PHP - How To Display Data From The Mysql Database To My Webpage
I have got connection to the the mysql database, how do I get the data from the database to display on the webpage
Similar Tutorialsim making a game and i need to show a users money but i dont know how help? 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-->
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CREATE TABLE product( php code
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Hi, I have a webpage that the super administrator log's into. Once the super administrator is logged in he/she can view his/her clients. The super admins clients also have their own clients, once the super admins clients login they can view their own clients details who register with them. What i am trying to do is have a link on that clients name (for the super admins clients) that will bring me to a new page where i can edit his/her details that is stored in the mysql database. I have used INPUT buttons for each clients row in a FORM which works......but i know this is not the right way to do it. This is the code i have used to display the clients with the input button: <?php $query = "SELECT ID FROM clients WHERE username = '$username'"; $result = mysql_query($query); if(mysql_num_rows($result)) { while($row = mysql_fetch_assoc($result)){ $userID = $row['ID']; $query = "SELECT * FROM users WHERE userID = '$userID'"; $result = mysql_query($query); $num=mysql_numrows($result); mysql_close(); ?> <table border="0" cellspacing="3" cellpadding="3" width="100%"> <tr> <th width="30%"><font face="Arial, Helvetica, sans-serif">Name</font></th> <th><font face="Arial, Helvetica, sans-serif">Username</font></th> <th><font face="Arial, Helvetica, sans-serif">Email Address</font></th> <th> </th> </tr> <?php $i=0; while ($i < $num) { $userID=mysql_result($result,$i,"userID"); $name=mysql_result($result,$i,"name"); $username=mysql_result($result,$i,"username"); $email=mysql_result($result,$i,"email"); ?> <tr> <td width="30%"><?php echo $name; ?></td> <td><?php echo $username; ?></td> <td><?php echo $email; ?></td> <td><form action="http://localhost/single-client.php?userID=<?php echo $userID; ?>&name=<?php echo $name; ?>" method="post" style="margin:0px;"> <input type="hidden" name="selectuser" id="selectuser" value="<?php echo $userID; ?>" /><input type="hidden" name="username1" id="username1" value="<?php echo $username; ?>" /><input type="hidden" name="name1" id="name1" value="<?php echo $name; ?>" /><input type="submit" name="submit" value="View Info" class="button" /></form></td> </tr> <?php $i++; }}} ?> </table> From the above code when i use the following code: ?userID=<?php echo $userID; ?>&name=<?php echo $name; ?> It doesn't make a difference .... it just displays the correct userID and name in the link. Once the INPUT button is clicked it will bring you to this page: <?php $query = "SELECT * FROM users WHERE userID = '".$_POST['selectuser']."' AND username = '".$_POST['username1']."'"; $result = mysql_query($query); if(mysql_num_rows($result)) { $userID = $_POST['selectuser']; while($row = mysql_fetch_assoc($result)){ $name = $row['name']; $userID = $row['userID']; $email = $row['email']; $username = $row['username']; $registered = $row['registered']; $last = $row['last']; ?> <?php echo $brokerID; ?> <table border="0" cellspacing="3" cellpadding="3" width="100%" summary="Client table"> <tr> <td><form action="http://localhost/single-client.php?userID=<?php echo $userID; ?>&name=<?php echo $name; ?>" method="post" style="margin:0px;"> <input type="hidden" name="selectuser" id="selectuser" value="<?php echo $userID; ?>" /><input type="hidden" name="username1" id="username1" value="<?php echo $username; ?>" /><input type="hidden" name="name1" id="name1" value="<?php echo $name; ?>" /><input type="submit" name="submit" value="Income" class="button" /></form></td> <td><form action="http://localhost/single-client.php?userID=<?php echo $userID; ?>&name=<?php echo $name; ?>" method="post" style="margin:0px;"> <input type="hidden" name="selectuser" id="selectuser" value="<?php echo $userID; ?>" /><input type="hidden" name="username1" id="username1" value="<?php echo $username; ?>" /><input type="hidden" name="name1" id="name1" value="<?php echo $name; ?>" /><input type="submit" name="submit" value="Bills" class="button" /></form></td> </tr> </table> <h2>Client Profile - <?php echo $_POST['name1']; ?