:32 AM, Sashikanth Gurram <sashi34u@xxxxxx
<mailto:sashi34u@xxxxxx>> wrote:
Hi,
Yes, the problem was solved, but It did not work fine when I used
the same code in my larger file. Now it makes sense.
right, just track down where you started sending the output, and
remember
if youre going to use header() calls in your scripts, that all of
them
must
come before sending any of the standard content.
Let me just repeat what you have said just to make sure that I did
not misread you.
So you say that the solution to this problem is to create another
php file with the image fetching header and just write an img tag
<img src="myimagescript.php?id=1234" /> in my original php file
(with the html tags).
what i explained in my first response is that youre mixing 2
different
approaches, and it was unclear what you were going for exactly.
if you
want
to have an image included in a page of html, then theres no need
for the
header() call (refer to my first response for the remaining details).
there
are however legitimate use cases for the use of header() & the
aforementioned image methods, i think between mine and some of the
other
posts on this thread, its explained clearly.
This is what I have understood.
Regarding the point you have mentioned ( If you set the content
type using header() to "image/jpeg", do not use HTML tags to
display your image!),
correct
I definitely need the HTML tags, because this application works
based on the user input. So unless there is not input through a
html form, it wont work.
right, then just configure your webserver such that you can first
access
the image directly via an http url, then integrate these links
into your
dynamic pages as i explained in my first response.
Thanks a lot for all the patient replies. All the suggestions led
me in a
positive direction. Finally, instead of using the header() in my
main PHP
file (with HTML tags), I have used it in a secondary file and
called it
using a tag<img src="imgtest.php">. It is working fine. But, the
image I
need to display is also dynamic and needs a user input. So, is
there any
way
in which I can transfer a particular variable (the user input) from my
main
php file (say A.php) to my secondary file containing the header ()
(say
B.php)
Yes. Use the Query String of your image-producing PHP script to pass
values. If you had an image tag like this:
<img src="imagescript.php?id=1234" />
Then you could grab the value of $_GET['id'] in your PHP script and
react accordingly.
Thanks a lot everyone, particularly Haliphax, Nathan, Virgilio and Bob.
I will try it and will come back to you.
You're very welcome. This page [1] may help you get started. It's a
bit dated, but the information still holds true today.
1. http://whn.vdhri.net/2005/10/how_to_use_the_query_string_in_php.html
Hello everyone,
Is there any way in which I can assign a variable to a query string?
Like for example, Let us say that there are two php files a.php and
b.php. I am currently using a image tag like <img
src="imgtest.php?id=Williams Hall" /> in a.php and am passing the value
to another variable using $build=$_GET['id']; in b.php. Now for this
purpose, I am using the img tag in the html part of my first file which
makes it kind of static. Now if I want to assign a variable (say
$building) to 'id' in my first file, and retrieve it using the variable
$build in the second file what is the best way to do it?
Will the command (written inside the PHP tags) echo ' <img
src="imgtest.php?id=$building />'; be of any help in performing the
task. I have actually tried this out, but it did not work for me (May be
my syntax is wrong although I have tried various combinations). So, is
there a better way or correct way of doing this (If what I have done is
wrong). I have tried to use $_session() command, but it is kind of
yielding me a header error.
Thanks,
Sashi
Either:
1. a.php
<img src="b.php?id=<?php echo $building; ?>" />
--or--
2. a.php
echo '<img src="b.php?id=' . $building .'" />';
Then, whicher you choose above: b.php:
$build = $_GET['id'];
--
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Sashikanth Gurram
Graduate Research Assistant
Department of Civil and Environmental Engineering
Virginia Tech
Blacksburg, VA 24060, USA
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