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PHP - Generating Dynamic Html Table While Displaying Mysql Data
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 $row_2["username"] . "<br>"; } ?> The above simply outputs all records of a user's friends (their usernames) in alphabetical order. The question of how I'd generate a new row each time a certain amount of columns have been met, however, is beyond me. Anyone know of any helpful resources that may solve my problem? Thanks in advance =) Similar TutorialsHello to all, I have problem figuring out how to properly display data fetched from MySQL database in a HTML table. In the below example I am using two while loops, where the second one is nested inside first one, that check two different expressions fetching data from tables found in a MySQL database. The second expression compares the two tables IDs and after their match it displays the email of the account holder in each column in the HTML table. The main problem is that the 'email' row is displayed properly while its while expression is not nested and alone(meaning the other data is omitted or commented out), but either nested or neighbored to the first while loop, it is displayed horizontally and the other data ('validity', 'valid_from', 'valid_to') is not displayed.'
Can someone help me on this, I guess the problem lies in the while loop? <thead> <tr> <th data-column-id="id" data-type="numeric">ID</th> <th data-column-id="email">Subscriber's Email</th> <th data-column-id="validity">Validity</th> <th data-column-id="valid_from">Valid From</th> <th data-column-id="valid_to">Valid To</th> </tr> </thead> Here is part of the PHP code:
<?php
while($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. So I have a jobs database with the following columns: id, jobtext, jobdate, and id. This is how it looks right now: http://prahan.com/jobs/display.html.php I have another table called author. In the authorid column in need the results of this query, SELECT name FROM author WHERE id = (SELECT authorid FROM job) , to be displayed for each row. I also want to be able to customize the header title for each column. Thanks in advance! Not really sure how to get the images I have stored in MySQL into a html form. I can call-up the text fields from the database but it cannot seem to find the index for the images. Here is my code:- <?php session_start(); mysql_connect("localhost","root","abc") or die ("Error! Cannot connect to database"); mysql_select_db("theimageworks") or die ("Cannot find database"); $query = "SELECT * FROM jobs"; $result = mysql_query($query) or die (mysql_error()); ?> <?php //display data in html table echo "<table>"; echo "<tr><td>Username</td><td align='center'>Message</td><td>Product Image</td></tr>"; while($row = mysql_fetch_array($result)) { echo "</td><td>"; echo $row['username']; echo "</td><td>"; echo $row['message']; echo "</td></tr>"; echo $row['image']; } echo "</table>"; ?> The error message I get is "Notice: Undefined index: image in....." Thanks in advance! Hi all, I'm new to php/myslq and I'm going crazy trying to figure this one out I'm building a personal calendar and I want to display the data into a HTML table. Code: [Select] //query the database $query = "SELECT * FROM tbl_events WHERE event_day=$day AND event_month=$month AND event_year=$year"; $query = mysql_query($query); //build the table echo '<table>'; for ($y = 0; $y < 6; $y++){ echo '<tr><td>'; //insert the data here echo '</td></tr>'; } echo '</table>'; When I store the events into the database, I assign a slot for each one depending on the hour. I don't want to use more than 6 events daily, hence the for loop. The problem I have is how to I insert the data into the designed <td>? In a particular day I could have only 2 events: event 1 - slot 2, event 2 - slot 6. I want to be able to enter each event into its own cell I hope I'm making myself clear enough. Sorry for any English mistakes if any. Thank you Hi all, I want to create an array from the output of a table. In the table cell I have some numbers where for each number I would like to relate it to a value in another table. However using the code below I get Array ( => "26","27","28","29","30","31" ) from the print_r($arr); meaning that $range ins't being treated as an array, and hence, the second query only gives one value. Code: [Select] $showIng = mysql_query("SELECT * FROM recipe WHERE name = '$dish'") or die(mysql_error()); $rowShowIng = mysql_fetch_assoc($showIng); $range = $rowShowIng['keyno']; $arr = array($range); print_r($arr); foreach ($arr as &$value) { $getIng = mysql_query("SELECT * FROM keywords WHERE keyid = '$value'") or die(mysql_error()); $rowGetIng = mysql_fetch_assoc($getIng); echo $rowGetIng['ingredient']; } So, I want to know how I can create the array from $range Thanks
