PHP - [php][mysql]display Dynamic Data From Database
im making a game and i need to show a users money but i dont know how help?
Similar TutorialsI have got connection to the the mysql database, how do I get the data from the database to display on the webpage 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
please help me the concept or code to retrive data from the database and display data according to the data from the dynamic list box option. Hello to all, I have problem figuring out how to properly display data fetched from MySQL database in a HTML table. In the below example I am using two while loops, where the second one is nested inside first one, that check two different expressions fetching data from tables found in a MySQL database. The second expression compares the two tables IDs and after their match it displays the email of the account holder in each column in the HTML table. The main problem is that the 'email' row is displayed properly while its while expression is not nested and alone(meaning the other data is omitted or commented out), but either nested or neighbored to the first while loop, it is displayed horizontally and the other data ('validity', 'valid_from', 'valid_to') is not displayed.'
Can someone help me on this, I guess the problem lies in the while loop? <thead> <tr> <th data-column-id="id" data-type="numeric">ID</th> <th data-column-id="email">Subscriber's Email</th> <th data-column-id="validity">Validity</th> <th data-column-id="valid_from">Valid From</th> <th data-column-id="valid_to">Valid To</th> </tr> </thead> Here is part of the PHP code:
<?php while($row = $stmt->fetch(PDO::FETCH_ASSOC)) { echo ' <tr> <td>'.$row["id"].'</td> '; while ($row1 = $stmt1->fetch(PDO::FETCH_ASSOC)) { echo ' <td>'.$row1["email"].'</td> '; } if($row["validity"] == 1) { echo '<td>'.$row["validity"].' month</td>'; }else{ echo '<td>'.$row["validity"].' months</td>'; } echo ' <td>'.$row["valid_from"].'</td> <td>'.$row["valid_to"].'</td> </tr>'; } ?>
Thank you. I'm using some code to create a select menu of a fieldname of data I have in a mysql database : Code: [Select] <?php mysql_connect('localhost' , 'dbname', 'password'); mysql_select_db('dbname'); $result=mysql_query("SELECT * FROM Persons"); if(mysql_num_rows($result)>0) { ?> <select name="Persons"> <?php while($rows=mysql_fetch_array($result)){ ?> <option value="<?php echo $rows['id']; ?>"> <?php echo $rows['FirstName']; ?></option> <?php } ?> </select> What I would like to do is to elevate this into a jump menu form so that if the user selects an item from my form they are taken to a results page showing the full row of data from the database. Basically a search form containing items from the database they can choose to see more details on. eg. In my example you select a persons name and then you are taken to a results page which displays the details of that person from the database. Problem is I don't know how to do this and have been trawling around for a couple of days to find a solution (sorry I'm new to php). I would appreciate some help or a working example would be great of : 1/ A working dynamic jump menu 2/ The page that would process the form 3/ The results page displaying the data I have selected. Thank you for your time... hi guys, im new to this forum I'm new also to php, I need help from you guys: I want to display personal information from a certain person (the data is on the mysql database) using his name as a link: example: (index.php) names 1. Bill Gates 2. Mr. nice Guy i want to click Bill Gates (output.php) Name: Bill Gates Country:xxxx Age: xx etc. How can i make this or how to learn this? I can not get the values from the javascript add row to go dynamically as a row into MySql only the form values show up as the form below as one row. I made it as an array, but no such luck, I have tried this code around a multitude of ways. I don't know what I am doing wrong, kindly write out the correct way.
