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PHP - Student Subject Positioning Based On Score Using Php And Mysql Database
Hello sir, My name is Soji. I'm working on a php project on result transcript processing. I got to a stage where i need to rank student based on their score (Subject Position) and also Overall Position. I have tried all i could but i still don't get it I have a table called Subject position where with field (id, studentregNo, subjectid, levelid, armsid, yearid, total). what i want is if there is 3 student in a class and the first student score 50 in english and the second student score 45 and the third score 40. i want the system to tell me that the first student position in English is 1st and the second student position in English is 2nd and third student position in English is 3rd. So all this will be applicable to all subjects that the students in a particular level are offering.
Moreover, I need Rank student in a subject in that class, And this will work with each subject for each student in that class. Similar TutorialsWe recently upgraded from PHP4 to PHP5 and the below script that was working perfectly in 4 has completely stopped working and I can't figure out why for the life of me. I'm not an experienced PHP programmer--I've done some forms, but this is the first time I've used a database. What needs to (and was) happen: A user enters their username in the form and gets a readout of their participation so far that month. The problem(s): I know that it's storing the variable 'user' because it echoes it back properly, but the database is no longer allowing me to select the row based on that variable. I know it's not that I can't connect to the database because if instead of '$user' I change the code to a username I know is in there, I get the proper readout. This all started as soon as I transferred over to PHP5--before that, no problems at all. The database information is all correct, I just took it out for privacy's sake. <form id="feedback" method="post" action="index.php"> <input name="user" type="text" value="Enter user name" size="20" maxlength="50" /><br /> <input name="send" id="send" type="submit" value="Submit" /> </form> <?php if (isset($_POST['user'])) { $_session['user'] = $_POST['user']; } ?> <p>You entered your username as: <strong><? echo $_session['user'];?></strong>. If this is not correct or you do not see your information below, please re-enter your username and click Submit again.</p> <?php $con = mysql_connect("database","username","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("database", $con); $result = mysql_query("SELECT * FROM March WHERE Username='$user'") or die ('Error: '.mysql_error ()); while($row = mysql_fetch_array($result)) { echo "<table border='0'>"; echo "<tr>"; echo "<td><strong>Username:</strong> </td><td>" . $row['Username'] . "</td></tr>"; echo "<tr><td><strong>Mar. 2 Discussion:</strong></td><td>" . $row['Mar2Q'] . "</td></tr>"; echo "<tr><td><strong>Mar. 2 Poll:</strong></td><td>" . $row['Mar2P'] . "</td></tr>"; echo "<tr><td><strong>Mar. 9 Discussion:</strong></td><td>" . $row['Mar9Q'] . "</td></tr>"; echo "<tr><td><strong>Mar. 9 Poll:</strong></td><td>" . $row['Mar9P'] . "</td></tr>"; echo "<tr><td><strong>Mar. 16 Discussion:</strong></td><td>" . $row['Mar16Q'] . "</td></tr>"; echo "<tr><td><strong>Mar. 16 Poll:</strong></td><td>" . $row['Mar16P'] . "</td></tr>"; echo "<tr><td><strong>March Participation To-Date:</strong></td><td>" . $row['Participation'] . "</td></tr>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> ANY help would be greatly appreciated! I've got a couple hundred people who use this on a regular basis and are starting to ask why it's not working. Not sure why this isnt working. Code: [Select] <?php session_start(); ?> <!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> <script src="http://js.nicedit.com/nicEdit-latest.js" type="text/javascript"></script> <script type="text/javascript">bkLib.onDomLoaded(nicEditors.allTextAreas);</script> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <LINK REL=StyleSheet HREF="inc/dyn_style.css" TYPE="text/css" MEDIA=screen /> <?php ?> <?php include('logic.inc'); mysqlConnect(); ?