PHP - Pick An Item From Database Based On A Number
Hi,
I need to write some code for this but unsure where to start. Basically i have a database of say 10 rows. I need to be able to input a number and some sort of fair algorithm will choose one of the rows. The number will be from 6 to 10 numbers long, and will need to pick the same result each time based on that number. Its for a lotto game Similar TutorialsI have five names in an array and also numbers 5,2,1,3 and what i want to do is php to look at the numbers and print out the name that corresponds to the number. so for my example the name it should echo out is 'Lee'
$names = array ("Stan", "John", "Dean", "Sam", "Lee"); here are the five names. The idea is to learn php with making a little fixture list with these five names. The problem i have is to make php read the number and then look up the array for the name that corresponds to the number. I hope this explains what I would like to do?
Hey, I have an array which stores a bunch of numbers lets say: Code: [Select] [0] = 1[1] = 6 Now a second array has: Code: [Select] [0] = 5 [1] = 6 What im trying to do - is find the first highest number from the second array that is not found in the first array.Both arrays are sorted already by the way using sort(). What is the best way to do this? I Have the following issue and I spend my whole day looking for it. I have a database with a simple admin where I add/delete values. The database structure is the following: id, zip, email The database is called zipdatabase Here is the mailing part: Code: [Select] //headers $headers = "From: <$Email>"; $headers .= "\r\nBcc: <$Bcc>\r\n\r\n"; // send email $success = mail($EmailTo, $Subject, $Body, $headers);The variable $Bcc must come from the database. I am really ignorant so I only got to a point on how to echo the results I need (and this works): Code: [Select] $query = "SELECT email FROM zipdatabase WHERE zip = '$ZIP'"; $result = mysql_query($query); while($row = mysql_fetch_array($result)) { echo "{$r['email']}, "; } ?> For a zip I get more than one result so when I tried using this Code: [Select] $query = "SELECT email FROM zipdatabase WHERE zip = '$ZIP'"; $result = mysql_query($query); while($row = mysql_fetch_array($result)) { $Bcc = "{$r['email']}, "; } echo "Bcc" ?> It only gives me one result. What I noticed is I have 2 issues: - I can't extract the BCC from the database - I don't know how to add more email addresses to the bcc field, I only managed to add more addresses to the $Emailto like: Code: [Select] $EmailTo = "address1@domain.com, address2@domain.com"; $Bcc = "address3@domain.com"; All these variables are defined by a webform and it works flawlessly except the database problem: Code: [Select] $Name = Trim(stripslashes($_POST['Name'])); $Phone = Trim(stripslashes($_POST['Phone'])); $ZIP = Trim(stripslashes($_POST['ZIP'])); $Email = Trim(stripslashes($_POST['Email'])); $Message = Trim(stripslashes($_POST['Message'])); // validation $validationOK=true; if (!$validationOK) { print "<meta http-equiv=\"refresh\" content=\"0;URL=contacterror.htm\">"; exit; } // prepare email body text $Body = ""; $Body .= "Name: "; $Body .= $Name; $Body .= "\n"; $Body .= "Phone: "; $Body .= $Phone; $Body .= "\n"; $Body .= "ZIP: "; $Body .= $ZIP; $Body .= "\n"; $Body .= "Email: "; $Body .= $Email; $Body .= "\n"; $Body .= "Message: "; $Body .= $Message; $Body .= "\n"; Sorry for my ignorance and hope someone here can help. The webform works perfectly except the bcc from the database part. I have somewhat of a dilemma. When someone clicks on a Buy Now button and subsequently follows necessary steps to complete process of purchase, when that transaction is completed, in my product table, I want to subtract 1 from whatever value is in the field. E.g. say product one is being purchased, prior to purchase in product table, there are 5 product one's in stock, when purchase is complete, subtract 1 from 5 (to get 4). Here is my code Code: [Select] <?php $title = "Like This Product, Buy It NOW!!!"; require ('includes/config.inc.php'); include ('./includes/header.html'); require (MYSQL); include ('./includes/main.html'); if($id = isset($_GET['prodID'])) { $query = "SELECT `prodID`, `product`, `prod_descr`, `image`, `price` FROM product WHERE `prodID`='{$_GET['prodID']}'"; $r = mysqli_query($dbc, $query); $showHeader = true; echo "<div id='right'>"; while($row = mysqli_fetch_array($r)) { if($showHeader) { //Display category header echo "<h1>" . "<span>" . "# " . "</span>" . $row['product'] . "<span>" . " #" . "</span>" . "</h1>"; echo "<div id='item'>"; // div class 'item' echo "<div class='item_left'>"; echo "<p id='p_desc'>"; echo $row['prod_descr']; echo "</p>"; echo "<p>" . "<span>" . "£" . $row['price'] . "</span>" . "</p>"; echo "</div>"; echo "<div class='item_right'>"; echo "<img src='db/images/".$row['image']."' />"; $showHeader = false; echo "</div>"; ?> <p> <form target="paypal" action="https://www.sandbox.paypal.com/cgi-bin/webscr" method="post"> <input type="hidden" name="cmd" value="_s-xclick"> <input type="hidden" name="hosted_button_id" value="7UCL9YCYYXL3J"> <input type="hidden" name="item_name" value="<?php echo $row['product']; ?>"> <input type="hidden" name="item_number" value="<?php echo $row['prodID']; ?>"> <input type="hidden" name="amount" value="<?php echo $row['price']; ?