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR...ransitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Dynamic Fields js/php to MySql need to submit dynamically to the database</title> <?php require ('database.php'); ?> <script type="text/javascript"> var counter = 1; var collector = ""; function addfields(indx) { var tbl = document.getElementById('table_id'); var newtr = document.createElement('tr'); counter = counter + indx; newtr.setAttribute('id','tr'+counter); newtr.innerHTML = '<td><input type="checkbox" name="checkb'+counter+'" id="checkb'+counter+'" value="'+counter+'" onclick="checkme('+counter+')"></td><td><input type="text" name="text1[]"></td><td><textarea name="textarea1[]"></textarea></td>'; tbl.appendChild(newtr); } function checkme(dx) { collector += dx+","; } function deletetherow(indx) { var col = collector.split(","); for (var i = 0; i < col.length; i++) { var remvelem = document.getElementById('tr'+col[i]); var chckbx = document.getElementById("checkb"+col[i]); if(remvelem && chckbx.checked) { var tbl = document.getElementById('table_id'); tbl.removeChild(remvelem); } } } </script> </head> <body> <form enctype="multipart/form-data" id="1" style="background-color:#ffffff;" action="<?php echo $_SERVER['PHP_SELF']; ?>"></form> <table id="table_id" > <tr id="tr1" class="trmain"> <td> </td> <td> <input type="text" name="text1[]"> </td> <td> <textarea name="textarea1[]"></textarea> </td> </tr> </table> <input type="button" value="Add" onClick="addfields(1);" /> <input type="button" value="Delete" onClick="deletetherow()" /> <input type="submit" value="Send" id="submit" name="submit"/> <?php if(isset($_POST['submit'])) { for ($i=0; $i < count($_POST['text1']); $i++ ) { $ced = stripslashes($_POST['text1'][$i]); $erg = stripslashes($_POST['textarea1'][$i]); } $bnt = mysql_query("INSERT INTO tablename (first, second) VALUES ('$ced', '$erg')")or die('Error: '. mysql_error() ); $result = mysql_query($bnt); } ?> </body> </html> Hi I am trying to display data from the table "event" in my database, I use the code below but it will not work and I cannot figure out why. CAn anyone help? CODE: <?php $host="localhost"; // Host name $username="root"; // Mysql username $password=""; // Mysql password $db_name="test"; // Database name $tbl_name="event"; // Table name mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $result = my_sql_query("SELECT * FROM event WHERE eventid = '1'"); while($row = mysql_fetch_array($result)) { $eventname= $row["eventname"]; $eventdate= $row["eventdate"]; echo "<b><u>Event Name:</b></u> $eventname" echo "<b><u>Event Date:</b></u> $eventdate<br>"; } ?> DISPLAY: Event Name: $eventname echo "Event Date: $eventdate "; } ?> Hi, I'm trying to make a dynamic html table to contain the mysql data that is generated via php. I'm trying to display a user's friends in a table of two columns and however many rows, but can't seem to figure out what is needed to make this work. Here's my code as it stands: Code: [Select] <?php //Begin mysql query $sql = "SELECT * FROM friends WHERE username = '{$_GET['username']}' AND status = 'Active' ORDER BY friends_with ASC"; $result = mysql_query($sql); $count = mysql_num_rows($result); $sql_2 = "SELECT * FROM friends WHERE friends_with = '{$_GET['username']}' AND status = 'Active' ORDER BY username ASC"; $result_2 = mysql_query($sql_2); $count_2 = mysql_num_rows($result_2); while ($row = mysql_fetch_array($result)) { echo $row["friendswith"] . "<br>"; } while ($row_2 = mysql_fetch_array($result_2)) { echo $row_2["username"] . "<br>"; } ?> The above simply outputs all records of a user's friends (their usernames) in alphabetical order. The question of how I'd generate a new row each time a certain amount of columns have been met, however, is beyond me. Anyone know of any helpful resources that may solve my problem? Thanks in advance =) how i want to display data from database to look like this : <table width="633" height="224" border="1"> <tr bgcolor="#999900"> <td width="45">Bil</td> <td width="121">Course_name</td> <td width="83">session</td> <td width="83">start_date</td> <td width="83">end_date</td> <td width="83">notes</td> <td width="89">pre-req</td> </tr> <tr bgcolor="#6A7AEA"> <td rowspan="2">1.</td> <td rowspan="2" bgcolor="#6A7AEA">Math</td> <td>1st session </td> <td>1 jan 11 </td> <td>6 jan 11 </td> <td rowspan="2"> </td> <td rowspan="2"><image icon that will link to the oter site> </td> </tr> <tr> <td bgcolor="#6A7AEA">2nd session </td> <td bgcolor="#6A7AEA">8 jan 11 </td> <td bgcolor="#6A7AEA">15 jan 11 </td> </tr> <tr> <td bgcolor="#0066CC">2.