How to connect mySQL to SVG using RaphaelJS? - php

I am trying to link a MySQL DB to an SVG image to dynamically change the SVG elements with Raphael JS.
I have a MySQL DB where I query using PHP and display the results in table form to an html page: (The script below works and displays the username and a picture only when the condition of the timestamp is met.)
<?php
mysql_connect("","","");
mysql_select_db("");
$res=mysql_query("select username, picture from 'table' WHERE status > UNIX_TIMESTAMP(NOW()) - 300");
echo "<table>";
if (!$res) {
die("Query to show fields failed");
}
$fields_num = mysql_num_fields($res);
echo "<h1>Table:Status</h1>";
echo "<table border='1'><tr>";
for($i=0; $i<$fields_num; $i++)
{
$field = mysql_fetch_field($res);
echo "<td>{$field->name}</td>";
}
echo "</tr>\n";
while($row=mysql_fetch_array($res))
{
echo "<tr>";
echo "<td>"; echo $row["username"]; echo "</td>";
echo "<td>"; ?> <img src=" <?php echo $row["picture"]; ?>" height="50">
<?php
How can I take the similar concept above of displaying the results in table form to an SVG image where the SVG elements will change/update only when the query condition is met?
Here is my sample SVG image with 5 elements:
<polygon fill="#B2B2B2" points="150.3,8.8 203.8,31.7 169.8,91.4 133.4,75.8 "/>
<circle id="circleT3" circle fill="#FFFFFF" cx="163.1" cy="53.6" r="7.3"/>
<circle id="circle3_1" circle fill="#CCCCCC" cx="184.5" cy="82.4" r="7.3"/>
<circle id="circle3_5" circle fill="#CCCCCC" cx="136.6" cy="27.2" r="7.3"/>
<circle id="circle3_4" circle fill="#CCCCCC" cx="166.4" cy="7.3" r="7.3"/>
Can someone point me to some sample code or tutorial? Or is there a better way to do this? Thanks.
EDIT:
In MySQL DB I have a column for username, password and timestamp. When a user logs into webpage the timestamp updates. The PHP code above is used to query who has logged within 5 minutes ago from current time.
What I would like to do with this information with SVGs is create a graphical representation of the login.
So each username will have their own SVG element (a circle) associated with them and when they log in/out, that SVG element (code above) will change color.
Right now I do not know how to link the username with my SVG elements so the SVG element will dynamically update like my table I query from MySQL when the timestamp changes.

The answer will depend on further information that isn't really available until the rest is written.
You could combine Snap (to modify existing inline SVG or create it) or Raphael (to create new SVG only, you can't use it to modify inline SVG), or another SVG library of choice (eg svg.js or jquery.svg maybe).
Assuming you already have something to use on the page, that is showing the logged in user, you could do something like in pseudocode...
loop user;
if( document.getElementById( userId ) ) Snap('#' + userId + '_image').attr({ fill: 'green' });
(The svg reference may be the same as the circles, but somewhere you would need some type of lookup to know which circle is which userid)
This assumes the svg is on the page. If its not, you could create it with
paper.circle(x,y,r).attr({ fill: 'green' });
If you want it dynamic (so status changes without a refresh), you may need to tie ajax calls to get status from the mysql db, but if you already have a user name displaying on the page, I'm assuming that is already taken care of.

I have successfully update the svg element color to when the user logs in, the corresponding circle will change color. So I have my svg code from illustrator. I then put in this script in my php file:
window.onload = function () {
if(document.body.innerHTML.toString().indexOf('username') > -1){
circle1_1.setAttribute("fill", "yellow");
};
};
Whenever a user logs in, the info is populated on the table in the html from the MYSQL query and the script looks to see if that username is on the page and if it is, change color of SVG element.
So it basically links the SVG element to any value/variable.
Not the prettiest code or logic out there but for anyone else doing something similar, enjoy.

Related

Changing and retaining specific a-tag background color once clicked in PHP

I've got a a-tag dynamically populated which extracts data from a database and multiplies to the number of records existing. So this to be looked at more like an item in a list.
echo "<a class='Keylist ".$wecb_color."' id='myIDhere' href='index.php?idd=".$record['categoryid']."'><div>".$record['descriptor']."</div></a>";
I'm trying to keep one of the a-tags background set on a different color when a user clicks. Behavior I'm trying to achieve is when a user scrolls and clicked on one link, that a-tag to remain selected depicting a different background color.
I've written snippet trying to do that.
$wecb_color ='';
if(isset($_GET['idd'])){
$_SESSION['link'] = $_GET['idd'];
echo "<br><br><br><br><br>".$_SESSION['link'];
if($_GET['idd'] == $_SESSION ['link']) {
$wecb_color = 'changetogreen';
}
}
The issue is this makes all the a-tag's which are generated having a background color changed to green instead the one that user clicks.
Hard coded CSS script reads as :- .changetogreen{background-color: green;}
Any thoughts appreciated.
It'd change the background color of all links because you're changing the whole class style.
Try this:
echo "<a class='"Keylist " style=" '.$wecb_style.' ;" id='myIDhere' href='index.php?idd=".$record['categoryid']."'><div>".$record['descriptor']."</div></a>";
So you can change the style of particular tag.
And PHP would be:
$wecb_style ='';
if(isset($_GET['idd'])){
$_SESSION['link'] = $_GET['idd'];
echo "<br><br><br><br><br>".$_SESSION['link'];
if($_GET['idd'] == $_SESSION ['link']) {
$wecb_style = 'background: green';
}
}

i want to retrieve data from MySQL db by displaying it using table, arrange it the same way it appear in the image

