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PHP - Side By Side Tables
I have a few tables in php and want them to be side by side instead of under each other like below. I have spent about 2 days trying to figure it out, Anyone got a clue?
This is the code im using <?php $con = mysql_connect("localhost","user","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("People", $con); $result = mysql_query("SELECT * FROM wang"); echo "<table border='1'> <tr> <th>Wang Total Report</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['got'] . "</td>"; echo "</tr>"; } echo "</table>"; $result2 = mysql_query("SELECT * FROM miah"); echo "<table border='1'> <tr> <th>Miah Total Report</th> </tr>"; while($row = mysql_fetch_array($result2)) { echo "<tr>"; echo "<td>" . $row['got'] . "</td>"; echo "</tr>"; } echo "</table>"; $result3 = mysql_query("SELECT * FROM vinc"); echo "<table border='1'> <tr> <th>Vinc Total Report</th> </tr>"; while($row = mysql_fetch_array($result3)) { echo "<tr>"; echo "<td>" . $row['got'] . "</td>"; echo "</tr>"; } echo "</table>"; $result4 = mysql_query("SELECT * FROM ketarie"); echo "<table border='1'> <tr> <th>Ketarie Report</th> </tr>"; while($row = mysql_fetch_array($result4)) { echo "<tr>"; echo "<td>" . $row['got'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Similar TutorialsI have a web portal which is scripted using client side scripting language i.e HTML. I want to create a login page in .php and want the other pages to be in .hmtl. Can i do so? I want to do this because i want the login password to be stored in a database. Also I want the password to get encrypted (by either MD5 javascript or OpenSSL ) when user enters it inside the login form. I cannot create the entire portal in php because my portal makes use of C code. And php doesn't have an interface with C language. Can a user get directed to a .html page after a secure login from login.php page ? My main aim is to secure the access to my web portal using a password. I tried to authenticate the login using javascript where the password was stored in an array. But i feel any one having moderate knowledge can easily break that password. Any help would be greatly appreciated!! I’m trying to write a post-panel where the user can see the preview of his post to the right of the text area he’s writing into. I tried the following into a file called writepost.php : <?php $text=(isset($_GET['message']))?$_GET['message']:''; $formatted_text=nl2br(stripslashes(htmlspecialchars($text))); echo '<form method="post" action="proceedtopost.php?>'. '<table>'. '<td style width="50%"><tr>'. '<fieldset> Write your post here : <br> <textarea cols="50" rows="12" '. 'id="message" name="message">'.$text.'</textarea>'. '</fieldset> <p> '. '<input type="submit" name="submit" formaction="writepost.php" value="Preview" /> '. '<input type="submit" name="submit" value="Sent" /> '. '</p>'. '</td>'. '<td>'. '<fieldset> Your post will appear as follows :<br><p> '. $formatted_text. '</fieldset>'. '</td></tr>'. '</table>'. '</form>';There are several things wrong with this code : 1) When the user hits the "Preview" button, I expect writepost.php to be reloaded (this is what happens), and the current content of the textarea to be stored in $_GET['message'] (this is not what happens). 2) Why does my browser output the preview part under the text area (or in other words outputs the HTML table as a single column of two cells), when I insist in my HTML code for the table to be displayed as a single row ? need a little help guys! I use the script below to display profile images, trouble is it shows 1 on top of the other, and i need it to double up 2 profile images on top of 2 profile images ect any ideas how i can do this. require("./include/mysqldb.php"); $con = mysql_connect("$dbhost","$dbuser","$dbpass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("$dbame", $con); $result = mysql_query("SELECT * FROM Search_profiles_up WHERE upgrade_one ='1' ORDER BY RAND() LIMIT 40"); print "<table width=\"293\" height=\"111\" border=\"0\"> <tr>\n"; while($row = mysql_fetch_array($result)) { print "<td width=\"142\"><img src=" . $row['search_small_image'] . " width=\"144\" height=\"169\" /></td>\n"; print " </tr>\n"; print " <tr> \n"; print "<td>" . $row['star'] . "</td>\n"; print " </tr>\n"; print " <tr>\n"; print "<td>" . $row['username_search'] . "</td>\n"; print " </tr>\n"; print " <tr> \n"; print "<td>" . $row['phone_search'] . "</td>\n"; print " </tr> \n"; } print "</table>"; mysql_close($con); ?