PHP - (fixing Code) Html Table + Php + Mysql
Hi everyone.
I've two MySQL tables (tbl_csv_input & tbl_qn_types). Table Details: 1. tbl_csv_input (qid, qname, qn_answer, qn_type) Primary Key = qid; Foreign Key = qn_type ref tbl_qn_types qn_type_id 2. tbl_qn_types (qn_type_id, qn_type_name, notes) Primary Key = qn_type_id Sample Data: tbl_csv_input 1 ; Ferrari is the fastest car? ; Yes||No ; T1 2 ; Which below features would you like to have in you... ; Navigator||Airbag||Seat Belt||Camera & Sensors ; T1 3 ; Which model would you prefer? ; CX||MX||SX||LX ; T2 4 ; Comments ; ; T3 tbl_qn_types T1; checkbox; This qn type is used for yes/no T2; radio; Multiple options but only one is correct T3; text area; Users enter input like comments The whole idea is to have a questionnaire displayed in a HTML table depending upon the question types (T1, T2, T3) In simple terms: You have a question and below it there are options. Some questions have check boxes and some have radio buttons and some have text areas. The problem I'm facing is with the column: qn_answer; and column = qn_type. I'm unable to make a loop inside a table which is already in a loop. Please see the attached files (which has the code I've written). TIA. [attachment deleted by admin] Similar TutorialsHello, I need some help. Say that I have a list in my MySQL database that contains elements "A", "S", "C", "D" etc... Now, I want to generate an html table where these elements should be distributed in a random and unique way while leaving some entries of the table empty, see the picture below. But, I have no clue how to do this... Any hints? Thanks in advance, Vero I know I'm doing it something right, but can someone tell me why only one table is showing up? Can you help me fix the issue? Heres my code: function showcoords() { echo"J3st3r's CoordVision"; $result=dbquery("SELECT alliance, region, coordx, coordy FROM ".DB_COORDFUSION.""); dbarray($result); $fields_num = mysql_num_fields($result); echo "<table border='1'>"; // printing table headers echo "<td>Alliance</td>"; echo "<td>Region</td>"; echo "<td>Coord</td>"; // printing table rows while($row = mysql_fetch_array($result)) { // $row is array... foreach( .. ) puts every element // of $row to $cell variable foreach($row AS $Cell) echo "<tr>"; echo "<td>".$row['alliance']."</td>\n"; echo "<td>".$row['region']."</td>\n"; echo "<td>".$row['coordx'].",".$row['coordy']."</td>\n"; echo "</tr>\n"; } echo "</table>"; mysql_free_result($result); } I have 2 rows inserted into my coords table. Just frustrated and ignorant to php. I'm getting this error when I use the following code. Could someone help me to fix this. Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource $result =sprintf("SELECT * FROM my_tbl WHERE username ='%s' || email= '%s'", mysql_real_escape_string($username), mysql_real_escape_string($email)); $resultfin = mysql_query($result); if (mysql_num_rows ($resultfin) > 0){ Thanks, Bickey. Hi, In 2005 I got help on this forum to write some php code that would return results based on the date input +-3 days. It worked at the time but then it stopped working and I left it until now. Im now updating the database and would love for this query to work again. If anyone can tell me why this code isnt returning anything I would be extremely grateful <?php $db = mysql_connect("localhost", "xxxxx", "xxxxxx"); mysql_select_db("xxxxxxx", $db); // code checking here // ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=windows-1252"> <title>Manx bird record search</title> </head> <body> <p align="center"><u><font face="Comic Sans MS" style="font-size: 16pt">Manx records date search</font></u></font></p> <p align="center"> </p> <form method="POST" action=""> <p align="center"> Input date (eg for 18th July type in 18-07 ): <input type="text" name="query"> <input type="SUBMIT" value="Search!"