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PHP - Giving One Variable More Weight Over Another (working With Chance & Percents)
I am trying to create a script that will choose from 3 different Variables. This is much like a Chance game.
Variable A has a 20% chance to be chosen Variable A has a 35% chance to be chosen Variable A has a 45% chance to be chosen How would I go about setting each of these variables chance to be chosen? Similar TutorialsI can't figure out how I'm messing up so badly. Anyway, I have:
$con=mysqli_connect("removedforsecurity");
$finderCleanup = mysqli_query($con, "SELECT * FROM Entries ORDER BY PID DESC LIMIT 1");
$finderCleanupArray = mysqli_fetch_array($finderCleanup);
$integerCleanup = $finderCleanupArray['PID'] - 40;
$lastPID = $finderCleanupArray['PID'];
$lastPIDreduced = $finderCleanupArray['PID'] - 8;
When I var_dump $lastPID, it returns INT with a number (it works correctly.)
Then, I have this:
function getResults($con,$lastPIDreduced)$con works correctly in the function, but $lastPIDreduced is "undefined." What is my sin? Thanks. Probably real basic but i cant get to grips with it for the life of me, can anyone correct this and explain how its done please so i gain a better understanding of php thanks. <?php if(SELECT mode FROM settings WHERE mode = 1){ ?> What is the error in this code? It's such a simple bit of code, but doesnt work... So make me look stupid by pointing out my coding error!!! Good luck!! $dbName = $_GET['dbIdent']; $dirArr = scandir(("uploadedPGN/gameDatabases/"), 0); foreach ($dirArr as $key => $value) { if (is_null($value) || $value=="." || $value==".." || $value=="...") { unset($dirArr[$key]); } } ///// works all the way to here \\\\\ echo '<form method="POST"><select name="process"><OPTION>Select Database: </OPTION>'; foreach($dirArr as $val2) { echo '<OPTION value="processPGN_NONANCL_DB.php?dbIdent='.$val2.'">'.$val2.'</OPTION>'; } echo '</select><input type="submit"></form>'; ////// so why is the little bit of code above causing me such a nightmare? grrr \\\\\ I have a script that adds points together based upon the placing. This is the actual script: Code: [Select] <? $points = 0; if($place === '1st') {$points = $points + 50;} elseif($place === '2nd') {$points = $points + 45;} elseif($place === '3rd') {$points = $points + 40;} elseif($place === '4th') {$points = $points + 35;} elseif($place === '5th') {$points = $points + 30;} elseif($place === '6th') {$points = $points + 25;} elseif($place === '7th') {$points = $points + 20;} elseif($place === '8th') {$points = $points + 10;} elseif($place === '9th') {$points = $points + 10;} elseif($place === '10th') {$points = $points + 10;} elseif($place === 'CH') {$points = $points + 50;} elseif($place === 'RCH') {$points = $points + 40;} elseif($place === 'TT') {$points = $points + 30;} elseif($place === 'T5') {$points = $points + 30;} elseif($place === 'Champion') {$points = $points + 50;} elseif($place === 'Reserve Champion') {$points = $points + 40;} echo "Total HF Points: $points"; ?>What it *should* do (my friend's script works the same way and it works) it starts at points = 0, than if there is a first place, it adds 50, and so forth until it reaches the end. It is included into a file, in this area: Code: [Select] <div class="tabbertab"> <h2>Records</h2> <? $query92 = "SELECT * FROM THISTABLE WHERE VARIABLE='$id' OR VARIABLE = '$name' ORDER BY ABS(VARIABLE), VARIABLE"; $result92 = mysql_query($query92) or die (mysql_error()); echo "<table class='record'> <tr><th>Show</th> <th>Class</th> <th>Place</th></tr> "; while($row92 = mysql_fetch_array($result92)) { $class = $row92['class']; $place = $row92['place']; $entries = $row92['entries']; $race = $row92['show']; $purse = number_format($row92['purse'],2); echo "<tr><td>$race</td> <td>$class</td> <td>$place</td></tr>"; } ?> <tr><td colspan='3'><div align='right'><? include('includes/points.php'); ?