> is it true that all the server side code runs before the client side code when accessing a webpage? if that is true, what about PHP code that is embedded inside HTML tags? wouldn't that result in an error a lot of times? Hi guys, I am wanting to do the same as Blogger.com (aka Blogspot) and Tumblr. And allow people to set their own domain names for "pages" or blog sites. So if I was to tell them to set their own CNAME (on domain) to domains.mydomain.com What would be the PHP part to make this work? Or how would I make this work if its not PHP? Like Blogger and Tumblr it would not be a redirect (to view) (unless I am wrong about this). Any and all help is much appreciated. (PS - Am running Apache/PHP and MySQL on dedicated server). This part of the question is based off of php so I'll put it here of course. The idea behind this is jquery/ajax is going to get passed an id (eventid) from a dropdown select box inside a form and it's going to run a query to find all the events in my database with that same id. After running the query it's going to list them out and find the highest number of another field (label) and its going to add one to its value and then what I would like for it to is send it back to the form in the jquery/ajax and and place that label inside of the existing input text field in the form. So I'm asking is if my code is presented and is doing what I"m attempting it to do on the php side and if so if I can take out the print tag of the input text field. $eventid = $_GET['eventid']; $result = mysqli_query("SELECT * FROM `events` WHERE `event_id` = '$eventid'"); $list = mysqli_num_rows($result); $label = $list + 1; print '<input type="text" name="label" class="text" readonly="readonly" value="' . $list . '" />'; Hi guys, I'm using this EOD thingi in order to print a table and get it PDF. I am using TCPDF and following their examples to get the PDF output. I need to print a php variable's value within the EOD. Can you please help me out. Below is the piece of code I'm running. Code: [Select] $tbl = <<<EOD <table cellspacing="0" cellpadding="1" border="1"> <tr> <td rowspan="4"><img src="images/m24.jpg"></td> <td align="right" width="150">From the desk of</td> <td>: $contat</td> </tr> <tr> <td align="right">E-Mail</td> <td>: $email</td> </tr> <tr> <td align="right">Telephone</td> <td>: $telephone</td> </tr> <tr> <td align="right">Mobile</td> <td>: $mobile</td> </tr> </table> EOD; the $variables are simply ignored. or even if I echo, there is no PHP value. Can you guys please shed some light on this issue. Pretty pleeeaaassseee I'm in the habit of verifying input client side. For example, there is a field to enter a number, and I ensure that it is within a particular range using Javascript. My backend PHP code does not do such checking... Is this bad? Should I be doing checking using my backend code and send a response back to the client to display a message? Admittedly, I avoid it, since it's a little extra work -- well more extra than just javascripts, IF-THEN-ALERT type statement. Hello guys. I was thinking on posting this under the mysql thread. but i think this can be solved by php.. anyway, i want to dump mysql data on the client side whenever a user logs out. how can i do this? thanks in advance. I am doing server side validations using php and jquery.Its validating properly.But problem is the erros messages are displaying on the top of the page.I want to display side of field.All of the validations store in array.How to display errors to side of the field. I'm trying to learn how to validate a form using the php server side method. Everything I search is very different and old. Anyone have knowledge of a good tutorial or source that I could use? None of my books have anything about it either! Thanks! I have a text box and a submit button. I want that when the submit button is pressed, the value on the text box will be checked if it matches the values that are on my server before proceeding, and if they don't match it should give an error report and stop. All I can make myself is the text box and the submit button, I have tried for hours to find a way to do the rest, but all I can find on the internet are vague explanations and information that a novice like me cannot use. It would be a shame to let me web site that I have worked hard on to go to waste because I cannot do this myself Thank you in advance hello, since php is run server side, can i launch a program (.exe) on the server from a link? I'm sorry if this is a silly question. When user logins to some other site, he gets a member page. Without login he gets a guest page. I want to write a code that takes member page and parses it, then