Hello, I have the problem that only one user is displayed in the table
require('connection/db1.php');
// Teilnehmerliste
$query = 'SELECT * FROM convoy_part WHERE user_convoy= :I';
$start = $bdd->prepare($query);
$start->execute(array(':I' => $_GET['id']));
//fetch
$result2 = $start->fetch();
// Zählung der Datensätze
$count = $start->rowCount();
<table class="table">
<thead>
<tr>
<th scope="col">User</th>
<th scope="col">Datum</th>
<th scope="col"><a class="btn btn-primary" href="convoy_user.php?id=<?php echo $result['id']; ?>&action=part" role="button">Teilnehmen</a></th>
</tr>
</thead>
<tbody>
<tr>
<th scope="row"><?php echo $result2['name']; ?></th>
<td><?php echo $result2['date']; ?></td>
</tr>
</tbody>
</table>
Friends I am new to php and i have to submit my coursework in php by 3rd dec, I stuck at one place where i have to upload multiple photo and one can see all the photo he has uploaded and can edit or delete that photo so i have done uploading now i am showing those pics in table by running loop and generating tr and td but now i have two buttons with each row edit and delete now when i clicked on one delete or edit that pic should be delete or give text box to edit description of pic, Please help me how to do that....... Ok... I need some help - I want to show a players balance in a game beside there name (Balance is in mysql database)I can do that but... - I also want to show if there online or offline at the same time( This is stored in a different database) I have the code which says whether they are online or offline <?PHP // Conect to the Mysql Server $connect = mysql_connect("localhost","scswccla_bukkit","********"); //connect to the database mysql_select_db("scswccla_bukkit"); //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 1"); //title echo "<font size='100'>NovaCraft Users</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='black'>|Online|<br><font color='green'>$users</font></font><br>"; endwhile; //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 0"); //title echo "<font color='black'>|Offline|</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='red'>$users</font><br>"; endwhile; ?> Here is the page: www.scswc.com/Offline_Users.php displaying that But I want to Create something like this: Nocvacraft Players |Online| Name:Player Balance:$20 |Offline| Name:Player Balance:$15 Here is what I have tried: <?PHP // Conect to the Mysql Server $connect = mysql_connect("localhost","scswccla_bukkit","**********"); //connect to the database mysql_select_db("scswccla_bukkit"); //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 1"); $query2 =mysql_query("SELECT * FROM iBalances WHERE player = $users"); //title echo "<font size='100'>NovaCraft Users</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query) $rows2 = mysql_fetch_array($query2)): $users = $rows['name']; $balance = $rows2['balance']; echo "<font color='black'>|Online|<br><font color='green'>Name:$usersBalance:$balance</font></font><br>"; endwhile; //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 0"); //title echo "<font color='black'>|Offline|</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='red'>$users</font><br>"; endwhile; ?> I know I am trying to use a variable before it is been set - but if I don't how I have tried this as well... <?PHP // Conect to the Mysql Server $connect = mysql_connect("localhost","scswccla_bukkit","**********"); //connect to the database mysql_select_db("scswccla_bukkit"); //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 1"); //title echo "<font size='100'>NovaCraft Users</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='black'>|Online|<br><font color='green'>Name:$users</font></font><br>"; endwhile; //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 0"); //title echo "<font color='black'>|Offline|</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='red'>$users</font><br>"; endwhile; $query = mysql_query("SELECT * FROM iBalances WHERE player = $users"); WHILE($rows = mysql_fetch_array($query)): $balance = $rows['balance']; echo "<font color='red'>$users $balance</font><br>"; endwhile; // ?