>"> <input type="hidden" name="currency_code" value="GBP"> <input type="image" src="https://www.sandbox.paypal.com/en_US/i/btn/btn_cart_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!"> <img alt="" border="0" src="https://www.sandbox.paypal.com/en_US/i/scr/pixel.gif" width="1" height="1"> </form> </p> <p> <form name="_xclick" action="https://www.paypal.com/cgi-bin/webscr" method="post"> <input type="hidden" name="cmd" value="_xclick"> <input type="hidden" name="business" value="me@mybusiness.com"> <input type="hidden" name="currency_code" value="GBP"> <input type="hidden" name="item_name" value="<?php echo $row['product']; ?>"> <input type="hidden" name="amount" value="<?php echo $row['price']; ?>"> <input type="image" src="http://www.paypal.com/en_US/i/btn/btn_buynow_LG.gif" border="0" name="submit" alt="Make payments with PayPal - it's fast, free and secure!"> </form> </p> <?php echo "</div>"; // End of div class 'item' $strSQL = "SELECT prodID, product, price, image FROM product ORDER BY RAND() LIMIT 1"; $objQuery = mysqli_query($dbc, $strSQL) or die ("Error Query [".$strSQL."]"); while($objResult = mysqli_fetch_array($objQuery)) { echo "<div class='love'>"; echo "<h6>Like this......you'll love this!!!</h6>"; echo "<ul>"; echo "<li>" . "<img src='db/images/" . $objResult['image'] . "' width='50' height='50' />" . "</li>"; echo "<br />"; echo "<li>" . "<a href='item.php?prodID={$objResult['prodID']}' title='{$objResult['product']}'>" . $objResult['product'] . "</a>" . " - " . "£" . $objResult['price'] . "</li>"; echo "</ul>"; echo "</div>"; } } } ?> <?php echo "</div>"; } include ('./includes/footer.html'); ?> How is this achievable please? This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=354366.0 did some searching on the internet and didnt find much. i have created an online classifieds website and want all created listings to expire and delete after a certain amount of days. honestly i have no idea where to start with this implementation. cookies were the only thing that even crossed my mind... any ideas? btw each listing is stored in a table called ob_listings Hi, I have a web site that sells both video and images. The full versions are located securely behind a paywall, however in a bid to get visitors to purchase them and to promote the members offering I have trailer videos of 30 seconds to one minute and four free images per imageset. I want to be able to restrict the number of images a visitor can view for free to 12 and the number of videos a visitor can play to 5. With a little graphic/overlay notifying the user when they have exceeded their free play/preview limit. I have been advised that a tracking cookie and php if statement would be the way to go to achieve this. I have got a cookie that is successfully logging views that is written to on a free preview link as an onclick function. I have tried writing some php if code that will have the normal thumbnail or video divs display as usual if the views recorded in the cookie are below 12 or five, respectively, and have a not allowed, please subscribe div that would show over the thumbnails or video divs should the number of views exceed the threshold. My PHP code does not seem to be working at all, I have ran it through an online syntax checker - corrected the mistakes and now it's supposed to be working but opts to display nothing instead of either if argument. Videos then become a whole other problem because at the moment I have no other way to track the view/visit other than to count the loading of the video summary page with the free player as the preview, whether or not the clip is played - ideally I would like the view to be counted when the video begins to play. But I have no idea how to write to the cookie from Flash (is that even possible?) Any help would be greatly appreciated. Hi All, First time posting here. I've googled the problem, but can't seem to find a response that's the same. All I want to do is have a list of id numbers and for each id number in the array, submit a MySQL query to retrieve information relating to the id number. When I execute the code below however, I end up with only the last item in the array being printed in the echo statement. Any clues? Thanks, Code: [Select] // get array of ids $ids = getIDs($ids); // loop through input list foreach ($ids as &$id) { getVarDetails($id); } function getVarDetails($local) { $con = mysql_connect('localhost:3306', 'root', '********'); if (!$con) { die('Could not connect: ' . mysql_error()); } // set database as Ensembl mysql_select_db("Ensembl", $con); $result = mysql_query("SELECT * FROM variations WHERE name = '$local' LIMIT 1"); $row = mysql_fetch_array($result) while($row = mysql_fetch_array($result)) { echo $row['name'] . " " . $row['id']; echo "<br />"; } // close connection mysql_close($con); } This topic has been moved to PHP Regex. http://www.phpfreaks.com/forums/index.php?topic=317036.0 Hi Everyone. i have a page which list all records. i then have several detail pages which i need to link via the id number. for ex. records: id number name 1 john smith 2 peter parker the detail page is layout out like so: View details page 1 | view details page 2 | view details page 3 Any help would be appreciated. Thanks. I don't have a lot of money and I'm launching a high traffic website, I intend to have a payment system linked to hosting service I am using so that in the event that my website is actually successful, it can expand accordingly.