</td> <td bgcolor="#0066CC">English</td> <td bgcolor="#0066CC">1st session </td> <td bgcolor="#0066CC">1 feb 11 </td> <td bgcolor="#0066CC">6 feb 11 </td> <td bgcolor="#0066CC"> </td> <td bgcolor="#0066CC"><image icon that will link to the oter site></td> </tr> <tr> <td rowspan="2" bgcolor="#6A7AEA">3.</td> <td rowspan="2" bgcolor="#6A7AEA">Science</td> <td height="29" bgcolor="#6A7AEA">1st session </td> <td bgcolor="#6A7AEA">8 march 11 </td> <td bgcolor="#6A7AEA">15 march 11 </td> <td rowspan="2" bgcolor="#6A7AEA"> </td> <td rowspan="2" bgcolor="#6A7AEA"><image icon that will link to the oter site></td> </tr> <tr> <td bgcolor="#6A7AEA">2nd session</td> <td bgcolor="#6A7AEA">16 march 11 </td> <td bgcolor="#6A7AEA">21 march 11 </td> </tr> </table> ** all the view data is called from database including the icon image thanks... hey guys so im trying to display data into text boxes that are fetched from database according to checkbox with value id. processing is located before <!DOCTYPE html>: if(isset($_POST['edit_event']) && isset($_POST['check'])) { require "connection.php"; foreach ($_POST['check'] as $edit_id) { $edit_id = intval($GET['event_id']); //i tried (int)$edit_id; $sqls = "SELECT event_name,start_date,start_time,end_date,end_time,event_venue FROM event WHERE event_id IN $edit_id "; $sqlsr = mysqli_query($con, $sqls); $z = mysqli_fetch_array($sqlsr); { }button and form opens: <form method="post" action="event.php"> <input type="submit" name="edit_event" value="Edit Event">this is the html where the data will be echoed: <div id="doverlay" class="doverlay"></div> <div id="ddialog" class="ddialog"> <table class="cevent"> <thead><tr><th>Update Event</th></tr></thead> <tbody> <tr> <td> <input type="text" name="en_" value="<?php echo $z['event_name']; ?>"> </td> </tr> <tr> <td> <input type="text" name="dates_" value="<?php echo $z['start_date']; ?>"> <input type="text" name="times_" value="<?php echo $z['start_time']; ?>"> </td> </tr> <tr> <td><input type="text" name="datee_" value="<?php echo $z['end_date']; ?>"> <input type="text" name="time_" value="<?php echo $z['end_time']; ?>"> </td> </tr> <tr> <td><input type="text" name="ev_" value="<?php echo $z['event_venue']; ?>"> </td> </tr> <tr> <td><input type="submit" name="update" value="Update Event" id="update"> <input type="submit" id="cancelupdate" name="cancel" value="Cancel" > </td> </tr> </tbody> </table> </div>this is the part which is populated by data from database where isset($_POST['check']) gets the 'check' from: echo "<tr> <td><input type='checkbox' name='check[]' value='$id'>$name </td> </tr>";</form> thanks in advance! Edited by noobdood, 19 May 2014 - 10:42 PM. This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=321339.0 Hi guys, Should be a simple 1. If i have the following at the top of the page: $page_views = $row['page_views'] + 1; mysql_query("UPDATE table SET page_views='$page_views'"); and then the following at the bottom of the page: echo $row['page_views']; Should I see the page views as 1 the first time the page is visited, 2 the second time the page is visited, and so on......? At the moment im seeing 0 on the first page visit, 1 on the second page visit, 2 on the third..... I had this problem before on another page i was working, and i simply solved it by displaying the mysql query above the echo similar to the code above. However now it does not seem to be working. Am i missing something really simple? lol Thanks I want to make a "hint" feature for an app ive been working on. Its a guessing game, and I want to be able to give out hints. Ok so say there is an answer in the database. Say the value is "firetruck". I want to retrieve that value, and obscure it, displaying only one or two letters that are in the word. Or, for a number, say 178, I want to display only one digit from it. Is this possible? Im sure it is, anything seems possible with PHP & Mysql. If so, how could I implement this? Trust me, I read the manuals. The manuals for both PHP and Mysql are so vast and a little advanced for my level.. then again im not complete novice and know the basics plus more of both. I'm not sure what I'm doing wrong here... This is my index <html <head> <title>Admin applications</title> </head> <body text="#000000"> <center> <?php include("connect.php"); echo "<table cellpadding='3' cellspacing='2' summary='' border='3'>"; echo "<tr><td>Real name:<br><br>"; echo $row['real_name']; echo "</td><td>Age:<br><br>"; echo $row['age']; echo "</td><td>In-Game Name:<br><br>"; echo $row['game_name']; echo "</td><td>Steam ID:<br><br>"; echo $row['steamid']; echo "</td><td>Agreement:<br><br>"; echo $row['agreement']; echo "</td><td>Will use vent:<br><br>"; echo $row['vent']; echo "</td><td>Activity:<br><br>"; echo $row['activity']; echo "</td><td>Why this person wants to be an admin:<br><br>"; echo $row['why']; echo "</td></tr></table>"; ?