> Can you use variables in mysql_query()?Is that why it isn't working? This is my first php script so if I need to give you more information for you to help me just tell me Thanks Here is database pictures iBalances users_online Hey guys have been trying to get this script to work for a while now, i am new to php and mysql so i am sure i am missing something simple. I have DB setup and need to pull data based on the key item code and get the following I want to get the fields item_code description allergy_statement useable_units region_availability order_lead_time ingredients for item_code 12-100 LITERALLY 12-100, no range, but like i said before i am really new to php and mysql. I have 1187 items that when a user clicks a link in search results it takes them to the product details page for that item code All that data is in my database just can't figure out how to get it out of the database. Is this even the right script to achieve that result. here is the code to get the data from database Code: [Select] <?php require_once('includes/mysql_connect_nfacts_ro.php'); $query = "SELECT item_code, description, allergy_statement, useable_units, region_availability, order_lead_time, ingredients " . "FROM products " . "WHERE item_code = '12-100' "; $resuts = mysql_query($query) or die(mysql_error()); ?> And need to display the data like so : Code: [Select] <td width="715" align="center" valign="top"> <h1>Product Details</h1> <h3>DISPLAY description HERE</h3> <table width="420" border="0"> <td class="ingreg"> </td> </table> <h5>Item Number</h5> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY ITEM_CODE HERE</td> </tr> </table> <h3>Ingredients:</h3> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY INGREDIENTS HERE</td> </tr> </table> <h4>Allergy Statement:</h4> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY Allergy Statement HERE</td> </tr> </table> <h4>Useable Units Per Package:</h4> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY Useable Units Per Package HERE</td> </tr> </table> <h4>Region Availability: </h4> <table width="420" border="0"> <tr> <td class="ingreg">&DISPLAY ITEM_CODE HERE</td> </tr> </table> <h4>Order Lead Time:</h4> <table width="420" border="0"> <tr> <td class="ingreg">&DISPLAY order lead time HERE</td> </tr> </table> <p> </p> <div align="right"></div></td> </tr> </table> how do i get data in database to display where i need it to? Can any one shine some light on this So i pull some records out of a mysql table and i want to display them in 5 even columns. I'm not entirely sure how to do the math & logic to accomplish this. The pull is simple $qry = "SELECT DIST_PART_NUM FROM $tablename"; $sql = mysql_query($qry) or die(mysql_error()); while($res = mysql_fetch_assoc($sql)) { // CREATE 5 even columns here. } so let's say i just retrieved 5,000 part numbers, i'd like to display then in a table of 5 columns with 1000 records per column. This is easy math, but i need the script to automatically figure out the #'s. Also the tricky part is that i dont want to display the part numbers like so 11111 22222 33333 44444 55555 66666 77777 88888 99999 00000 but rather 11111 44444 77777 22222 55555 88888 33333 66666 99999 00000 the remainder if there is one can go in the last column or whatever is easier. I'd tried googling this, but it's not easy to phrase what i'm looking for. Thanks for the help. PS: I'm not looking to copy and paste code, if possible please explain your way so that i can learn the logic. im making a game and i need to show a users money but i dont know how help? Hi guys. I'm trying to build a games site for a friend. Currently, the front-end (HTML/CSS) of the site is done. Now, I want to make a way for him to easily add games to the site. Ideally, I'd like to make a database with the following columns: ID, Name, Category, Link, Thumbnail_Link. So, those would be the ID, name of the game, the category, a link to the game, and a link to the 50x50 thumbnail image respectively. Then, using PHP, I'd like to call the first x number (not sure what it will be yet, let's say 50) and make format it as a grid in the following way: There's the thumbnail image followed by the game name, and they're all a clickable link to the game URL. Is this possible? How would I go about doing this? I've already set up a database for a login module to the site, so each page has already opened a connection to the MySQL database. However, I've only ever done basic PHP for mail forms and am otherwise extremely new to it, and am especially new to MySQL. Could anyone walk me through how to do this or even give me a quick example script to work off of? Thanks, any of your time is greatly appreciated! I am querying my database to show the visit statistics for a particular week and it shows the number of visits for the countries, but does not display the country name.
I have proved that the MySQL works by going into phpMyAdmin and pasting the query into SQL query tab, replacing the POST with 1, for week 1.
I can't see why it is not displying the country.