I want to know ahead of time about bandwith usage, I have a 3TB limit per month which may seem like a lot but what if thousands of users are accessing/using files that are between 2 to 10 MB each? Uploading, viewing, scrolling, text...
This calculation also applies to storage
I haven't really given this much thought, I'm just curious, I have a lot to do and I would like to start amassing information ahead of time
Thank you for any help
Hello, I had 8 select boxes(dropdowns) in a form. Its like a search kind of implementation in the website. I don't know what the user selects, he may choose different select boxes, any number of select boxes. So, based upon the user selection, I need to generate the data. The data fields to be generated would be almost same. So, How would I know what the user selection is and how many select boxes has been selected and how could I generate different data based upon selected boxes. I had to make a small note abt this, that the data generation fields may change for some of the user select combinations, but most of the result fields would be same. so, can you please help me out, how to do, how to make different combination data generations, because, i had 8 fields, i had to make 8! combinations, that would result in a big code. i have 8 division (div), i want to display 4 rows in 4 division and the remain 4 rows in the next 4 division here is my code structure for carousel
<div class="nyie-outer"> second row third row
fourth row fifth row sixth row seven throw eighth row
</div><!--/.second four rows here-->
sql code
CREATE TABLE product( php code
<?php how can i echo that result in those rows
I have a script that seems to work well to insert a bookmark into a users database when he/she is logged into the system but I am having a hard time figuring out how I would go about making a work-a-round for having an item selected before being logged in, and inserted after they have logged in or registered. For example, I would like a user to be able to select an Item to add to bookmark whether that user is logged in/registered or not and if they are not, they would be greeted with a login/registration form and after successful login the add bookmark script would be initiated on the item previously selected. What I've got this far: Simple form to add bookmark: <form name="bm_table" action="add_bms.php" method="post"> <input type="text" name="new_url" value="http://" /> <input type="submit" value="Add Bookmark"/> </form> Then I have the add bookmark script: BEGIN php $new_url = $_POST['new_url']; try { check_valid_user(); //cannot get past this part since it ends the script....code below if (!filled_out($_POST)) { throw new Exception('Form not completely filled out.'); } // check URL format if (strstr($new_url, 'http://') === false) { $new_url = 'http://'.$new_url; } // check URL is valid if (!(@fopen($new_url, 'r'))) { throw new Exception('Not a valid URL.'); } // try to add bm add_bm($new_url); echo 'Bookmark added.'; // get the bookmarks this user has saved if ($url_array = get_user_urls($_SESSION['valid_user'])) { display_user_urls($url_array); } } catch (Exception $e) { echo $e->getMessage(); } END php Checking valid user - the portion I cannot get past in the above script: function check_valid_user() { // see if somebody is logged in and notify them if not if (isset($_SESSION['valid_user'])) { echo "Logged in as ".$_SESSION['valid_user'].".<br />"; } else { // they are not logged in do_html_heading('Problem:'); echo 'You are not logged in.<br />'; do_html_url('login.php', 'Login'); do_html_footer(); exit; } } How would I go about modifying the script so that a user could fill in the form (later it would be a link...obviously they probably wouldn't be filling in a form that is log-in specific - but same concept I think) Thanks in advance for the help! tec4 How do I call the last ID number from a table in MySql and echo it? This is what I have so far: Code: [Select] $query = mysql_query("SELECT idnum FROM ready_aclassft"); Maybe im comeing at it all wrong. Like the title says. I have 4 of the same number in my database, but I only want my website to display one of the 4. How can I do something like this? Hi there, I think this is a big question but I'd appretiate any help you can provide!! I have a list of items and subitems in a table that looks like this: id parent_id title 1 0 House Chores 2 1 Take Out Trash 3 1 Clean Room 4 0 Grocery List 5 4 Eggs 6 4 Produce 7 6 Lettuce 8 6 Tomato 9 4 Milk I want to display it like this: (+) House Chores: > Take Out Trash > Clean Room (+) Grocery List: > Eggs (+) Produce > Letutce > Tomato > Milk So basically each entry in the table has an unique id and also a parent id if it's nested inside another item. I "sort of" got it figured out in one way, but it doesnt really allow for nested subgroups. I'd like to know how would y'all PHP freaks to this Also taking suggestions for the javascript code to expand/collapse the tree !! Thank you! How to text a phone number in database as well as record recieved textmessages into mysql database. Any help here? i've been googling for ages. I am designing a mmorpg guild site, and decided to make a coordinate database. A user will enter their own coords, x and y, and the 10 closest coords to them will be displayed. I need help doing this. My table is: id coordx coordy alliance region I am using $_GET to retrieve the user's input. so $_GET['mex'] $_GET['mey'] Thanks in advance for your help. |