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PHP - Moved: Creating Dynamic Images
This topic has been moved to Miscellaneous.
http://www.phpfreaks.com/forums/index.php?topic=342341.0 Similar TutorialsHello guys. I'm kinda new to PHP, I've been understanding everything so far. I have trouble with "imagestring" What I want is to query a row in MySQL. I want to display some of that info in the image. It's not working properly. Here's the code: <?php $imagepath="css/images/dog.jpg"; $image=imagecreatefromjpeg($imagepath); $imgheight=imagesy($image); $color=imagecolorallocate($image, 255, 255, 255); include("config.php"); $result = mysql_query("SELECT * FROM playerinfo WHERE Level > 0 ORDER BY Level DESC", $connect); while($myrow = mysql_fetch_row($result)) { // Not working properly. imagestring($image, 5, 50, $imgheight-50, $myrow[0], $color); } // This line works like it should: //imagestring($image, 5, 50, $imgheight-50, "Using it like this works!", $color); header('Content-Type: image/jpeg'); imagejpeg($image); ?> Any help will be GREATLY appreciated it. Thanks! This topic has been moved to mod_rewrite. http://www.phpfreaks.com/forums/index.php?topic=357949.0 Hi Everyone I found some code to decrease the writing of $_POST variables especially on big forms, but I cannot seem to get it to work: I am trying to get the $_POST variables from this: Code: [Select] <?php $callsign = $_POST['mf_callsign']; $surname = $_POST['mf_surname']; $firstnames = $_POST['mf_firstnames']; $knownas = $_POST['mf_knownas']; $rsaid = $_POST['mf_rsaid']; $birthdate = $_POST['mf_birthdate']; //more code [code] to this (see //Create $_POST Variables section) [code] <?php //Initailise Database first mysql_connect ("localhost", "***", "***") or die ('I cannot connect to the database because: ' .mysql_error()); mysql_select_db ("***"); ========================= //Create $_POST Variables foreach($_POST as $inputKey=>$inputValue): //find all form fields starting with 'mf_' if((strstr($inputKey,'mf_') === true)): $inputKey = mysql_real_escape_string($inputValue); endif; endforeach; ========================= //Then do the Insert into the Table $query="INSERT INTO personal_details (callsign, surname, firstnames, knownas, rsaid, birthdate)Values ('$mf_callsign', '$mf_surname', '$mf_firstnames', '$mf_knownas', '$mf_rsaid', '$mf_birthdate')"; mysql_query($query) or die ('Error Inserting Data into Database'); //If Insert Successful, Goto next Form header("Location: contactdetails.html"); die; ?> The //Create $_POST Variables section is not correct as it is not setting the variables correctly. If I go back to the old way and use that ($callsign = $_POST['mf_callsign'] it inserts no problem. What Am I doing wrong? Regards Allen Hello,
I have this very frustrating problem I'm trying to make a dynamic table with only 5 columns per row. So every 5 items, I need a new row. I have tried many different examples with no success. What is the best way to approach this? Here is what I am working with, this doesn't show what I have tried, just what I am working with at the moment which of course, just outputs it one column per row. http://www.mesquitew.../inc-legend.php
// Lets parse the data
$entries = simplexml_load_file($data);
if(count($entries)):
//Registering NameSpace
$entries->registerXPathNamespace('prefix', 'http://www.w3.org/2005/Atom');
$result = $entries->xpath("//prefix:entry");
foreach ($result as $entry):
$event = $entry->children("cap", true)->event;
// Set the alert colors for the legend
include ('../inc-NWR-alert-colors.php');
// Lets creat some styles for the list
$spanStyle = "background-color:{$alertColor};border:solid 1px #333;width:15px;height:10px;display:inline-block;'> </span><span style='font-size:12px;color:#555;";
$legend .= "<table>";
$legend .= "<tr>";
$legend .= "<td> <span style='$spanStyle'> $event</span></td>";
$legend .= "</tr>";
$legend .= "</table>";
endforeach;
endif;
echo $legend;
-Thanks!
Hi there,
I have a table in a MySQL database where I keep a list of user privileges. I am trying to create variables where the name of variable matches the privileges in the table.
This is also known as variable variables (I think).
EDIT (17/07/2014 04:02 PM): This might be a better way to describe what I'd like, so if the value from the table is admin_panel I'd like to dynamically create a variable with that name.
I have created a code so far, but all I seem to be getting is a list of Notice errors telling me that the variable is undefined. (I have supplied a list of errors a bit further down the post).