> <script type="text/javascript" src="bbeditor/ed.js"></script> <link rel="stylesheet" type="text/css" href="dyn_style.css" /> <title> Social </title> <script type="text/javascript"> function changeTitle(title) { document.title = title; } </script> </head> <body> <?php $inc = 'new_story.php'; $view = 'Newest '; $by = 'added'; $where = " "; $where2 = " "; $order = "ASC"; $gen = "All"; $rat = 'All'; $blerg = ""; $sort = 'newest'; //---------------------------------------------------------------------- if(isset($_GET['sub'])) { $sort = $_GET['sort']; switch($sort) { case "Most Popular"; $by = 'views'; $view = 'Most Popular '; $order = 'DESC'; break; case "Most Reviewed"; $view= 'Most Reviewed '; $by = 'reviews'; $order = 'DESC'; break; case "Newest"; $by = 'added'; $view = 'Newest'; $order = 'ASC'; break; } $genre = mysql_real_escape_string($_GET['cat']); $rating = mysql_real_escape_string($_GET['rat']); if($gen == 'All') { $where = " "; $blerg = ""; } else { $where = "WHERE cat='$gen'"; } if ($rat == "All") { $where2 = ' '; $blerg = 'AND'; } else { $where2 = $blerg ." rating = '$rat' "; } } //---------------------------------------------------------------------- ?> <?php serch(); ?> <form action="story.php" method="get"> <label id='inline'> Order By: </label> <select name='sort'> <option selected='yes' label='Currently Selected' > <?php echo $view; ?> </option> <option> Newest </option> <option> Most Popular</option> <option> Most Reviewed </option> </select> <input type='hidden' value='spec_view' name='p' /> <label id='inline'> Genre/Catagory: </label> <select name='gen'> <option selected='yes' label = 'Selected Genre - <?php echo $gen; ?>'> <?php echo $gen; ?> </option> <option> All </option> <option> Fantasy </option> <option> Adventure </option> <option> Science Fiction</option> <option> Drama</option> <option> Fable </option> <option> Horror</option> <option> Humor</option> <option> Realistic Fiction </option> <option> Tall Tale</option> <option> Mystery </option> <option> Mythology </option> <option> Poetry </option> <option> Shorty Story </option> <option> Romance </option> </select> <label id='inline'> Rating: </label> <select name='rat'> <option selected='yes' label = "Selected Genre - <?php echo $rat; ?>"> <?php echo $gen; ?> </option> <option> All </option> <option> C </option> <option> C13 </option> <option> YA </option> <option> A </option> </select> <input type='submit' value='Go!' name = 'go' /> </form> <?php $query = " SELECT * FROM story_info ORDER BY $by $order $where $where2 "; echo $query; $select = mysql_query($query) or die(mysql_error()); while($rows = mysql_fetch_assoc($select)) { $viewsdb = $rows['views']; $titledb = $rows['title']; $userdb = $rows['user']; $catdb = $rows['cat']; $ratdb = $rows['rating']; $id_db = $rows['story_id']; $sumdb = shorten($rows['sum']); echo "<h3><a href='?p=page&id=$id_db'> $titledb </a> </h3>"; echo "<div id='fun_info'>"; echo "$sumdb <br />"; echo "By <a href='?p=profile&user=$userdb'> $userdb </a> <br /> "; echo "$viewsdb Views | Rated: $ratdb | Catagory: <a href='?p=cat_view&gen=$catdb'> $catdb </a> </div>"; } ?> </div> </body> </html> In my app, the users will take a multiplication test with 10 problems... I'm not sure how I would post the grade that the user got to the database. When the user finishes the test and hits score it should post their Name/Date & Time/ and Score to the database. I have the name and date & time working. Just not sure how I would approach the score. This is my application... It is small and simple to run in browser. Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Lab 5</title> </head> <body> <?php require_once('database.php'); define ('ROWS', 3); define ('COLS', 3); define ('MAX_NUMBER', 12); date_default_timezone_set('America/New_York'); if (isset($_POST['btn_score'])) { $result_name=$_POST['result_name']; $sql = "INSERT INTO results (result_name, result_score, result_date_time) VALUES ('$result_name', 100, NOW());"; $db->exec($sql); //print_r ($_POST); $time1 = $_POST['ts']; $time1_object = new DateTime($time1); $now = new DateTime(); $time_span = $now->diff($time1_object); $minutes = $time_span->format('%i'); $seconds = $time_span->format('%s'); $seconds+= $minutes * 60; echo "It took $seconds seconds to complete the test<hr />"; foreach ($_POST as $problem => $answer) { if ($problem <> "btn_score" && $problem <> "ts") { //echo "$problem -- $answer <br />"; $problem = explode('_', $problem); $num1 = $problem[2]; $num2 = $problem[3]; $right = $num1 * $num2; if ($answer != $right) { echo "$num1 * $num2 = $answer , The right answer is $right<br />"; } } } } ?