<?php
$con = mysqli_connect("localhost","root","","shady");
$query = "SELECT * fROM hall1 ";
$result = mysqli_query($con,$query) or die(mysqli_error());
echo "<form action='food.php' method='post'>";
echo "<table >";
$size = 0;
while($row = mysqli_fetch_array($result))
{
$imagewidth=200;
if($size > 900)
{
echo"<tr>";
}
echo'<td width='.$td3=100 .'px > </td>' ;
echo'<td width='.$td1=200 .'px> <label><input type="radio" name="radii" value='.$row[1].' checked> <img src="'.$row['image'].'" width="200" height="200" style="margin-top:10px;"> </label></td>' ;
echo"<td width=".$td2=200 ."px> Name &nbsp: " .$row[1] ."<br> Size &nbsp &nbsp : ".$row[2] ."Person <br> Price &nbsp &nbsp: ".$row[3] ." SDG <br> See More </td>";
$size+= $imagewidth+$td1+$td2+$td3;
if($size > 900)
{
echo"<tr>";
}
}
echo"</table>";
echo "<BR><BR><BR><CENTER><input type='submit' name='radiii' value='next' width='200PX' height='200PX' ></a> </CENTER></b>";
echo "</form>";
}
?>
There are many problems with your code:
It's unclear what the variable size does. I'm willing to bet it's unnecessary
You are using deprecated code. The width attribute on HTML elements is no longer supported, and if you insist on using it, you should only enter the width in pixels without the px like so: <td width="200">. The <center> tag is also deprecated.
Your <label> tag is pretty much useless. What are you trying to achieve with it?
It's a bad practice to set the widths on all rows, just set the widths on the first row's cells. Even better, do not use any style directly on the HTML tags and using CSS rules instead.
Lastly, i believe it would be much more readable code to NOT echo HTML and instead use the following approach to "plug in" your variables from PHP into HTML code :
<? $a = "Some variable"; $b = 123; $c = "http://example.com/img/1.png"; ?>
<p><strong>My variable a: <?=$a?></p>
<p><strong>My variable b: <?=$b?></p>
<p><strong>My img: <img src="<?=$c?>"></p>
In short, wrap your PHP logic in <? and ?> (be sure to have short_open_tag = On in your PHP settings or you'll need to use the alternative opening tag <?php like you did.
The write plain old HTML without all the echo. Whenever you want to echo something you simply write a new PHP opening tag like so :
<? echo $a ?>
OR, the shorthand version
<?=$a?>
Now your real problem is about designing the page. And i think you are starting in the wrong order. I would suggest you
1- Build a design that works using some IDE like Adobe Dreamweaver or a freeware alternative. With that tool, make the design for 1 item (1 iteration of your while loop).
The next step is to see how you're gonna repeat it. You seem to want a Horizontal list that wraps around lines at a certain width. One way you would do that is that you would wrap all the list items with a div element for which you will set a fixed width. And inside you will have the code you have for each item repeated for all items. If your outer div has a width of 900px and each element consumes 300px in width and you have 9 items, you would have a 3x3 grid.
Here is a summary code of what i explained. In your PHP file you would have:
<div class="grid">
<? while($row = .... ){ ?>
<div class="item">
<!-- The code for the image, label and whatever information you want to have for each item -->
</div>
<? } ?>
</div>
And a CSS file which you would link in your PHP file, containing:
.grid {
width: 900px;
}
.item {
width: 300px;
}
If the information your are displaying come from user input you should sanitize it and have some sort of logic that breaks long text, either with CSS or PHP.
In short, you seem to be doing all the math in PHP so that you jump into a new line once the total width exceeds your desired grid width. This is a design issue and not a logic / programming issue. Therefore you should tackle this with CSS and not PHP.
I strongly suggest you either follow a video tutorial (there are plenty online) or read some documentations on HTML, CSS, JavaScript, PHP and MySQL in that order.
Here is a link to get you started: Mozilla Developer Network