></div></td></tr> </table> </div> This is the code that is relevant. When ended here, it echoes the last place that appears in the results (such as a 5th place echoing 30 points). When I move it to be included in the while loop, it shows Total Points: 50 Total Points: 25 Total Points: 10 (depending on the results displayed on that page). What am I doing wrong? Hello everyone, I can get Test 2 to successfully operate the if statement using a variable variable. But when I try the same method using a session variable (Test 1) the if statement is not executed. Please could you tell me why the if statement in Test 1 is not being executed? Code: [Select] <?php # TEST 1 $_SESSION[test_variable] = "abcd"; $session_variable_name = "_SESSION[test_variable]"; if ($$session_variable_name == "abcd") { echo "<br>line 373, abcd<br>"; } # TEST 2 $test_variable = "efgh"; $test_variable_name = "test_variable"; if ($$test_variable_name == "efgh") { echo "<br>line 379, efgh<br>"; } ?> Many thanks, Stu Hi there Does anyone know where can i get php/mysql calculation weight & shipping cost tutorial ? Thank you in advanced! Hi guys, I have a problem with the code on below. When I input the value into $name and $email method, the page come into blank page. <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbusername'); define('DB_PASSWORD', 'mydbpassword'); define('DB_DATABASE', 'mydbname'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $name = clean($_GET['name']); $email = clean($_GET['email']); $comments = clean($_GET['comments']); $type = clean($_GET['type']); if($name == '') { $errmsg_arr[] = 'name or member ID missing'; $errflag = true; } if($email == '') { $errmsg_arr[] = 'email address ID missing'; $errflag = true; } else { } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $insert = array(); if(isset($_GET['name'])) { $insert[] = 'name = \'' . clean($_GET['name']) .'\''; // echo "tested"; } if(isset($_GET['email'])) { $insert[] = 'email = \'' . clean($_GET['email']) . '\''; } if(isset($_GET['type'])) { $insert[] = 'type = \'' . clean($_GET['type']) . '\''; } if(isset($_GET['comments'])) { $insert[] = 'comments = \'' . clean($_GET['comments']) . '\''; } if (count($insert)>0) { $names = implode(',',$insert); if(isset($name)) { $name = $_GET['name']; $headers = "From: "-f .$name."@myemail.com"; $to = "myname@mymail.com"; $subject = $type; $message = $comments . '; $header .= "MIME-Version: 1.0\l\n"; $add = "-f". $name ."@myemail.com"; mail($to, $subject, $message, $header, $add); echo "Thank you for sent us your feedback"; } else { if(isset($email)) { $email = $_GET['email']; $headers = "From: "$email"; $to = "myname@mymail.com"; $subject = $type; $message = $comments . '; $header .= "MIME-Version: 1.0\l\n"; $add = "-f". $email ."; mail($to, $subject, $message, $headers, $add); } echo "Thank you for sent us your email"; } } } ?> There must be the problem coming from this: if(isset($name)) { $name = $_GET['name']; $headers = "From: "-f .$name."@myemail.com"; $to = "myname@mymail.com"; $subject = $type; $message = $comments . '; $header .= "MIME-Version: 1.0\l\n"; $add = "-f". $name ."@myemail.com"; mail($to, $subject, $message, $header, $add); echo "Thank you for sent us your feedback"; } else { if(isset($email)) { $email = $_GET['email']; $headers = "From: "$email"; $to = "myname@mymail.com"; $subject = $type; $message = $comments . '; $header .= "MIME-Version: 1.0\l\n"; $add = "-f". $email ."; mail($to, $subject, $message, $headers, $add); } I am not sure where the problem is, so please could you help me?? Hi, I have a variable named $siteName stored in the server.php file in the data directory data/server.php Now in the header.php page I want to retrieve the $siteName variable so it can be used in the header information. This is what I have but it isnt working. server.php Code: [Select] <?php $siteName = 'My+Test+Site'; ?><?php $adminEmail = 'pwithers2009@hotmail.co.uk'; ?><?php $sendmailLoc = '/usr/sbin/sendmail'; ?><?php $imgdir = 'http://www.tropicsbay.co.uk/images/'; ?><?php $imgdirbase = '/home/tropicsb/public_html/images/'; ?