> </center> </body> </html> and this is my database connect <?php $database="admin"; mysql_connect ("localhost", "root", "waygan914"); @mysql_select_db($database) or die( "Unable to select database"); mysql_query("SELECT * FROM applications"); ?> The database table "applications" has 8 fields, and 2 records, but when I view the page i get the table but no data: I am managing a shop website which is using php and mysql for data. The website has a section that will display the related products with the one that you are watching. Now my problem is that if there are more than 5 items it will continue to display all the items in one row with the consequent that the page will not display well. I need to find a way to begin a new row after the 5th item so it will display 5 items in each row. this is my current code that is responsible for showing the related items. There is also a picture attached with the problem. while ($row = mysql_fetch_array($retd)) { $code = $row["code"]; $name = $row["name"]; echo("<td width=150 align=center>"); echo ("<a href=../products/info.php?scode=$code><img src=pictures/$code.gif border=0 alt=Item $name</a>"); echo ("<br><a href=../products/info.php?scode=$code><span class=fs13>$name</span></a>"); echo("</td>"); } Does anyone know how can I do this? This is actually a page to edit the drama details. Database- 3 tables 1. drama dramaID drama_title 1 friends 2. drama_genre drama_genreID dramaID genreID 1 1 2 2 1 1 3 1 3 3. genre genreID genre 1 comedy 2 romance 3 family 4 suspense 5 war 6 horror <?php $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = ''; $dbname = 'drama'; $link = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql'); $id = $_GET['dramaID']; $link2 = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql'); $sql2 = "SELECT * from drama , drama_genre , genre WHERE drama.dramaID='".$id."' AND drama_genre.dramaID='".$id."' AND drama.dramaID = drama_genre.dramaID AND drama_genre.genreID = genre.genreID"; $status2 = mysqli_query($link2,$sql2) or die (mysqli_error($link2)); $link3 = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql'); $sql3 = "SELECT * from genre "; $status3 = mysqli_query($link3,$sql3) or die (mysqli_error($link3)); <form method='Post' action='doUpdate.php' enctype="multipart/form-data"> <?php while ($row2 = mysqli_fetch_assoc($status2)) { ?> <td height="1"></td> <td><select name="gdrop"> <option value="<?php echo $row2['genreID'];?>"><?php echo $row2['genre'];?></option> <?php while ($row3 = mysqli_fetch_assoc($status3)) { ?> <option value="<?php echo $row3['genreID'];?>"><?php echo $row3['genre'];?></option> <?php } mysqli_close($link3); ?> <input type='hidden' id='drama_genreID' name='drama_genreID' value = "<?php echo $row2['drama_genreID']; ?>"/> </select> <input type="hidden" id="dramaID" name="dramaID" value = "<?php echo $row['dramaID'];?>"/> <input type="submit" Value="Update"/> The result was there were 3 dropdown menus but only the first dropdown menu has all 6 genres from my database and also the genre that belongs to the drama. I'm also wondering how I can bring all the drama_genreIDs to my save(doupdate.php) page and update all 3 of them because it seems like only the last dropdown menu's data is saved. And also how can I display only 6 genres instead of 7 with the genre that belongs to the drama , being set as the default selection. Hi, here's my problem: I am trying to make a simple online buying website and I want to display a table with all the fields for each item. So I got that part down which is to just use mysql_fetch_assoc("SELECT * FROM myTable") and use the html table tags stuff, but now I want to display my images in the table, so here's my code to display my mysql database table in html's table tag along w/ php: <html> <head> <title>My Online buying website project</title> </head> <body> <?php mysql_connect("localhost","root"); mysql_select_db("myTable"); $imagesArray=array("Apple_iPhone3GS.jpg","Apple_iPhone4.jpg","product3.jpg","product4.jpg","product5.jpg"); $result=mysql_query("SELECT Name, Manufacturer, Price, Description, SimSupport FROM myTable"); if(mysql_num_rows($result))//if there is at least one entry in bellProducts, make a table { print "<table border='border'>"; print "<tr> <th>Name</th> <th>Manufacturer</th> <th>Price</th> <th>Description</th> <th>SimSupport</th> </tr>"; //NB: now output each row of records while($row=mysql_fetch_assoc($result)) { extract($row); print "<tr> <td>$Name</td> <td>$Manufacturer</td> <td>$Price</td> <td>$Description</td><td>$SimSupport</td> </tr>"; }//END WHILE }//END IF ?> </table> </body> </html> *So how do I go about adding my images in this table? |