> <h1>Lab 5</h1> <form name="lab5" method="post" action="lab5b.php"> <?php $now = new DateTime(); //echo $now->format('Y-m-d H:i:s'); echo "<input type='hidden' name='ts' value='" . $now->format('Y-m-d H:i:s') . "'>"; ?> <table border="1" cellspacing="5" cellpadding="5"> <?php $no_of_problems = 0; for ($row=0; $row<ROWS; $row++) { echo "<tr>"; for ($col=0; $col<COLS; $col++) { $num1 = mt_rand(1,MAX_NUMBER); $num2 = mt_rand(1,MAX_NUMBER); echo "<td>$num1 * $num2 </td>"; echo "<td><input type='text' size='2' name=${no_of_problems}_mult_${num1}_${num2}></td>"; $no_of_problems++; } echo "</tr>"; } $colspan = 2 * COLS; echo "<tr><td colspan=$colspan align='right'><input type='submit' value='Score' name='btn_score'></td></tr>"; ?> </table> <br> <br> <label for="result_name">Student Name:</label> <input type="text" id="result_name" name="result_name" /><br /> </form> <br> <br> </body> </html> Hi All I am kinda new to php.... I disparately need help with php coding. I am entering in the website using student id. If student id does not exist in mysql database, it gives me error. That works fine. But If I try to echo StudentID on 2nd page, it is not displaying anything. Second problem is I want to display student first and last name using StudentID. But it is not displaying anything using StudentID. Why? I have been trying to solve it, but no success Following is the code for both problems - if(!$db_selected) { die("Can not use".DB_NAME.':'.mysql_err()); } @$Stud_ID = $_POST['Stud_ID']; ?> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <link rel="stylesheet" type="text/css" href="ACSDP.css" /> <title>Student Main Page</title> <head><h1>Welcome to Undergraduate Student Main Page</h1> </head> <body> <form name ="mainpage" method='POST'> <table> <tr> <td><label id="Stud_ID2" for="Stud_ID"> Student ID <?php echo "is:" ." ". @$Stud_ID; ?> </label></td> <td> <label id="degree">Degree: Computer Science</label></td> </tr> <tr> <td> <label id="name"> Student Name: <?php $query2 =mysql_query("SELECT Stud_Lastname, Stud_Firstname FROM Student WHERE Stud_ID='$Stud_ID'") or die('wrong query'.mysql_error()); while($row = mysql_fetch_array($query2)) { echo $row['Stud_Lastname'] . " " . $row['Stud_Firstname']; } If have made those statements bold. please help me...... At the moment I am creating a search function for my website. The approach I have in mind is a pseudo-PHP database. To give an example: A HTML form will submit the results to a PHP file. HTML FORM - Colour: Black PHP RESULT PAGE - if ($_POST['color'] == 'Black') {readfile("./products/black/*.html");} HTML FORM - Price: <$50 PHP RESULT PAGE - if ($_POST['Price'] == '<$50') {readfile("./products/less50/*.html");} The problem here is if there is an item that is black and costs less than $50, then its going to be listed twice. There is probably some code I can write to ommit the listing of duplicate entries, but it is probably going to be messy, so I am wondering if its better to use a centralized MySQL database, rather than a pseudo-PHP database? I've never used MySQL and don't know much about it and this is my first real attempt at using PHP. Hi, I have a project where i want to send the sms to the customers at the particular time calling from the database, how to do these in php? Thanks, 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 I have two tables. Table Name:Users Fields: User_name user_email user_level pwd 2.Reference Fields: refid username origin destination user_name in the users table and the username field in reference fields are common fields. There is user order form.whenever an user places an order, refid field in reference table will be updated.So the user will be provided with an refid Steps: 1.User needs to log in with a valid user id and pwd 2.Once logged in, there will be search, where the user will input the refid which has been provided to him during the time of order placement. 