Image Array, Display One Image, With 'Next' Button

Oct 25, 2007

I need help building an array and have it display the $row[i] based on a $_GET variable number in the URL- I want to show one image from an album and have a 'NEXT' button so that I can cycle though all the images in that album.

This is a very small/simple gallery, all images will be in one table as shown below. Code:


Select Button W/ Image And Have That Image Display On Another Page?

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Would like to be able to click on a radio button that represents an image. Once selected and submitted, have that image display on another page. I have an idea, but need some guidance.

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i doubt whether this is possible, if it is possible i ll be delighted . I have a form of php array , with a submit button passing the values , this submit button is so ordinary , so i want to change the look and feel of button , where i have a button image for it , but i cant fix it , ths submit button in php is this,

$form['submit'] = array('#type' => 'submit', '#value' => t('Create Account'), '#weight' => 40);

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I would be grateful if anybody could help me out with this.

I have a table which stores the filepath of a set of images, e.g col filepath stores values like: ./phpimages/image3.jpg. Note that my images are stored in folder 'phpimages'

Now i want to loop through all the rows in my table and display thumbnails of the images.

Here is my code:

*************************Display all records from images table***************/
//create an instance is Image
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I would like to display one image at a time on a webpage. I used a query to get a set of images from the database into an array. I have tried along these lines:

$my=& JFactory::getUser();
$db=& JFactory::getDBO();
$a=& $my->id;
$query="SELECT id FROM jos_phocagallery_categories WHERE accessuserid='$a'";


I changed echo line above because the line was not allowed After this code I've tried to add an input button that calls "next($results)" and then replace $image, but it seems the whole query is beeing implemented from the beginning again. So I can't get the pointer of the array to step forward. Not even in an echo.

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when I upload image ..the php will resize the image into two type size and put into two folder "folder for images" and "folder for thumb" and size for all this type of image are...images = 100x100 thumb = 80x80 with same name file.

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$query = "SELECT * FROM upload";
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I came across following problem. I have a table where I store user profile images, but I want to display default image if user has no image uploaded. what I do is I look into "profile_images" table where $owner_id=$user_id. I need to figure how to do something like this:

This is not a real code, just the logic

(Select all from "profile_images" where $owner_id=$user_is) if user has uploaded image (so table is not empty) set image file as user profile picture else if table is empty set image with id=0 as user image (so set default image as user profile image)  

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I know that an if statement would do the trick, but I can't do "if's" inside of the "echo" is it possible to accomplish this?  How? Code:

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The code that prints the data is as follows

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Code: [Select]{$key_image_value} is actually the code that prints the code from the database. but when i try to envelop it with an <img> tag I get errors and it does not display the image.

This is what I have already tried:

Code: [Select]<img src="{$key_image_value}">

It would be lovely to have the actual time signatures print out in all their glory!!

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Code: [Select]
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I need to call it in another function and see its output. How can i do that? I want something like:

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$dbLink = new mysqli('localhost', 'root', '', 'gallery');
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if i use "<input type="submit" name="btnsearch">" it works.. so how do i use image button?

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I am building a site with PHP and MySQL and it makes extensive use of forms. I have one such form and I want to make the submit button an image.I have tried

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if (isset($_POST['submit_absence'])) {
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submit_absence is the name of the submit image. PHP Code:

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If I use the above code, nothing is ever set and the query never executes. if I use..
PHP Code:

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I'm using a jQuery plugin called jQtransform [URL] which skins HTML form elements like the input box and submit buttons. All is well until I need to use an image as the submit button. I tried the following CSS code but the original button still appears, and not the image.


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#search {
width: 32px;
height: 32px;
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** HTML code looks like it has been processed by the jQuery plugin when viewed in Chrome's 'Inspect Element' feature. The above is the original HTML code as seen when you select 'View Page Source' in Chrome.

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