Here is the code:
<?php
include('connect_visits.php');
doDB7();
$WVisit_data="SELECT WeekNo15.WNo, WeekNo15.WCom, Countries.Country, Countries.CID, ctryvisits15.CVisits
FROM ctryvisits15
LEFT JOIN Countries
ON ctryvisits15.country = Countries.CID
LEFT JOIN WeekNo15
ON ctryvisits15.WNo = WeekNo15.WNo
WHERE ctryvisits15.WNo = '{$_POST['WeekNo']}'
ORDER BY ctryvisits15.CVisits DESC";
$WVisit_data_res = mysqli_query($mysqli, $WVisit_data) or die(mysqli_error($mysqli));
$display_block ="
<table width=\"20%\" cellpadding=\"3\" cellspacing=\"1\" border=\"1\" BGCOLOR=\"white\" >
<tr>
<th>Country</th>
<th>Visits</>
</tr>";
while ($WV_info = mysqli_fetch_array($WVisit_data_res)){
$Ctry = $WV_info['country'];
$Visits = $WV_info['CVisits'];
//add to display
$display_block .="
<tr>
<td width=\"10%\" valign=\"top\">".$Ctry."<br/></td>
<td width=\"5%\" valign=\"top\">".$Visits."<br/></td>
";
}
mysqli_free_result($WVisit_data_res);
mysqli_close($mysqli);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<!--
Design by Free CSS Templates
http://www.freecsstemplates.org
Released for free under a Creative Commons Attribution 2.5 License
Name : Yosemite
Description: A two-column, fixed-width design with dark color scheme.
Version : 1.0
Released : 20091106
-->
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta name="keywords" content="" />
<meta name="description" content="" />
<meta http-equiv="content-type" content="text/html; charset=utf-8" />
<title>1066 Cards 4U - Stats for country</title>
<link href="style.css" rel="stylesheet" type="text/css" media="screen" />
</head>
<body>
<div id="wrapper">
<div id="menu">
<ul>
<li class="current_page_item"><a href="index.php">Home</a></li>
<li><a href="Links.html">Links</a></li>
<li><a href="Verse_Menu.html">Verses</a></li>
<li><a href="Techniques.html">Techniques</a></li>
<li><a href="blog.php">Blog</a></li>
<li><a href="Gallery.html">Gallery</a></li>
<li><a href="contact.html">Contact</a></li>
<li><a href="AboutUs.html">About Us</a></li>
<li><a href="stats1.html">Stats</a></li>
</ul>
</div><!-- end #menu -->
<div id="header">
<div id="logo">
<h1><a href="http://www.1066cards4u.co.uk">1066 Cards 4U</a></h1>
</div><!-- end #wrapper -->
</div><!-- end #header -->
<div id="page">
<div id="page-bgtop">
<div id="page-bgbtm">
<div id="content">
<h3>Statistics for Week Commencing <? echo $WkCom; ?> in 2015</h3>
<div id="table">
<?php echo $display_block; ?></div>
</div><!-- end #content -->
</body>
</html>
Can you help please?
Edited by rocky48, 07 January 2015 - 07:33 AM. Hi all! I've been recently messing with PHP and MySQL to attempt to create a search function for my tables (too advanced, so I toned it down to learning to just display data from my table). Anyways, I'm watching a YouTube video on how to do it, and I'm copying his code exactly, but I'm still getting an error. Can anyone help? Thanks. Edit: Sorry, my error was snuggled down below the code. Anyways, my error is just my while echo appearing in and out of my tables on the page. My code:
<?php
//Make connection
mysql_connect('localhost', 'root', 'test');
// select db
mysql_select_db('testdatabase');
$sql="SELECT * FROM employees";
$records=mysql_query($sql);
?>
<html>
<head>
<title>Employee Data</title>
</head>
<body>
<table width="600" border="1" cellpadding="1" cellspacing="1">
<tr>
<th>ID</th>
<th>First name</th>
<th>Surname</th>
<th>Age</th>
</tr>
<?php
while($employees=mysql_fetch_assoc($records)){
echo "<tr>";
echo "<td>".$Employee['ID']."</td>";
echo "<td>".$Employee['First name']."</td>";
echo "<td>".$Employee['Surname']."</td>";
echo "<td>".$Employee['Age']."</td>";
echo "</tr>";
}
// end while
?>
</table>
</body>
</html>