3.Now User is able to view all the details for any refid 3.Up to this we have completed. Query: Now we need to retrieve the details based on the user logged in. For eg: user 'USER A' has been provided with the referenceid '1234' during the time of order placement user 'USER B' has been provided with the referenceid '2468' during the time of order placement When the userA login and enter the refid as '2468' he should not get any details.He should get details only for the reference ids which is assigned to him. Alright, I've been assigned a project at work. I did not develop the application and the individual who did used CodeIgnited framework and mysql as the db.
Here's the problem, I'm not given much OT to do this and in our meeting the best way to proceed was to replicate the database for different parts of the organization. Basically we are a subsidiary and have been using an application that other groups within the organization want to use. Usually I would reconfigure the db schema and add org ids and in the user table add the appropriate organization to go to. However, they are not giving me enough time to do that.
So what I'm thinking is to just create a copy of the database we use (just the structure) and create a new database.
What I want to know is how to use mysql to check to see if a user exists in one database and if they don't then to go on to the next database. I understand this is a very sloppy way to do it, but it's the way we are moving forward.
I found the code to connect to the db in CodeIgnitor... how can I connect to a database, check to see if the user exists, then close that db connection and try the next database?
/**
* Select the database
*
* @access private called by the base class
* @return resource
*/
function db_select()
{
return @mysql_select_db($this->database, $this->conn_id);
}
Thanks in advance.
I have two tables. Table Name:Users Fields: User_name user_email user_level pwd 2.Reference Fields: refid username origin destination user_name in the users table and the username field in reference fields are common fields. There is user order form.whenever an user places an order, refid field in reference table will be updated.So the user will be provided with an refid Steps: 1.User needs to log in with a valid user id and pwd 2.Once logged in, there will be search, where the user will input the refid which has been provided to him during the time of order placement. 3.Now User is able to view all the details for any refid 3.Up to this we have completed. Query: Now we need to retrieve the details based on the user logged in. For eg: user 'USER A' has been provided with the referenceid '1234' during the time of order placement user 'USER B' has been provided with the referenceid '2468' during the time of order placement When the userA login and enter the refid as '2468' he should not get any details.He should get details only for the reference ids which is assigned to him. Hi php freaks i am new to php im doin a small project to handle tickets created by customers now i want to change the font color based on the status for example: ID + Name + Product + Status ------------------------------------------------------------------------------------------------------------------------------------ 1111 + abcdef + bat + processing ----------------+--------------------------------------+------------------------------------------+------------------------------ i can connect to database and fetch it the only place i need help is changing the any help would be very helpful thanks in advance hey guys, I am having a small problem and I can't wrap my head around it. I want users to be able to upload up 10 photos in the database and what I want to do is check the database to see how many pics they have already and if they have 10, just show the photo with action items but if they have 7 for example, show the 7 pictures and three upload fields, if they have 4, show the pics and 6 fields and so on... let me know what you guys think and some help. Thanks in advance guys, you always come through... Hello,
I want to show checkbox is checked when there is entry of that id in a table in my database.
