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']; -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
