Re: Retrieving Image Location in PHP from MySQL
| From: | Sashikanth Gurram | Date: | Tue, 10 Mar 2009 01:14:21 +0000 |
| Subject: | Re: Retrieving Image Location in PHP from MySQL | ||
| References: | 1 2 3 4 5 6 7 8 9 10 11 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-289731@lists.php.net to get a copy of this message | ||
haliphax wrote:
On Mon, Mar 9, 2009 at 10:52 AM, Sashikanth Gurram <sashi34u@vt.edu> wrote: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.haliphax wrote: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)wrote:On Mon, Mar 9, 2009 at 10:24 AM, Sashikanth Gurram <sashi34u@vt.edu>Nathan Nobbe wrote:<mailto:sashi34u@vt.edu>> 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.On Mon, Mar 9, 2009 at 7:32 AM, Sashikanth Gurram <sashi34u@vt.edu
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 -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Sashikanth Gurram Graduate Research Assistant Department of Civil and Environmental Engineering Virginia Tech Blacksburg, VA 24060, USA