I have 2 tables page and access_level. I am getting data from page table and displays it in <ul><li> tag with checkbox to select all or only few. After selecting the checkbox, i will store only selected checkbox value in access_level table along with table id. Page link and page name details will be stored in page table.
Now if i want to edit , i should display all the pages which is there in page table and i should also mark checked to those which are already stored in access_level table.
I am using LEFT OUT JOIN, It displays all the pages. But it is not displaying the check mark to those which are already selected.
Here is my code
<?php $s1 = mysql_query("SELECT pages.page_id, pages.code, pages.page, pages.href, access_level.aid, access_level.page_id as pgid, access_level.department, access_level.position, access_level.active FROM pages LEFT OUTER JOIN access_level ON pages.page_id=access_level.page_id WHERE access_level.department=".$department." AND access_level.position=".$position." AND pages.code='sn'") or die(mysql_error());
while($s2 = mysql_fetch_array($s1)) { ?>
<tr><td><li><?php echo $s2['page']; ?> </td><td><input type="checkbox" name="sn[]" value="<?php echo $s2['page_id']; ?>"
<?php if($row['pgid'] === $s2['page_id']) echo 'checked="checked"';?> />
here is my pages table
pages.JPG 26.55KB
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access_level
access_level.JPG 19.09KB
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In access_level table i do not have page ids 8 and 9. But i want to display that also from pages table and for 1 to 7 and 10 i should display check mark.
How i can achieve this? Please Help
Maybe some of the great coders here can help this noob out. So here is what I have so far: Code: [Select] //set age criteria for deletion $age = 60; //get current date $datenow = date("Y-m-d"); //set the range we want to delete $delete_range = $datenow - $age; //get old user_id from users table $oldusers_users = mysql_query ("SELECT user_id FROM users WHERE lastvisit < $delete_range "); //get user_id from images table that correspond to users table $oldusers_images = mysql_query ("SELECT user_id FROM images WHERE $oldusers_users=user_id.images "); //find folders that correspond to the usernames and delete $oldusers_files = mysql_query ("SELECT username FROM users WHERE lastvisit < $delete_range "); //print out username //$result = mysql_($oldusers_files); $foldername = mysql_result($oldusers_files, 0); $sigspath = "sigs/"; unlink($sigspath . $foldername . ".gif/index.php"); rmdir($sigspath . $foldername . ".gif"); //now delete user_id's from database mysql_query("DELETE * FROM users WHERE user_id=$oldusers_users"); mysql_query("DELETE * FROM images WHERE user_id=$oldusers_images"); unset($oldusers_users, $oldusers_images, $oldusers_files, $foldername, $sigspath ); mysql_close($link); Right now it is only returning the first entry, not the entire list that meets the criteria. It does delete that one file though but does not remover the rows from the database. I am sure the database stuff is jacked up I am really new to that part. What am I doing wrong here or is there a better way to do this perhaps Hi im new to php and I need help making webpage that queries a mysql database based on a 3 check boxes and displays results on the same page or on another page. The table being queried has 4 columns, name, gps, wifi, bluetooth. So for example a row in the table would be like, samsung galaxy s2, yes, yes, yes. The idea is for it to be a website that will display phones according to their features. So the idea is depending on if the boxes were ticked the samsung galaxy would be displayed as a result. So i need some help understanding how to make this. Some1 gave me the code below in attempt to help me (im not sure it works or not) but im not sure how fully use it, ie what pages i need to make and how i create the connection to the mysql database, and how to use the query that they wrote to display the results thanks code: Code: [Select] <form action="?do=filter" method="post"> <table cellspacing="0" cellpadding="3" border="1"> <tr> <td>GPS<input type="checkbox" name="gps" value="checked"></td> <td>Wifi<input type="checkbox" name="wifi" value="checked"></td> <td>Bluetooth<input type="checkbox" name="bluetooth" value="checked"></td> </tr> <tr><td><input type='submit' name='filter' value='Filter'></td></tr> </table> </form> </html> <?php function filterMe($filter){ if(isset($_POST[$filter])){ return "Yes"; }else{ return "No"; } } if(isset($_POST['filter'])){ echo "Gps - " . filterMe('gps'); echo " Wifi - " . filterMe('wifi'); echo " Bluetooth - " . filterMe('bluetooth'); } ?