Edited by thwikehu1990, 19 January 2015 - 01:33 PM. I have a field in my table that gets populated with a custom value from a subscription form, but when the data gets inserted into the table, it adds some text I do not want to output. The field is for the user's name, but when the form saves it, it saves the name, plus the text #2# (this is from a Wordpress plugin and I can't figure out how to remove the #2# when I insert the value there either). Is there a way to remove this #2# text when I display the field's value in MySQL and PHP? I have a table in a mysql database with 5 columns, id, Name, Wifi, Bluetooth, GPS, with rows that are for example 1, Galaxy S2, yes, yes, yes. So basically i want to build a form that has check boxes (3 checkboxes for wifi bluetooth and GPS respectively) that once selected will query the database depending on which check boxes are selected. I have made the form but need to know what to put in the filter.php to make the results be displayed accordingly. Ideally if anyone knows how i would want it so the results were shown on the same page which i believe u need to use ajax to happen but any help on how to show the results will be greatful. the code for the form i have made is as follows: Code: [Select] <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <title>test</title> </head> <body> <form action="filter.php" method="post"> <INPUT TYPE=CHECKBOX NAME="option[]" VALUE="Wifi" id="r1">Wifi <INPUT TYPE=CHECKBOX NAME="option[]" VALUE="Bluetooth" id="b1">Bluetooth <INPUT TYPE=CHECKBOX NAME="option[]" VALUE="GPS" id="g1">GPS <input type="submit" name="formSubmit" value="Submit" /> </form> </body> </html> im not sure on how to make the filter.php page but i do have this code for displaying the all the data but i want to kno how to make it only display the data from the choices on the checkboxes Code: [Select] <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <title>pls work</title> </head> <body> <?php function h($s) { echo htmlspecialchars($s); } mysql_connect("localhost", "root", "") or die (mysql_error()); mysql_select_db("project") or die (mysql_error()); $result= mysql_query('SELECT * FROM test') or die('Error, query failed'); ?> <?php if (mysql_num_rows($result)==0) { ?> Database is empty <br/> <?php } else { ?> <table> <tr> <th></th> <th>Name</th> <th>Wifi</th> <th>Bluetooth</th> <th>GPS</th> </tr> <?php while ($row= mysql_fetch_assoc($result)) { ?> <tr> <td> <a href="uploaded-images/<?php h($row['Name']); ?>.jpg"> <img src="uploaded-images/<?php h($row['Name']); ?>.jpg" alt="test"/> </a> </td> <td><a href="textonly.html"><?php h($row['Name']); ?></a></td> <td><?php h($row['Wifi']); ?></td> <td><?php h($row['Bluetooth']); ?></td> <td><?php h($row['GPS']); ?></td> </tr> <?php } ?> </table> <?php } ?> </body> </html> Hi all, I'm trying to display a form based on a dropd own selection. The drop down is propagated from a database query. I have the drop down list displaying, but when I click submit, I can't get the result to display. Code: [Select] $sql="SELECT id, company_name FROM webprojects ORDER BY company_name ASC"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $id=$row["id"]; $company_name=$row["company_name"]; $options.="<OPTION VALUE=\"$id\">".$company_name; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>PHP Dynamic Drop Down Menu</title> </head> <body> <form id="form1" name="form1" method="post" action="selected.php"> <SELECT NAME=id> <OPTION VALUE=0>Choose <?=$options.="<OPTION VALUE=\"$id\">".$company_name.'</option>';?> </SELECT> <label> <input type="submit" name="submit" id="submit" value="Submit" /> </label> </form> </body> </html> Any help is greatly appreciated Hello,
I'm not sure if I'm allowed to post the code, as I didn't write it. But, if you could offer some ideas on where I'm going wrong, I'd be very appreciative!
I basically have a feature on my homepage that displays "Hello _username_!" upon login, but the username is missing. So, I'm presuming that the functionality for pulling the username from the table is somehow failing.
Any ideas?
