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PHP - Creating Dynamic Images
Hello 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! Similar TutorialsThis topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=342341.0 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. This topic has been moved to mod_rewrite. http://www.phpfreaks.com/forums/index.php?topic=357949.0 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. :/ Folks, I need help (Php code ) to generate a Dynamic Text on a Base Image. What i want to do is, to make this Image as header on my Site and to make this Header Specific to a Site, i want to Add the Domain Name on the Lower Left of the Image. Got the Idea? Here is the Image link: Quote http://img27.imageshack.us/i/shoppingheader1.jpg/ PHP Variable that holds the Domain name is: $domain All i need the Dynamic PHP Codes that i can put on all my sites to generate this Text on Image (Header) Dynamically... May Anyone Help me with this Please? Cheers Natasha T. Hi all I need to combine these two scripts: Firstly, the following decides which out of the following list is selected based on its value in the mySQL table: <select name="pack_choice"> <option value="Meters / Pack"<?php echo (($result['pack_choice']=="Meters / Pack") ? ' selected="selected"':'') ?>>Meters / Pack (m2)</option> <option value="m3"<?php echo (($result['pack_choice']=="m3") ? ' selected="selected"':'') ?>>Meters / Pack (m3)</option> <option value="Quantity"<?php echo (($result['pack_choice']=="Quantity") ? ' selected="selected"':'') ?>>Quantity</option> </select> Although this works OK, I need it also to show dynamic values like this: select name="category"> <?php $listCategories=mysql_query("SELECT * FROM `product_categories` ORDER BY id ASC"); while($categoryReturned=mysql_fetch_array($listCategories)) { echo "<option value=\"".$categoryReturned['name']."\">".$categoryReturned['name']."</option>"; } ?> </select> I'm not sure if this is possible? Many thanks for your help. Pete I have this script from http://lampload.com/...,view.download/ (I am not using a database) I can upload images fine, I can view files, but I want to delete them. When I press the delete button, nothing happens
http://www.jayg.co.u...oad_gallery.php
<form>
<?php $dir = dirname(__FILENAME__)."/images/gallery" ; $files1 = scandir($dir); foreach($files1 as $file){ if(strlen($file) >=3){ $foil = strstr($file, 'jpg'); // As of PHP 5.3.0 $foil = $file; $pos = strpos($file, 'css'); if ($foil==true){ echo '<input type="checkbox" name="filenames[]" value="'.$foil.'" />'; echo "<img width='130' height='38' src='images/gallery/$file' /><br/>"; // for live host //echo "<img width='130' height='38' src='/ABOOK/SORTING/gallery-dynamic/images/gallery/ $file' /><br/>"; } } }?> <input type="submit" name="mysubmit2" value="Delete"> </form>
any ideas please?
thanks
Hello! If you search Google, you'll notice the URL: http://www.google.com/webhp?hl=en#hl=en&source=hp&q=php+freaks&aq=f&aqi=g10&aql=&oq=&gs_rfai=CtdaTdtdRTNnPD5HuzASZtMWiCgAAAKoEBU_Q0ZT5&pbx=1&fp=19d754eee0b4f223 You can copy that URL anywhere you like... and the user will still see the same results. So basically the URL dynamically does an action, and accesses the database. How on do I make a URL like this? Meaning when you change the URL parameter values, it request a slightly different database query? Does that make sense? Similarly with NexTag.com: http://www.nextag.com/serv/main/buyer/ProductCompare.jsp?search=camera&page=0&node=500001&psort=%2FDigital-Cameras--zzcameraz500001zB6z5---html&zipcode=&cptitle=657166355&cptitle=656751324&cptitle=620051906&cptitle=705150048 That was a Dynamically made URL (I selected from check boxes which products to compare) and it makes that URL so anyone can see those products. Any ideas on how to do this? More examples on the URL: http://www.cars.com/go/compare/modelCompare.jsp?myids=9721,11439 (i select the cars, it generates that URL. Notice the IDs 9721,11439) Hello everyone, I found a tutorial on creating a pdf with php and I just tried to create a basic pdf to see if it worked and modify it from there. Here is the link to the tutorial: http://www.astahost.com/info/tfclt-create-pdf-php-create-fly-pdf-web-server.html Well, after uploading the following php file to the server and testing it, I got this error: Fatal error: Call to undefined function pdf_new() in /home/caresa6/public_html/BubblyBounce.com/admin/pdftest.php on line 2 Here's the code that produced the error: Code: [Select] <?php $mypdf = PDF_new(); PDF_open_file($mypdf, ""); PDF_begin_page($mypdf, 595, 842); $myfont = PDF_findfont($mypdf, "Times-Roman", "host", 0); PDF_setfont($mypdf, $myfont, 10); PDF_show_xy($mypdf, "Sample PDF, constructed by PHP in real-time.", 50, 750); PDF_show_xy($mypdf, "Made with the PDF libraries for PHP.", 50, 730); PDF_end_page($mypdf); PDF_close($mypdf); $mybuf = PDF_get_buffer($mypdf); $mylen = strlen($mybuf); header("Content-type: application/pdf"); header("Content-Length: $mylen"); header("Content-Disposition: inline; filename=gen01.pdf"); print $mybuf; PDF_delete($mypdf); ?