> All you need to do is use a query something like SELECT name,gps,wifi,bluetooth FROM `product` WHERE `gps`='".filterMe('gps')."' AND `wifi`='".filterMe('wifi')."' AND `bluetooth`='".filterMe('bluetooth')."' This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=349871.0 Hi! Can anyone please help me form script for doing a MySQL query based on variables in a bbcode? BBCode could be like this: [datatype::dataid] Explode datatype and dataid intro separate strings and using them in querys. Example of string containing bbcodes: Code: [Select] $body = "Lorem ipsum dolor sit amet, consectetur adipiscing elit. Morbi fringilla, enim eget dictum sodales, enim velit iaculis dolor, sed congue orci diam id ipsum. Integer ullamcorper elit ac arcu condimentum id elementum arcu tempus. Ut aliquam commodo eros, mollis tristique magna adipiscing eu. [video::3456] Duis congue turpis quis eros porta eu sollicitudin mi porttitor. Ut ac nisi id risus imperdiet aliquet. Suspendisse lacinia sem id ipsum blandit egestas. Mauris id vulputate ligula. In in quam urna. Nam purus augue, laoreet vel sodales nec, rutrum non augue. [video::6789] Phasellus facilisis commodo arcu, vel dapibus tortor pretium eu. Integer elit nisi, condimentum ut vehicula eget, ullamcorper non quam. Nullam suscipit tempor consequat. Vestibulum sollicitudin ante tristique neque vestibulum a fermentum nulla venenatis. In eget nulla nulla."; I found something close he http://stackoverflow.com/questions/2801228/php-bbcode-with-sql-selection I used search function, could not find a subject that addressed my need. I hope I am not repeating a well answered question.
I have two MySql fields I want to hyperlink, but not as text links, but with an image.
Here is truncated code, if more deatail is need, I can post. Both links work fine now, but show the person's email name and google map link text - looks unproffessional and I think the email links may be able to be harvested by spammers.
The E-mail link:
echo "<td><a href=mailto:".$row['email'] . ">" .$row['email'] . "</a></td>";
The Google map link:
echo "<td><a target='_blank' href=".$row['location_map_link'] . ">" .$row['location_map_link'] . "</a></td>";
Can we use images in this code?
I am not a PHP coder!!
Ron Hi Guys. Hopefully someone can help me with this...New to coding and pretty lost on this. I have a Mysql database which is displaying results to my webpage with no problems. However I would like to be able to add a combo box to my webpage that would update the mysql database results based on the combo box selection. For example if Ford is chosen from the combo box, the webpage would refresh and show all the results for Ford in the webpage. Can someone please help me? Here is the code I have at the moment that works just fine. But results of the database are based on the WHERE statement. Quote <?php $con = mysql_connect("server","database","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("a5525005_cars", $con); $result = mysql_query("SELECT * FROM `cars` WHERE Makel='Ford'"); echo "<table class='ex1' border='0' width='113%' style=text-align:center; cellpadding='6' cellspacing='0'> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr style=font-family:verdana;font-size:80%;>"; echo "<td width=13%>" . $row[""] . "<img src=\"" . $row["Photo"] . "\"></a>"; echo '<td width="14%"><a class="mylink" href="' . $row['URL'] . '">' . $row['Model'] . '</a></td>'; echo '<td width="5%"><a class="mylink" href="' . $row['URL'] . '">' . $row['Year'] . '</a></td>'; echo '<td width="4%"><a class="mylink" href="' . $row['URL'] . '">' . $row['Fuel'] . '</a></td>'; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> |
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