> </form> <div align="center"> <table border=1 cellpadding="3" height="14"><tr> <th bgcolor="#CCCCFF" width="200"><span style="font-weight: 400"> <font face="Comic Sans MS">Bird Name</font></span></th> <th bgcolor="#CCCCFF" width="70"><span style="font-weight: 400"> <font face="Comic Sans MS">Date</font></span></th> <th bgcolor="#CCCCFF" width="200"><span style="font-weight: 400"> <font face="Comic Sans MS">Location</font></span></th></tr> <?php list($day, $month) = explode('-', $query); $query = '2000-'.$month.'-'.$day; //the variable query should now have a date formatted for the mysql query $result = mysql_query("SELECT *, DATE_FORMAT(Date, '%m%d') AS mmdd from Records WHERE DATE_FORMAT(Date, '%m%d') BETWEEN DATE_FORMAT(DATE_SUB('$query', INTERVAL 3 DAY), '%m%d') AND DATE_FORMAT(DATE_ADD('$query', INTERVAL 3 DAY), '%m%d')ORDER BY mmdd", $db) or die(mysql_error()); // keeps getting the next row until there are no more to get while ($row = mysql_fetch_array($result)){ // stores the date in dd-mm format as variable $date2 $date2 = date('j M', strtotime($row['Date'])); echo "<tr><td>".$row['Bird']."</td><td>".$date2."</td><td>".$row['Location']."</td></tr>"; } ?> </table> </div> <p align="center"> </p> <p align="center" style="margin-top: 0; margin-bottom: 0"><font size="2">All records from Manx Ornithological Societies annual "Peregrine" reports 1994-2003.</font></p> <p align="center" style="margin-top: 0; margin-bottom: 0"><font size="2">For details on how to purchase the "Peregrine" reports please click on the "Contact me" link of the website.</font></p> <p align="center" style="margin-top: 0; margin-bottom: 0"> </p> </body> </html> require_once 'phpSimpleHtmlDomClass.php'; $html = '<div> <div class="man">Name: madac</div> <div class="man">Age: 18 <div class="man">Class: 12</div> </div>' $name=$html->find('div[class="man"]', 0)->innertext; $age=$html->find('div[class="man"]', 1)->innertext; $cls=$html->find('div[class="man"]', 2)->innertext; wanna get a text from each div class="man" but it didn't work because there is a missing closing div tag on 2nd line of html code. please help me to fix this. 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 I am having so much trouble with this. I want to create a html table that looks like this: Code: [Select] <table id="datatable"> <thead> <tr> <th></th> <th>test</th> <th>test 2</th> </tr> </thead> <tbody> <tr> <th>2011-03-17</th><td>1</td><td>0</td> </tr> <tr> <th>2011-03-18</th><td>3</td><td>2</td> </tr> <tr> <th>2011-03-19</th><td>1</td><td>0</td> </tr> </tbody> </table> out of this mysql result: opens | name | date 1 | test | 2011-03-17 3 | test | 2011-03-18 2 | test 2 | 2011-03-18 1 | test 2 | 2011-03-19 Here is my mysql code: Code: [Select] SELECT count( o.campaign_id ) AS opens, c.name, date(o.created_at) as date FROM opens o LEFT JOIN campaigns c ON c.id = o.campaign_id WHERE (o.created_at between '2011-03-17 00:00:00' and '2011-03-19 23:59:59') GROUP BY date,c.name I also attached a sql dump of my database. I was wondering how does one go about showing results from SELECT query in columns in a html table. I have a list of products in a table, and would like to show them on the page in 4 columns. I have done many searches on google to try and find the sulution, but the majority of what im finding instead is about displaying a table from phpmyadmin as a table in html. If its a large operation to do this, I would be very happy if someone could poiint me in the direction of a tutorial maybe. Here is the code I have so far to display the products, but for some reason, it only show 1 row instead of all the rows from my table. Code: [Select] <?php $dbhost = "localhost"; $dbuser = "user"; $dbpass = "pass"; $dbname = "dbname"; mysql_connect ($dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error()); mysql_select_db($dbname) or die(mysql_error()); $result = mysql_query("SELECT * FROM mcproducts"); while($row = mysql_fetch_array($result)) { $products_local_id = $row['products_local_id']; $productname = $row['product_name']; $thumburl = $row['image_from_url']; $productlink = $row['product_local_url']; $thumbnail = $row['product_image_small']; $currencysymbol = $row['product_currency']; $price = $row['product_price']; $flagicon = $row['product_country_from']; } ?