PHP display image from link

I am trying to get my image to display when clicked via the link. However when I click on the link it finds the image id as displayed in the url link ok but does not display any image. Can you please help?
<?php
//sets up thisPage
$pageSize=10;
if (isset($_POST["thisPage"])) $thisPage=$_POST["thisPage"];
else $thisPage=1;
//selects all distinct expenses that have been uploaded
$dbQuery="SELECT * FROM images WHERE user_id = '$userID' ";
$dbResult=mysqli_query($db_connection, $dbQuery) or die(mysqli_error($db_connection));
echo "<table cellspacing=\"5\" class=\"recordsTableBG\"> <thead
class=\"recordsTableHeader\">";
echo '<tr> <th>ID</th><th>Amount</th><th>Description</th><th>Filename</th>
<th>Project ID</th><th>Status</th></tr></thead>';
echo '<tr class="alternateRowColor">';
'<tr>';
while ($dbRow=mysqli_fetch_array($dbResult)){
echo "<img src = 'uploaded/$image' width = '200' height = '200'>";
// display row with expense
echo '<td>'. $dbRow['id'] .'</td>';
echo '<td>'. $dbRow['user_id']. '</td>';
echo '<td><a href='.$_SERVER['PHP_SELF'].'?imageid='.$dbRow['id'].'>
Click here to view image</a></td>';
}
echo "</table>";
echo "</form>";
?>
<!-- add submitt button
close form -->
</div>
You have a mixture of two approaches going on here, and that's where the confusion lies.
First, notice your
<img src='uploaded/$image' width = '200' height = '200'>
The browser is going to see something like
<img src='uploaded/picture_of_cat.jpg' width='200' height='200'>
and render a 200px by 200px image of a cat. I'm assuming that's your thumbnail. (As an aside, I notice that it's not enclosed in <td></td> even though it's in the table row).
In another part of the table, you have
<a href='.$_SERVER['PHP_SELF'].'?imageid='.$dbRow['id'].'>Click here to view image</a>
which is going to render as something like
<a href='http://example.com/index.php?imageid=1234'>Click here to view image</a>
When the user clicks that link, it's going to make a GET request to the server with imageid equal to 1234. What is the server going to do with that imageid? In the code you've posted, nothing.
You have two choices:
if you want to keep the link as it is, you'll have to write some code that will take the imageid value, find the appropriate image, and return it to the browser as image data - that means setting the appropriate headers and sending the data back as binary data.
the simpler way to do it would be to replace the URL in the link with the same one you have in your <img> tag - when the user clicks on it, the server will just return the image.

Php image customization

I have an image gallery displaying all photos from my file directory with php. I have defined what type of file to use etc but the main line of code is as below:
echo '<img src="'.$dir.'/'.$file.'" alt="'.$file.'" width="600px" />';
I know by changing the width=" " above I can alter how many images I have in a row but if i had a drop-down box, with options stating how many columns i want the photos displayed in and a button to refresh the page to show the columns option i have selected.
So for example: I select a drop down option "15 columns" and hit the refresh button, the page should then refresh and the images be redisplayed in columns of 15. How would i go about this?
You should be using HTML Tables for it. Anyhow here is how you should do it.
$allowed_per_line = $_GET['columns'];
$counter = 0;
while($image = fetch...) {
$counter++;
echo "<img src='{$dir}/{$file}' alt='{$file}' />";
if($counter==$allowed_per_line) {
echo '<br />';
$counter=0;
}
}

Script to move to next image

I am writing a script to display images on my site. The images are stored on the server and the source (path) is on a mySQL database. I want to be able to move to the next (or previous picture) by invoking two php functions I have made to do so. The code I have written so far is:
<?php
require "db_connection.php";
$query="SELECT source FROM photos";
$result=mysql_query($query);
$count=0;
$numberofpics=mysql_num_rows($result);
$image=mysql_result($result, $count, "source");
$num=1;
function slideshowForward() {
$num=$num+1;
if($num==($numberofpics+1)) {
$num=1;
}
$count=$count+1;
$image=mysql_result($result, $count, "source");
}
function slideshowBack() {
$num=$num-1;
if ($num==0) {
$num=$numberofpics;
}
$count=$count-1;
$image=mysql_result($result, $count, "source");
}
?>
The html portion to display the images is:
<!-- FORWARD AND BACK FUNCTIONS-->
<a class="back" href="http://mywebsite.com/discoverandrank.php?function=slideshowBack()"> <img src="graphics/back.png"/></a>
<a class="next" href="http://mywebsite.com/discoverandrank.php? function=slideshowForward()"><img src="graphics/forward.png"/></a>
<!--DISPLAY MIDDLE PHOTO-->
<div id="thepics">
<img src="http://www.mywebsite.com/<?php echo $image; ?>" name="mypic" border=0 height="300" width="500"></img>
</div>
I'm pretty sure that the php script is incrementing/decrementing the count currently for the image, but I think the problem might be because the html (specifically img src="....") part is not re-evaluated when the count of the image increases?
Mmm there are two things to point out here:
I think that passing a function name as a parameter on an action is not a good idea, instead you should user something like: "http://mywebsite.com/discoverandrank.php?op=next&current=10" and then evaluate the op and current value to take some action.
For gettting the next picture from database you could use something like this:
"Select source FROM photos LIMIT "current+1" , 1"
For gettting the previous picture from database you could use something like this:
"Select source FROM photos LIMIT "current-1" , 1"
This way you dont need to keep counters and stuff, just have to check if those queries return any data.
Hope this helps!

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