></h2> <table border="0" cellspacing="3" cellpadding="3" width="100%"> <tr> <th><font face="Arial, Helvetica, sans-serif">Name</font></th> <th><font face="Arial, Helvetica, sans-serif">Username</font></th> <th><font face="Arial, Helvetica, sans-serif">Email Address</font></th> <th><font face="Arial, Helvetica, sans-serif">Registered</font></th> <th><font face="Arial, Helvetica, sans-serif">Last</font></th> </tr> <tr> <td><?php echo $name; ?></td> <td><?php echo $username; ?></td> <td><?php echo $email; ?></td> <td><?php echo $registered; ?></td> <td><?php echo $last; ?></td> </tr> </table> <?php } } else { echo "<p>Error</p>"; } ?> This page displays the correct user info. Is there a way to do it differently instead of using FORMS and INPUT buttons? and rather using links? Any help would be greatly appreciated. i have a register and login page where the users can perform register and login task and their detail gets saved in the database. my requirement is that when the user enters his/her account, they should be redirected to their profile page, where they can view all their profile details. Till now i have been able to make the user login and dislay a part of their profile page, for this purpose the code that i am using is
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<?php Hi, I want to extract weather for 5 cities from www.wunderground.com/global/SB.html and put the field data such as colombo, temprature, humidity and conditions in to a mysql table so I can display the weather for this 5 cities and automatically set a cronjob to update it everyday. Issue is I am unable to extract the specific data from this page. Please help. thanks 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. 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? 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: $eventdate "; } ?> how i want to display data from database to look like this : <table width="633" height="224" border="1"> <tr bgcolor="#999900"> <td width="45">Bil</td> <td width="121">Course_name</td> <td width="83">session</td> <td width="83">start_date</td> <td width="83">end_date</td> <td width="83">notes</td> <td width="89">pre-req</td> </tr> <tr bgcolor="#6A7AEA"> <td rowspan="2">1.</td> <td rowspan="2" bgcolor="#6A7AEA">Math</td> <td>1st session </td> <td>1 jan 11 </td> <td>6 jan 11 </td> <td rowspan="2"> </td> <td rowspan="2"><image icon that will link to the oter site> </td> </tr> <tr> <td bgcolor="#6A7AEA">2nd session </td> <td bgcolor="#6A7AEA">8 jan 11 </td> <td bgcolor="#6A7AEA">15 jan 11 </td> </tr> <tr> <td bgcolor="#0066CC">2.</td> <td bgcolor="#0066CC">English</td> <td bgcolor="#0066CC">1st session </td> <td bgcolor="#0066CC">1 feb 11 </td> <td bgcolor="#0066CC">6 feb 11 </td> <td bgcolor="#0066CC"> </td> <td bgcolor="#0066CC"><image icon that will link to the oter site></td> </tr> <tr> <td rowspan="2" bgcolor="#6A7AEA">3.</td> <td rowspan="2" bgcolor="#6A7AEA">Science</td> <td height="29" bgcolor="#6A7AEA">1st session </td> <td bgcolor="#6A7AEA">8 march 11 </td> <td bgcolor="#6A7AEA">15 march 11 </td> <td rowspan="2" bgcolor="#6A7AEA"> </td> <td rowspan="2" bgcolor="#6A7AEA"><image icon that will link to the oter site></td> </tr> <tr> <td bgcolor="#6A7AEA">2nd session</td> <td bgcolor="#6A7AEA">16 march 11 </td> <td bgcolor="#6A7AEA">21 march 11 </td> </tr> </table> ** all the view data is called from database including the icon image thanks... hey guys so im trying to display data into text boxes that are fetched from database according to checkbox with value id. processing is located before <!DOCTYPE html>: if(isset($_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. 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 This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=321339.0 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 there. I have this simple code which displays 5 results. How can i grab each element separately instead of displaying all the results at once. Thanks:) Code: [Select] <?php $query = mysql_query("SELECT product_name, product_price FROM products WHERE product_type = 'laptop' LIMIT 5"); $numrows = mysql_num_rows($query); if ($numrows != 0) { while ($row = mysql_fetch_assoc($query)) { $product_name = $row['product_name']; $product_price = $row['product_price']; echo $product_name . '<br />'; echo $product_price . '<br /><br />'; } } ?> 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? Hi,
I wish to find out is there any possible that I can delete some data inside my php website but inside my sql database, the record will still at there?
Thank you.
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