> <html> <head> <link href="style/stylesheet.css" rel="stylesheet" type="text/css" /> </head> <body> <div class="displaybox"> <div class="productimage"> <a href="<?php echo $productlink; ?><?php echo $products_local_id; ?>"><img src="<?php echo $thumburl; ?><?php echo $thumbnail ?>" width="150" height="150"></a> </div> <div class="productdescription"> <div class="pro_name"> <a href="<?php echo $productlink ?><?php echo $products_local_id; ?>"><?php echo $productname; ?></a> </div> <div class="pro_description"> </div> <?php if ($flagicon=="Ireland") { $flagicon = "<img src=\"flags/ireland.jpg\">"; } elseif ($flagicon=="UK") { $flagicon = "<img src=\"flags/uk.jpg\">"; } else echo ""; ?> <div class="pro_description"><?php echo $flagicon; ?><?php echo $currencysymbol ?> <?php echo $price ?></div> </div> </div> </body> </html> Many thanks, DB
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>
Hi, I have this script which I would like to use to build an html form from a MySQL table. The <input text................ writes successfully to file but none of the other form types are written to file when using the elseif command Also, I would like this script to carry out the file write foreach row in the table It seems complicated and i am not sure if I am going about it in the right way, but here goes Code: [Select] //// Create Forms from MySQL $result = mysql_query("SELECT * FROM forms"); while($row = mysql_fetch_assoc($result)){ $field_label = $row['field_label']; $column_name = $row['column_name']; $field_type = $row['field_type']; if ($row['field_type'] = 'Text') { $stringData = "$field_label<input type=text name=$column_name>\n"; fwrite($fh, $stringData); } elseif ($row['field_type'] = 'Text Area') { $stringData = "$field_label<input type=textarea name=$column_name></textarea>\n"; fwrite($fh, $stringData); } elseif ($row['field_type'] = 'Select Menu') { $stringData = "$field_label<select name=$column_name></select>\n"; fwrite($fh, $stringData); } elseif ($row['field_type'] = 'Checkbox') { $stringData = "$field_label<input type=checkbox name=$column_name></textarea>\n"; fwrite($fh, $stringData); } fclose($fh); As always, any help is much appreciated 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! Hello guys, I am a new programmer and i am building a new website. I would like to give me your advice for the following problem: I want to build a webpage, in which there will be a <div id="book-content"> ...............</div> part. Inside this div i would like to dynamically display pages from a book. Each page will have text, scripting code blocks, blocks with the output of each scripting code, and images. There will be a bar on the left of the webpage in which the user can select which page of the book he wants to load. For example...My web page will look something like this... Code: [Select] <HTML> <HEAD> </HEAD> <BODY> <DIV ID="PAGE-HEADER"> //LOGO OF THE WEB PAGE </DIV> <DIV ID="PAGE-MENU"> //PAGE MENU </DIV> <DIV ID="BOOK-INDEX" WITH FLOAT:LEFT > //HERE A WILL SHOW THE CHAPTERS OF THE BOOK AND THE PAGES OF EACH CHAPTER </DIV> <DIV ID="BOOK-PAGE"> //THE PAGE SELECTED FROM THE PREVIOUS 'BOOK-INDEX' MENU WILL BE DISPLAYED HERE WITH A MYSQL QUERY ****** </DIV> </BODY> </HTML> Then in a mysql database, i would like to have records with a text field, that