> header.php Code: [Select] <?php include("data/server.php"); $siteName = $_GET['siteName']; echo ' <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>$siteName Admin Panel</title> <link rel="stylesheet" href="admin.css" type="text/css" /> </head> <div id="pagehead"> <div id="navigation"> <img alt="" src="images/leftNav.gif" height="32" width="4" id="leftNav" /> <img alt="" src="images/rightNav.gif" height="32" width="4" id="rightNav" /> <div id="globalLink"> <a href="admin.php?cmd=manage&username=admin&password=$adminpw" id="gl1" class="glink" onmouseover="ehandler(event,menuitem1);">Users</a><a href="admin.php?cmd=editTemplate&username=admin&password=$adminpw" id="gl2" class="glink" onmouseover="ehandler(event,menuitem2);">Templates</a><a href="admin.php?cmd=mysqlBackup&username=admin&password=$adminpw" id="gl3" class="glink" onmouseover="ehandler(event,menuitem3);">Database</a><a href="admin.php?cmd=paymentLog&username=admin&password=$adminpw" id="gl4" class="glink" onmouseover="ehandler(event,menuitem4);">Payment</a><a href="admin.php?cmd=profileFields&username=admin&password=$adminpw" id="gl5" class="glink" onmouseover="ehandler(event,menuitem5);">Setup</a><a href="admin.php?cmd=changeAdminpw&username=admin&password=$adminpw" id="gl6" class="glink" onmouseover="ehandler(event,menuitem6);">Preferences</a><a href="admin.php?cmd=logout&username=admin&password=$adminpw" id="gl7" class="glink" onmouseover="ehandler(event,menuitem6);">Logout</a></div> </div></div> <div id="pagelayout"> <img alt="" src="images/leftCurve.gif" height="6" width="6" id="left" /> <img alt="" src="images/rightCurve.gif" height="6" width="6" id="right" /> <div id="pageName"> <h2>$siteName Admin Panel<h2> </div>'; ?> Any ideas on how to do this, it seems to work ok in one of my other scripts. Thanks I've been baffled by this for 2 days now and cannot figure it out after exhaustive searches. I'd like to think I'm doing this correctly, but cannot get it to work.
I'm trying to use a variable within a query WHERE statement, and it shows 0 results. If I directly hardcode the text instead of using the variable, it works. The variable is pulling from a $_GET, and if I echo that variable, it is showing the correct text.
Here's my code:
$Domain = $_GET['Domain']; $result = mysql_query(SELECT Code, Title, Domain, Status FROM tablename WHERE Domain="$Domain" ORDER BY Code');If I swap out "$Domain" for direct text, like "ABC", it works. I have tried swapping out the quotes and single quotes throughout the statement, removing the quotes around $Domain, concatenating the statement separately....all yield erros or the same result. And as stated, if I echo $Domain, it shows "ABC" (or whatever it's supposed to show), so i know it's pulling correctly from the $_GET. Anyone know what I'm doing wrong? i've been programming in PHP for years, and have done a substantial amount of work on applications of this nature. this problem has me stumped, not because i can't fix it (i did), but because i have no idea what the problem is. there are hundreds of lines of code involved here, so i'll break it down into a post-friendly format. take this for example, and forgive any typos. it's late, and i've been beating my head against this for over two hours... =\ this is from my form: Code: [Select] /* ... numerous form fields being passed as $_REQUEST arrays */ <input type="hidden" name="option_id[]" value="<?php print $query_result->option_id; ?>" /> /* a couple hundred more lines */ here's the DB update handler: Code: [Select] if (!empty($_REQUEST['option_name'])) { foreach ($_REQUEST['option_name'] as $k => $v) { if ($v != '') { $option_id = $_REQUEST['option_id'][$k]; $option_name = $_REQUEST['option_name'][$k]; $option_price = $_REQUEST['option_price'][$k]; $option_desc = htmlentities($_REQUEST['option_desc'][$k], ENT_QUOTES); if (!$option_id = '') { $sql_options = "UPDATE table SET" . " option_name = '" . $option_name . "', option_price = '" . $option_price . "', option_desc = '" . $option_desc . "' WHERE option_id = '" . $option_id "'"; if (!$query_function($sql_options)) { $error = true; } } else { $sql_options = "INSERT INTO table (option_name, option_price, option_desc)" . " VALUES ('" . $option_name . "', '" . $option_price . "', '" . $option_desc . "')"; if (!