Thanks
Daisy
Hello, I have a simple feedback form which asks for name, location and a simple comment about the business. I have inserted data into the database successfully but I'm not sure on the best way to display the data from the table on the same page as the form (after redirection). Here is my code: FORM: Code: [Select] <form id="feedback" name="feedback" action="php/phpcustom.php" method="POST"> <fieldset> <legend>Gardenable.com Feedback</legend> <p><label for="fname">Name:</label><input type="text" size="30" maxlength="40" id="fname" name="fname" /></p> <p><label for="loc">Location:</label><input type="text" size="30" maxlength="40" id="loc" name="loc" /></p> <p><label for="com">Comments:</label><textarea cols="40" rows="6" maxlength="300" id="com" name="com"></textarea></p> <p><input type="submit" name="send" id="submitbutton" value="Submit" /><input type="reset" name="reset" value="Reset" /></p> </fieldset> </form> ACTION SCRIPT on submit: Code: [Select] <?php session_start(); $dbhandle = mysql_connect('localhost', 'root', '') or die("Unable to connect to MySQL"); $selected = mysql_select_db("commentdatabase",$dbhandle) or die("Could not select the database"); $name = $_POST['fname']; $loc = $_POST['loc']; $com = $_POST['com']; $sql = "INSERT INTO userinfo (name, location, comment) VALUES ('{$name}','{$loc}','{$com}')"; if(!mysql_query($sql, $dbhandle)) { die('Error: ' . mysql_error()); } $dbselect = "SELECT * FROM userinfo"; if(!mysql_query($dbselect, $dbhandle)) { die('Error: ' . mysql_error()); } header('Location: ../contact.htm'); mysql_close(); ?> //This is underneath the FORM CODE: Code: [Select] <div id="feedbackDiv"> <?php $dbhandle = mysql_connect('localhost', 'root', '') or die("Unable to connect to MySQL"); $selected = mysql_select_db("commentdatabase",$dbhandle) or die("Could not select the database"); $result = mysql_query("SELECT * FROM userinfo"); while ($row = mysql_fetch_array($result)) { echo "ID:".$row{'name'}." Name:".$row{'location'}."Year: ". $row{'comment'}."<br />"; } ?> </div> It is displaying a tiny part of php code rather than the returned data. This is what i being displayed: "; } ?> The data is inserting correctly but i'm unsure as to whether or not I need to pass the values in the url string in the header function? Thank you for any information you can provide me. Regards, BuNgLe hi, I am displaying mysql data in a table form using php. The first attached pic shows how is it looking with the current code. As you can see there are heading on the left and top of the table. So user can make booking on a specific day (heading is on the top for days) and for specific duration of time(headings are on the left). As in first pic, there has been a booking on Wed at 10 o clock, but the problem is, it is displaying duration just for half an hour. As this duration is from 10 o clock to 11 o clock. So what I want to do is in the second pic. Please view the second pic. In the second pic I want to display the highlighted area in blie because all this time is booked or reserved. Here is my current code. Code: [Select] <?php require("dbconnect/dbconnect.php"); $building = $_POST['building']; $block = $_POST['block']; $d_m = $_POST['d_m']; $choose_one = $_POST['choose_one']; function get_data($table, $choose_one) { $bool = 1; $get_days = mysql_query("SELECT * FROM days"); $get_times = mysql_query("SELECT * FROM time"); $days = mysql_num_rows($get_days); //$row_day_id = mysql_fetch_assoc($get_days); //$day_id_db = $row['dayid']; echo "<table cellspacing='0' border='0' cellpadding='5'> <tr> <th scope='col' width='100'></th>"; while($row_time = mysql_fetch_assoc($get_times)) { while($bool <= $days) { $row_days = mysql_fetch_assoc($get_days); $dayname = $row_days['days']; echo " <th scope='col' width='100'>$dayname</th> "; $bool++; } $day_id = 1; $timeid = $row_time['timeid']; $time = $row_time['time']; echo " <tr> <th scope='row' width='100' height='50'>$time</th>"; while($day_id <= $days) { $get = mysql_query(" SELECT * FROM $table WHERE dayid='$day_id' AND timeid_from='$timeid' AND d_m='$choose_one' ") or die(mysql_error()); if(mysql_numrows($get) > 0) { $row = mysql_fetch_assoc($get); $block = $row['block']; $apt = $row['apartment_no']; $room = $row['room_no']; $d_m = $row['d_m']; $timeid_to = $row['timeid_to']; echo " <td width='110' height='50' bgcolor='lightblue'> <b>Block:</b> $block<br /> <b>Apartment:</b> $apt<br /> <b>Room:</b> $room </td> "; //echo "</tr>"; } else { echo " <td width='110' height='50' bgcolor='green'></td> "; } $day_id++; } //end of second inner while loop echo "</tr>"; } //end of outer while loop echo "</table>"; } if($building) { if($block) { if($d_m) { if($choose_one) { if($building == 'mik') { if($block == 'Block A' || $block == 'Block B') { get_data('reserve_mik_ab', $choose_one); } else if ($block == 'Block C' || $block == 'Block D') { get_data('reserve_mik_cd', $choose_one); } } if($building == 'p') { get_data('reserve_p',$choose_one); } } else { echo "Choose one!"; } } } } ?> Please help! |
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