Here is the code:
<?php
$host = "localhost";
$account = "***";
$password = "****";
$dbname = "****";
$connect = mysql_connect($host,$account,$password) or die("Unable To Connect");
$db = mysql_select_db($dbname,$connect) or die("Unable To Select DB");
$perm_query = "SELECT * FROM `privileges`";
$permission_query = mysql_query($perm_query);
while($row = mysql_fetch_array($permission_query))
{
$rows[] = $row;
}
foreach($rows as $row)
{
${$row['privilege']};
}
?>
The list of errors:
Notice: Undefined variable: admin_panel in C:\xampp\htdocs\DynamicVariables.php on line 20
Notice: Undefined variable: create_user in C:\xampp\htdocs\DynamicVariables.php on line 20 Notice: Undefined variable: edit_user in C:\xampp\htdocs\DynamicVariables.php on line 20 Notice: Undefined variable: delete_user in C:\xampp\htdocs\DynamicVariables.php on line 20 Notice: Undefined variable: create_group in C:\xampp\htdocs\DynamicVariables.php on line 20 Notice: Undefined variable: edit_group in C:\xampp\htdocs\DynamicVariables.php on line 20 Notice: Undefined variable: delete_group in C:\xampp\htdocs\DynamicVariables.php on line 20 Notice: Undefined variable: view_log in C:\xampp\htdocs\DynamicVariables.php on line 20 Notice: Undefined variable: log_settings in C:\xampp\htdocs\DynamicVariables.php on line 20 Notice: Undefined variable: password_change in C:\xampp\htdocs\DynamicVariables.php on line 20 Thanks Edited by chrisrulez001, 17 July 2014 - 10:05 AM. i've to populate google charts with dynamic data. the data looks llike. +----+-------+---------+-------+-----------+---------+------+ | id | name | physics | maths | chemistry | biology | sst | +----+-------+---------+-------+-----------+---------+------+ | 1 | Name1 | 10 | 25 | 35 | 42 | 62 | | 2 | Name2 | 80 | 45 | 45 | 45 | 25 | | 3 | Name3 | 63 | 25 | 63 | 36 | 36 | | 4 | Name4 | 82 | 36 | 75 | 48 | 42 | | 5 | Name5 | 45 | 45 | 78 | 25 | 24 | | 6 | Name6 | 36 | 36 | 15 | 75 | 36 | | 7 | Name7 | 99 | 45 | 24 | 24 | 45 | | 8 | Name8 | 45 | 85 | 85 | 85 | 96 | +----+-------+---------+-------+-----------+---------+------+ i have to create google charts based on this such that # namewise - such that when i select a name it displays all subject marks of that particular name in a column chart #markswise - when i select a subject, it displays all the names with marks in that particular subject. conisdiering that data may be added and i've to accumulate that also, for namewise i did
// chart.php
<?php
include("connection.php");
$query = "SELECT name FROM csv GROUP BY name DESC";
$statement = $connect->prepare($query);
$statement->execute();
$result = $statement->fetchAll();
?>
<!DOCTYPE html>
<html>
<head>
<title>Google charts</title>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<script src="https://code.jquery.com/jquery-1.12.4.js"></script>
</head>
<body>
<br /><br />
<div class="container">
<div class="panel panel-default">
<div class="panel-heading">
<div class="row">
<div class="col-md-9">
<h3 class="panel-title">Student Wise Marks Data</h3>
</div>
<div class="col-md-3">
<select name="name" class="form-control" id="name">
<option value="">Select Student</option>
<?php
foreach($result as $row)
{
echo '<option value="'.$row["name"].'">'.$row["name"].'</option>';
}
?>
</select>
</div>
</div>
</div>
<div class="panel-body">
<div id="chart_area" style="width: 1000px; height: 620px;"></div>
</div>
</div>
</div>
</body>
</html>
<script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
<script type="text/javascript">
google.charts.load('current', {packages: ['corechart', 'bar']});
google.charts.setOnLoadCallback();
function load_student_data(name, title)
{
var temp_title = title + ' '+name+'';
$.ajax({
url:"fetch.php",
method:"POST",
data:{name:name},
dataType:"JSON",
success:function(data)
{
drawStudentwiseChart(data, temp_title);
}
});
}
function drawStudentwiseChart(chart_data, chart_main_title)
{
var jsonData = chart_data;
var data = new google.visualization.DataTable();
data.addColumn('number', 'Physics');
data.addColumn('number', 'Maths');
data.addColumn('number', 'Chemistry');
data.addColumn('number', 'Biology');
data.addColumn('number', 'SST');
$.each(jsonData, function(i, jsonData){
var Physics = jsonData.Physics;
var Maths = jsonData.Maths;
var Chemistry = jsonData.Chemistry;
var Biology = jsonData.Biology;
var SST = jsonData.SST;
data.addRows([[Physics,Maths,Chemistry,Biology,SST]]);
});
var options = {
title:chart_main_title,
hAxis: {
title: "Subjects"
},
vAxis: {
title: 'Percentage'
}
};
var chart = new google.visualization.ColumnChart(document.getElementById('chart_area'));
chart.draw(data, options);
}
</script>
<script>
$(document).ready(function(){
$('#name').change(function(){
var name = $(this).val();
if(name != '')
{
load_student_data(name, 'Student wise marks data');
}
});
});
</script>
and in order to fetch data
// fetch.php
<?php
//fetch.php
include('connection.php');
if(isset($_POST["name"]))
{
$query = "
SELECT * FROM csv
WHERE name = '".$_POST["name"]."'
