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:

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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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<?php 
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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
$objImage = new Image;
$result = $objImage -> listImage();
$num_rows = mysql_num_rows($result);
//echo $num_rows."records in database";
code....

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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'";

[code]....

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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i'm getting an error when I load my php code in a browser. Here's my code snippet:

<?php
mysql_connect("localhost","root");
mysql_select_db("something");
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[Code]....

**I can't get it to display the first image (eg: Apple_iPhone3GS.jpg), I made sure that I have the image in same directory.

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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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This snippet downloads the image, but I want to display it instead:

$query = "SELECT * FROM upload";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
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[Code]....

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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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What is the best way to 'disable' it to prevent accidental double clicking.

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The code that prints the data is as follows

Code: [Select]{BEGIN key_image_fieldcolumn}<TD class=borderbody vAlign=middle {$key_image_style}>{$key_image_value} </TD>{END key_image_fieldcolumn}

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="http://www.gtdb.org/images/key_signatures/{$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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$display .= "<image scr="$image_url "  />";

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.....
...
..
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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:

function test(){
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<?php
$dbLink = new mysqli('localhost', 'root', '', 'gallery');
if(mysqli_connect_errno()) {
die("MySQL connection failed: ". mysqli_connect_error());
}
if(isset($_GET['id']) && is_numeric($_GET['id'])) {
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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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can anyone tell me or direct me to a link that shows how to use an image
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<head>
</head>
<body>

[Code].....

When submitting the form, I wold expect to be able to retrieve the value of getStarted, but that is not the case in IE.

Here are the URL's that are submitted:

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Code:
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<br>
<table class="menu" width=150 cellpadding=0 cellspacing=0 align=center>
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if (isset($_POST['submit_absence'])) {
                mysql_query($sqlInsert,$conn) or die('Error, query failed');    
}

submit_absence is the name of the submit image. PHP Code:

echo '<input type="image" name="submit_absence" src="images/submit.JPG"></input>'

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

echo '<input type="submit" name="submit_absence" value="Submit Request"></input>'

then it gets set and the query executes. I need to be able to use the image rather than the lame HTML submit button.

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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.

CSS:

/* this is the submit button */
#search {
width: 32px;
height: 32px;
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[Code]....

HTML:

<input type="submit" name="search" value="Search" id="search" />

** 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.

What should I do to replace the submit button with my own image? I'm not too good with jQuery...

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[Code]...

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<table>
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