$query_function($sql_options)) { $error = true; } } } } } the above code doesn't post to the database because the $option_id variable returns a null value. however, if i replace the $option_id variable where i build the query string with $_REQUEST['option_id'], it works just fine. Code: [Select] /* in relevant part */ $sql_options = "UPDATE table SET" . " option_name = '" . $option_name . "', option_price = '" . $option_price . "', option_desc = '" . $option_desc . "' WHERE option_id = '" . $_REQUEST['option_id'] . "'"; needless to say i was infuriated by having spent a couple of hours to come to this conclusion. i only used the variables in the first place because i need to expand the function that this lives inside and i don't want to have to type $_REQUESTs over and over. the only thing i can think is that it might be a type issue. the data is coming out of the mysql table from an INT field and being placed into the value for the hidden field straight from the row collection. would forcing a variant data type by not strongly typing my variable have caused this problem? i haven't tested the theory because i'm still too ticked off to open my code editor. i'm bouncing this off the community and posting my experience in the hope that it might help someone who comes after. so my situation is something like this , i'm trying to fetch user details based on `id` that isset is getting, but some how the `variable that contains the $_GET value doesn't work` in query but when i put an static value to pdo query then it works and show the result. i have checked by doing `var_dump` of variable `$user` before query and it shows the correct value but not working in query. Below is the code i'm working with: Hello Everyone, I am new to forum and could use some help with some php code that isn't working. I am very new to php/html/javascript and all of what I have learned, I learned from forums like this one so first....thank you! I am trying to assign a value from a php variable to the value of my form element. I'm sure there must be a way to do this but I can't seem to get the syntax right. here is my code... first I set the value of $loginname elsewhere in the script like so... <?php $loginname =strtolower(htmlspecialchars(strip_tags($_GET["loginname"]))); ?> This part works fine.. Then I try to set the value of my hidden text field inside the form to the value of $loginname to be passed to a javascript program. Everything works except that the value passed ends up being <?echo and not the expected user name inside of $loginname. <?php echo '<form name ="currentactivity" Id="currentactivity" action="<?php'.htmlspecialchars($_SERVER['PHP_SELF']).'?>" method="post">'; echo '<fieldset><legend><b>Your Current Activity Information</b></legend>'; echo '<input type="text" name="loginnm" style="visibility: hidden" value="<?php echo $loginname;?>">'; echo "<label for='myactivities'>Activity Name:</label>"; echo "<select name='myactivities' Id='myactivities' onchange=\"showdetails(this.form)\" value=''>"; echo "<option value = 'Select an activity'>Select an activity</option>"; for ($i = 1; $i <=$rowcount; $i++) { echo"<option value=$row[activity_name]>$row[activity_name] </option>"; $row = mysql_fetch_array($result); } echo "</select>"; echo '</fieldset>'; echo '</form>'; ?> Please note..the rest of the code is working perfectly, it is just this one value I can't seem to get. Any help you can give will be greatly appreciated. Hi Guys, I'm a PHP newbie and I'm having some trouble creating a function that I can call where I can perform a MYSQL SELECT querey which returns an associative array, and be able to pull a value from that array and return it to the call. My code is below. Everything seems to work up until where I try to set the $average variable. If I return $row instead of $round_val, I see me array as "Array ( [user_id] => 64 [AVG(twos_made)] => 5.0000 )" It seems as though my my functions arugment (twos_made - which is my $column_name var) is not getting passed through or something. Any help would be greatly appreciated! Thanks! Code: [Select] function avg_of($column_name) { global $connect_db; $query = "SELECT user_id, AVG($column_name) FROM table GROUP BY user_id"; $result = mysql_query($query, $connect_db); $row = mysql_fetch_assoc($result); $average = $row['AVG($column_name)']; $round_val = round($average, 1); return $round_val; } hi guys. i am trying to set a variable to the session global variable. I've initialized the session with session_start(); at the very beginning of my website, and then i try finding if a session variable has been set like so if(isset($_SESSION['user'])) { print 'user is logged in'; } but this works in reverse. i have not set any session variables, but asking if it is set, results in an affirmative answer. so i am thinking this is because of register_globals. because i went to read about register_globals, and it says that if register_globals is turned off, i cannot use any other varibale except members of the session array. i dont know what that is yet, but a question befo in the php manual it says register_globals is deprecated, and it appears i need to enable it to set other variables as session variables besides it's associative array. if this is true, how to enable register_globals and allow other variables to be used as session variables? This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=321230.0 This works: Code: [Select] form action="http://localhost/mysite/css/screen.css" method="post"> This does not work: Code: [Select] <?php $link_1 = "http://localhost/mysite/css/screen.css" ; $link_2 = '"' . "http://localhost/mysite/css/screen.css" .'"' ; ?> <form action=<?php echo $link_1;?> method="post"> <form action=<?php echo $link_2;?> method="post"> Why do neither of the PHP variables work? Thanks! Hello everyone. I am wondering how can i give different id's to a html form? Code: [Select] <?php ... if ($numrows!=0) { // fetching data from the database to variable while ($row = mysql_fetch_assoc($query)) { $view_jam = $row['jamname']; $view_recipe = $row['jamrecipe']; $view_rating = $row['jamrate']; $view_owner = $row['owner']; $view_comments = $row['comments']; $view_count = $row['count']; $view_id = $row['id']; $view_count2 = $row['count2']; echo "Jam name: <b>" . $view_jam . "</b><br />"; echo "Jam recipe: " . $view_recipe . "<br />"; echo "Owner: " . $view_owner . "<br />"; echo "Comments: " . $view_comments . "<br />"; echo "Jam Rating: " . $view_rating . "<br />"; [b]echo '<form action="main.php?id=jams" method="POST">[/b] Choose rating <input type="text" name="jamrate" /> <input type="submit" name="submit1" value="Rate"> </form><br /><br />'; ... ?>As you can see, the form is inside a loop. I am trying to build a rating system. When there is only one row in a database then everything goes smoothly, but if there are more, then the rating applies to all rows. I suppose i need to give different id's to the form. How? Thank you Hello Everyone, I am working on a project where I have to go in and grab the names of the fields in a table. I got it to work and for it to show me the field names but it also displays a lot more than what I would like for it to do. I keep getting the following error. It is repeated over my screen like 15 times. Quote Warning: mysql_field_name() [function.mysql-field-name]: Field 5 is invalid for MySQL result index 7 in C:\xampp\htdocs\CMS\admin\includes\get_table.php on line 44 This is the section of code that it is referring to. Line 44 is Quote $tablecolumn .= "<p>" . mysql_field_name($fieldnamesquery_result, $table_field) . "<input type='text' name='" . mysql_field_name($fieldnamesquery_result, $table_field) . "' id='" . mysql_field_name($fieldnamesquery_result, $table_field) . "'/>"; Code: [Select] $tablecolumn = "<form name='insert_table' id='insert_table' action='get_database.php' method='post' > <fieldset> <legend>table information</legend>"; for ($table_field = 2; $table_field < $fieldnamesquery_result; $table_field++) { $tablecolumn .= "<p>" . mysql_field_name($fieldnamesquery_result, $table_field) . "<input type='text' name='" . mysql_field_name($fieldnamesquery_result, $table_field) . "' id='" . mysql_field_name($fieldnamesquery_result, $table_field) . "'/>"; <-----This is line 44 } $tablecolumn .= " <p> <input type='submit' name='submit_info' id='submit_info' /> </p> </fieldset> </form> "; echo $tablecolumn; |
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