> Can anyone help me figure out what to do from here? I have some PHP knowledge, but I'm treading in uncharted waters with this, so I am totally lost. Thanks in advance. trying to be able to store client information in mysql db for reference later in an admin area. The code I have so far allows me to list the client names, but once I "select" the name I want it to show the rest of the database information stored for that user (email, height, weight, phone number, etc...) which it currently is not doing. Any and all help will be greatly appreciated, tired of struggling through this blindly. Code: [Select] <? // Connect database mysql_connect("localhost","",""); mysql_select_db("mydb"); if(isset($select)&&$select!=""){ $select=$_GET['select']; } ?> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> </head> <body> <form id="form1" name="form1" method="get" action="<? echo $PHP_SELF; ?> "> Client Name : <select name="select"> <option value="">--- Select ---</option> <? // Get records from database (table "users"). $list=mysql_query("select * from users order by id asc"); while($row_list=mysql_fetch_assoc($list)){ ?> <option value="<? echo $row_list['id']; ?>" <? if($row_list['id']==$select){ echo "selected"; } ?>><? echo $row_list['name']; ?></option> <? } ?> </select> <input type="submit" name="Submit" value="Select" /> </form> <hr> <p> <? if(isset($select)&&$select!=""){ // Get records from database (table "users"). $result=mysql_query("select * from users where id='$select'"); $row=mysql_fetch_assoc($result); ?> Information about <strong><? echo $row['name']; ?></strong> client...</p> <p><? echo $row['email']; ?><br> ........................................<br> ........................................ <? // End if statement. } // Close database connection. mysql_close(); ?> </p> </body> </html> I'm having trouble getting the dynamic data from my <select> menus to write into my MYSQL database. Can anyone see what I'm doing wrong here? First post btw The output looks like this, which is obviously wrong: Code: [Select] <html> <head> </head> <link rel="stylesheet" type="text/css" href="./css/newuser.css" /> <body> <?php session_start(); require 'default.inc.php'; ?> <?php if (isset($_POST['amount'])): $host = 'localhost'; $user = 'user'; $pass = 'password'; $conn = mysql_connect($host, $user, $pass); if (!$conn) { exit('<p>Unable to connect to the database server</p>'); } if (!@mysql_select_db('spikesusers')) { exit('<p>Unable to locate the database</p>'); } $locationname = $_POST['donor']; $donorid = mysql_query("SELECT id FROM donors WHERE locationname='$locationname'"); $amount = $_POST['amount']; $year = $_POST['year']; $type = $_POST['type']; $typeid = mysql_query("SELECT id FROM donationtype WHERE type='$type'"); $player = $_POST['player']; //$playerid = mysql_query("SELECT id FROM players WHERE $sql = "INSERT INTO donations SET donorid='$donorid', amount='$amount', yearofdonation='$year', typeid='$typeid'"; mysql_query($sql); ?> <div class='standard'> <h1>Donation Management</h1> <?php if ($sql) { echo "New donation added "; echo "<p></p>"; echo "<a href=managedonations.php>Back to donation management</a>"; exit(); } else { echo "Error adding new donation"; echo "<a href=adddonation.php>Try again</a>"; exit(); } ?></div> <?php else: $host = 'localhost'; $user = 'user'; $pass = 'pass'; $conn = mysql_connect($host, $user, $pass); if (!$conn) { exit('<p>Unable to connect to the database server</p>'); } if (!@mysql_select_db('spikesusers')) { exit('<p>Unable to locate the database</p>'); } $donor=@mysql_query('SELECT id, locationname FROM donors'); ?> <div class='standard'> <h1>Donation Management</h1> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <label>Donor: <select class="text" name="donor"> <option value=new>Add a new donor...</option> <?php while ($donors=mysql_fetch_array($donor)) { $donorname=$donors['locationname']; echo "<option value='<?php echo $donorname;?>'>$donorname</option>"; //echo "<option value='hi'>hi</option>"; } ?> </option> </select> </label><br /> <label>Amount: <input class="text" type="text" name="amount" class="text" /></label><br /> <label>Year of donation: <select class="text" name="year"> <option value='2011'>2011</option> <option value='2010'>2010</option> <option value='2009'>2009</option> </select> </label><br /> <?php $player=@mysql_query('SELECT id, firstname, lastname FROM players'); ?> <label>Player: <select class="text" name="player"> <option value="player" selected="selected"></option> <?php while ($players=mysql_fetch_array($player)) { $playerfirstname=$players['firstname']; $playerlastname=$players['lastname']; echo "<option value=player>$playerfirstname $playerlastname</option>"; } ?> </select> </label><br /> <?php $type=@mysql_query('SELECT id, type FROM donationtype'); ?> <label>Donation type: <select class="text" name="type"> <?php while ($types=mysql_fetch_array($type)) { $donationtype=$types['type']; echo "<option value=type>$donationtype</option>"; } ?> </select> </label><br /> <input type="submit" value="SUBMIT" class="buttons"/> <input type="button" name="Cancel" value="CANCEL" onclick="window.location = 'managedonations.php'" class="buttons"/> </form> </div> <?php endif; ?> </body> </html> |
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