ORDER BY id ASC
";
$statement = $connect->prepare($query);
$statement->execute();
$result = $statement->fetchAll();
foreach($result as $row)
{
$output[] = array(
'Physics' => $row["Physics"],
'Maths' => $row["Maths"],
'Chemistry' => $row["Chemistry"],
'Biology' => $row["Biology"],
'SST' => $row["SST"],
);
}
echo json_encode($output);
}
?>
it diplays a blank page. however when i play with singular values i.e one subject at a time it displays fine Edited March 19, 2019 by zetastreakI want to create an image using pre-existing image files. I want to use more than one file, otherwise I would use the imagecreatefromjpg function. What function should I use to create an image using a for statement to determine how many of each image is being displayed in the new image? Hello all I have a requirement to create images mostly made up of rectangles circles and squares etc I have discovered imageline but is there anything a bit more exspansive? Hi All, i am trying to Generate a image dynamically. Up to here i am able to do it . But when i give any echo statement with this its not working.. Here is my code pls help me in this <?php // Create a 55x30 image $im = imagecreatetruecolor(79, 79); $unknow = imagecolorallocate($im, 0, 255, 255); $white = imagecolorallocate($im, 255, 255, 255); // Draw a white rectangle imagefilledrectangle($im, 4, 4, 75, 25, $unknow); imagefilledrectangle($im, 4, 29, 75, 50, $unknow); imagefilledrectangle($im, 4, 54, 75, 75, $unknow); // Save the image echo 'hi hello'; header("Content-type: image/png"); //imagepng($im, './imagefilledrectangle.png'); imagepng($im); imagedestroy($im); ?> Thanks in advance. So, We are trying to create a animation using the gifs in a directory by using the first 10 images. We just cannot figure out how to do the animation.. needs to have a 3 second delay between each frame Heres the code at the moment. Which creates the starting images and also sorts the images in the directory by time.. <?php $time = date("YmdHis"); $handle_data = file_get_contents('http://www.bungie.net/Stats/Halo3/Nightmap.ashx'); $img = new Imagick(); $img->readImageBlob($handle_data); $img->writeImage('nightmap/'.$time.'.gif'); foreach (glob('nightmap/*.gif') as $f) { # store the image name $list[] = $f; } sort($list);# sort is oldest to newest, $gif_0 = array_pop($list); $gif_1 = array_pop($list); $gif_2 = array_pop($list); $gif_3 = array_pop($list); $gif_4 = array_pop($list); $gif_5 = array_pop($list); $gif_6 = array_pop($list); $gif_7 = array_pop($list); $gif_8 = array_pop($list); $gif_9 = array_pop($list); ?> Hi all, Could someone help me add a thumbnail script to the below that works on scaling it down to 200px x 133px. I guess it is not that hard. <?php $destination='aircraft/'.$reg."1.jpg"; $temp_file = $_FILES['image']['tmp_name']; move_uploaded_file($temp_file,$destination); ?> Thanks This is so hard. :/ I have the following code: Code: [Select] <?php $signature = @imagecreate(350,60); $image_generate = imagecolorallocate($signature,84, 84, 84); $red = imagecolorallocate($signature,255,36,0); imagefilledrectangle($signature, 0, 1, 150, 30, $red); header("Content-type:image/png"); imagepng($signature); ?> How can I get the rectangle in the center? http://zchighscores.99k.org/image.php I don't really know what to change to get it in the right coordinates. :/ This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=306771.0 This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=333152.0 This topic has been moved to Application Design. http://www.phpfreaks.com/forums/index.php?topic=342882.0 This topic has been moved to Application Design. http://www.phpfreaks.com/forums/index.php?topic=356776.0 This topic has been moved to PHP Applications. http://www.phpfreaks.com/forums/index.php?topic=333014.0 This topic has been moved to PHP Regex. http://www.phpfreaks.com/forums/index.php?topic=351100.0 This topic has been moved to mod_rewrite. http://www.phpfreaks.com/forums/index.php?topic=321416.0 This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=342456.0 |
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