will contain for example the followng: <h1> Chapter 1: bla bla </h1> <p> in this chapter we will speak about bla bla bla.... </p> <div id="code"> int main() { int x,y; x=2; y=3; x=x+y; } </div> <p> this will outpout the following:</p> <div id="code output"> x=5! </div> I would like to get this html code from the database, and then show it in the <div id=BOOK-PAGE> div. But i dont want to use php eval(). Also, if i store the code to a file and then include it, i will have too may files(equal to the book's number of pages etc 100). Any ideas? I have the following code that searches my database and displays results in a table: $fields = array("field1", "field2", "field3") $cols = implode (', ', $fields); $result= mysql_query (" SELECT $cols FROM tablename WHERE ................... "); if (!$result) {die('Could not search database: ' . mysql_error());} if($result) { if(mysql_num_rows($result) == 0) { return "Sorry. No records found in the database"; } else { $table = "<table border='1' cellpadding='5' cellspacing='5'>"; while($arr = mysql_fetch_array($result, MYSQL_ASSOC)) { $table .= "\t\t<tr>\n"; foreach ($arr as $val_col) { $table .= "\t\t\t".'<td>'.$val_col.'</td>'."\n"; } $table .= "\t\t</tr>\n"; } $table .= "</table>"; echo $table; } mysql_free_result($result); } As you can see each of the MySQL table fields specified by $fields is displayed in a new column in the html table. I want to change this so that e.g. "field3" is displayed in a new row instead. So, instead of the html table looking like: | "field1-result1" | "field2-result1" | "field3-result1" | | "field1-result2" | "field2-result2" | "field3-result2" | | "field1-result3" | "field2-result3" | "field3-result3" | I want it to look like: | "field1-result1" | "field2-result1" | | "field3-result1" | | | "field1-result2" | "field2-result2" | | "field3-result2" | | | "field1-result3" | "field2-result3" | | "field3-result3" | | I guess this is quite straightforward, but I can't work it out! Pls help! Thanks. I'm not sure if my problem is with the PHP, MySQL, HTML, or all of the above, but I've used a tutorial from ScriptPlayground (http://scriptplayground.com/tutorials/php/Printing-a-MySQL-table-to-a-dynamic-HTML-table-with-PHP/) to print a table to HTML. My problem isn't that I couldn't get it to work, as you can see from my site, http://dollapal.com/offerlist.php I've used the code on my site, trying to display a list of offers, with links to each, but I actually have two questions: Is there a way to EXCLUDE columns of the table? I would rather not show the 'id' or 'points' columns, since they don't offer any worth to the user. I would also need to hopefully add a dollar sign before the 'pay' quantities if possible. I would also like use the 'title' field as a hyperlink to the URL listed in the 'URL' field, instead of simply printing the URL. Are there any simple ways to do this and not have it look terrible? I have included a copy of the script that I'm working with so far. Thank you in advance! Hello I have an array with data from `mysql` that I would like to output it in a table using twig. The image is an example of want i want to achieve but without any luck. `print_r` of the array data Array ( [Administrator] => Array ( [0] => Array ( [RoleName] => Administrator [PermissionName] => Catalog-View [PermissionId] => 1 ) [1] => Array ( [RoleName] => Administrator [PermissionName] => Catalog-Edit [PermissionId] => 2 ) [2] => Array ( [RoleName] => Administrator [PermissionName] => Catalog-Delete [PermissionId] => 3 ) ) [Moderator] => Array ( [0] => Array ( [RoleName] => Moderator [PermissionName] => Catalog-View [PermissionId] => 1 ) ) ) The `HTML` code: <table> <tr> <thead> <th>Controller - Action</th> {% for permission in permissions %} {% for item in permission %} <th>{{item.RoleName}}</th> {% endfor %} {% endfor %} </thead> </tr> {% for permission in permissions %} {% for item in permission %} <tr> <td>{{item.PermissionName}}</td> <td>{{item.PermissionId}}</td> </tr> {% endfor %} {% endfor %} </table> OUTPUT: <table> <tbody> <tr></tr> </tbody> <thead> <tr> <th>Controller - Action</th> <th>Administrator</th> <th>Administrator</th> <th>Administrator</th> <th>Moderator</th> </tr> </thead> <tbody> <tr> <td>Catalog-View</td> <td>1</td> </tr> <tr> <td>Catalog-Edit</td> <td>2</td> </tr> <tr> <td>Catalog-Delete</td> <td>3</td> </tr> <tr> <td>Catalog-View</td> <td>1</td> </tr> </tbody> </table> Later Edit MySQL Query: SELECT t3.PermissionName, t1.PermissionId, t2.RoleName FROM tbl_user_role_perm AS t1 INNER JOIN tbl_user_roles AS t2 ON t1.RoleId = t2.RoleId INNER JOIN tbl_user_permissions AS t3 ON t1.PermissionId = t3.PermissionId MySQL Dump: -- Dumping structure for table tbl_user_permissions CREATE TABLE IF NOT EXISTS `tbl_user_permissions` ( `PermissionId` int(11) NOT NULL AUTO_INCREMENT, `PermissionName` varchar(50) NOT NULL, `PermissionDescription` varchar(100) NOT NULL, PRIMARY KEY (`PermissionId`) ) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=latin1; -- Dumping data for table tbl_user_permissions: ~2 rows (approximately) DELETE FROM `tbl_user_permissions`; /*!40000 ALTER TABLE `tbl_user_permissions` DISABLE KEYS */; INSERT INTO `tbl_user_permissions` (`PermissionId`, `PermissionName`, `PermissionDescription`) VALUES (1, 'Catalog->View', 'View Catalog Method'), (2, 'Catalog->Edit', 'Edit Catalog Method'), (3, 'Catalog->Delete', 'Delete Catalog Method'); /*!40000 ALTER TABLE `tbl_user_permissions` ENABLE KEYS */; -- Dumping structure for table tbl_user_role CREATE TABLE IF NOT EXISTS `tbl_user_role` ( `UserRoleId` int(10) NOT NULL AUTO_INCREMENT, `UserId` int(10) NOT NULL, `RoleId` int(10) unsigned NOT NULL, PRIMARY KEY (`UserRoleId`), KEY `FK_tbl_user_role_tbl_user_roles` (`RoleId`), KEY `UserId` (`UserId`) ) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1; -- Dumping data for table tbl_user_role: ~2 rows (approximately) DELETE FROM `tbl_user_role`; /*!40000 ALTER TABLE `tbl_user_role` DISABLE KEYS */; INSERT INTO `tbl_user_role` (`UserRoleId`, `UserId`, `RoleId`) VALUES (1, 13, 22), (2, 14, 22); /*!40000 ALTER TABLE `tbl_user_role` ENABLE KEYS */; -- Dumping structure for table tbl_user_roles CREATE TABLE IF NOT EXISTS `tbl_user_roles` ( `RoleId` int(10) unsigned NOT NULL AUTO_INCREMENT, `RoleName` varchar(50) NOT NULL, `CreatedDate` datetime NOT NULL, `ModifiedDate` datetime DEFAULT NULL ON UPDATE CURRENT_TIMESTAMP, PRIMARY KEY (`RoleId`) ) ENGINE=InnoDB AUTO_INCREMENT=29 DEFAULT CHARSET=utf8; -- Dumping data for table tbl_user_roles: ~7 rows (approximately) DELETE FROM `tbl_user_roles`; /*!40000 ALTER TABLE `tbl_user_roles` DISABLE KEYS */; INSERT INTO `tbl_user_roles` (`RoleId`, `RoleName`, `CreatedDate`, `ModifiedDate`) VALUES (22, 'Administrator', '2014-10-28 09:53:08', NULL), (23, 'Moderator', '2014-10-28 09:53:13', NULL), (24, 'Admin', '2014-10-28 12:22:05', '2014-10-28 12:22:06'), (25, 'User', '2014-10-29 15:10:36', '2014-10-29 15:10:37'), (26, 'SuperUser', '2014-10-29 15:10:45', '2014-10-29 15:10:46'), (27, 'Accountant', '2014-10-29 15:10:53', '2014-10-29 15:10:54'), (28, 'God', '2014-10-29 15:11:02', '2014-10-29 15:11:02'); /*!40000 ALTER TABLE `tbl_user_roles` ENABLE KEYS */; -- Dumping structure for table tbl_user_role_perm CREATE TABLE IF NOT EXISTS `tbl_user_role_perm` ( `RoleId` int(10) unsigned NOT NULL, `PermissionId` int(10) unsigned NOT NULL, KEY `RoleId` (`RoleId`), KEY `PermissionId` (`PermissionId`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; -- Dumping data for table tbl_user_role_perm: ~3 rows (approximately) DELETE FROM `tbl_user_role_perm`; /*!40000 ALTER TABLE `tbl_user_role_perm` DISABLE KEYS */; INSERT INTO `tbl_user_role_perm` (`RoleId`, `PermissionId`) VALUES (22, 2), (22, 1), (23, 1), (22, 3); /*!40000 ALTER TABLE `tbl_user_role_perm` ENABLE KEYS */; /*!40101 SET SQL_MODE=IFNULL(@OLD_SQL_MODE, '') */; /*!40014 SET FOREIGN_KEY_CHECKS=IF(@OLD_FOREIGN_KEY_CHECKS IS NULL, 1, @OLD_FOREIGN_KEY_CHECKS) */; /*!40101 SET CHARACTER_SET_CLIENT=@OLD_CHARACTER_SET_CLIENT */;Can you help me to make the coding required so that the correct column have the correct permissions? thank you 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! I'm trying to do something that I thought was very simple about 2 weeks ago :-( I want to put a form on my site and link it to a database so when a user types a surname into the form they can search the db and the page will only display the surnames that match the search criteria. I got the db set up using phpmyadmin in about 2 minutes flat, but keep getting error messages when I write the php. I'm currently working with the below script, which keeps giving me the error 'unexpected T_string' on line 176 I've searched every forum and help site I can find, and the mysql and php manuals are just mind-boggling. I'm sure I'm making some really amateur mistake, but I'd really appreciate help with this! thanks all! <html> <head> <title>SEARCH RECORDS</title> </head> <body> <FORM NAME ="Search" METHOD ="POST" ACTION = "test3"> <INPUT TYPE = "TEXT" VALUE ="surname" NAME = "surname"> <INPUT TYPE = "Submit" Name = "Search" VALUE = "Search"> </FORM> </body> </html> <? $user_name = "*****"; $password = "*****"; $database = "*****"; $server = "localhost"; $db_handle = mysql_connect($server, $user_name, $password); $db_found = mysql_select_db($database, $db_handle); if ($db_found) { $SQL = "SELECT * FROM *****" WHERE surname=$_POST['surname']; $result = mysql_query($SQL); while ($db_field = mysql_fetch_assoc($result)) { print $db_field['Grave'] . "<BR>"; print $db_field['Surname'] . "<BR>"; print $db_field['Forenames'] . "<BR>"; print $db_field['Death_Date'] . "<BR>"; print $db_field['Birth_Year'] . "<BR>"; } mysql_close($db_handle); } else { print "Database NOT Found "; mysql_close($db_handle); } ?> 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 =) Im inserting HTML into a database, and then outputting it on a PHP page. Its an iframe code, so when I output it, it shows the iframe. I need it to just display the HTML code. How can I do this? I thought it would be something simple but I can find anyway to do it. Help very appreciated!! Thanks I have a mysql table which will store users email addresses (each is unique and is the primary field) and a timestamp. I have added another column called `'unique_code' (varchar(64), utf8_unicode_ci)`. What I would very much appreciate assistance with is; a) Generating a 5 digit alphanumeric code, ie: 5ABH6 b) Check all rows the 'unique_code' column to ensure it is unique, otherwise re-generate and check again c) Insert the uniquely generated 5 digit alphanumeric code into `'unique_code'` column, corresponding to the email address just entered. d) display the code on screen. What code must I put and where? **My current php is as follows:** Code: [Select] require "includes/connect.php"; $msg = ''; if($_POST['email']){ // Requested with AJAX: $ajax = ($_SERVER['HTTP_X_REQUESTED_WITH'] == 'XMLHttpRequest'); try{ if(!filter_input(INPUT_POST,'email',FILTER_VALIDATE_EMAIL)){ throw new Exception('Invalid Email!'); } $mysqli->query("INSERT INTO coming_soon_emails SET email='".$mysqli->real_escape_string($_POST['email'])."'"); if($mysqli->affected_rows != 1){ throw new Exception('You are already on the notification list.'); } if($ajax){ die('{"status":1}'); } $msg = "Thank you!"; } catch (Exception $e){ if($ajax){ die(json_encode(array('error'=>$e->getMessage()))); } $msg = $e->getMessage(); } } |