shows user some parts of it user will select which parts to see. but when i use "file_get_contents", output becomes a guest page, i suppose it is because my php server takes it without user's login data. Is it possible to take member page and parse it ? Because i don't want user to give his login information to my site. Thank you very much for your help. I hope I'm posting in the correct forum. If not, I'm super sorry! Anyways, I need help with a website I'm working on. We have been asked to redesign our "Apply Online" page. My supervisor has asked that I find the correct code to make an upload button that will allow users to upload their resumes to our server, and send a copy to the specific branch they indicate (we have 17 branches). Could any of you point me in the correct direction for this code? I've seen several sites for Uploads to servers, but I'm worried this isn't exactly what we are looking for. We have a website that allows the user to upload his/her picture so we use the ftp_put() command. I use Code: [Select] <?php $path = getcwd(); echo $path; ?> This is the output: Code: [Select] /home/vol10/000a.biz/a000b_7363341/htdocs/gankgame to determine the absolute path where my website is stored. My website "CAN" connect to the ftp server but it can't upload files. Uploading always fail. I can't determine the right path. "My code is attached" Hi, I have this script which does what it is meant to do. I have assigned session_id() to the variable $sid so I can use it when logging in to a users account, however I get this error. How do I get around this? Warning: Unknown: Your script possibly relies on a session side-effect which existed until PHP 4.2.3. Please be advised that the session extension does not consider global variables as a source of data, unless register_globals is enabled. You can disable this functionality and this warning by setting session.bug_compat_42 or session.bug_compat_warn to off, respectively. in Unknown on line 0 As this is going to be installed on many servers, I do not want to have to edit the php.ini file on each one so as not to get this error. Is there an alternative way to assign the session_id() as so $sid = session_id(); so this doesnt happen? Code: [Select] <div id="pageNav"> <div id="sectionLinks"> <a href="admin.php?cmd=manage1&username=admin">Manage</a> <a href="admin.php?cmd=dashboard&username=admin">Dashboard</a> <a href="admin.php?cmd=msgcenter&username=admin">Message Center </a> <a href="admin.php?cmd=manage&username=logins">Logins</a> </div> </div> <div id="content"> <div class="page"> <table width="100%" border="1" align="center"> <td bgcolor="#99FF66"><div align="center"><span class="style3">Login</span></div></td> <td bgcolor="#99FF66"><div align="center"><span class="style3">Name</span></div></td> <td bgcolor="#99FF66"><div align="center"><span class="style3">Registration Date </span></div></td> <td bgcolor="#99FF66"><div align="center"><span class="style3">Approved </span></div></td> <td bgcolor="#99FF66"><div align="center"><span class="style3">Reset Password </span></div></td> <td bgcolor="#99FF66"><div align="center"><span class="style3">Delete</span></div></td> </tr> <?php session_start(); $sid = session_id(); session_register('sid'); include_once("data/mysql.php"); $mysqlPassword = (base64_decode($mysqlpword)); $db = mysql_connect("$localhost", "$mysqlusername", "$mysqlPassword") or die ("Error connecting to database"); mysql_select_db("$dbname", $db) or die ("An error occured when connecting to database"); $result = mysql_query("SELECT * FROM members"); while($row = mysql_fetch_assoc($result)){ echo "<tr><td><a href=templates/members/home.php?username=".$row['username']."&sid=$sid>".$row['username']."</a></td> <td>".$row['firstname']." ".$row['lastname']."</td> <td>".$row['registration_date']."</td> <td><a href=admin.php?cmd=approval&username=".$row['username']."&approved=".$row['approved'].">".$row['approved']."</a></td> <td><a href=templates/members/changepw.php?username=".$row['username']."&sid=$sid>Change Password</a></td> <td><a href=templates/members/delete.php?username=".$row['username']."&sid=$sid>Delete</a></td></tr>"; } ?> </div> </div> </div> <tr> <td> </td> <td> </td> <td> </td> <td> </td> <td> </td> <td> </td> </tr> </table> </div> </div> </div> Many Thanks Paul Basically what I am trying to achieve is making sure any $page called originates from my server, but if no $page is defined, to include a default page I have set up. This is my code, and I cannot figure out what's wrong with it. Code: [Select] <?php $path = 'pages/'; $extension = '.txt'; if ( preg_match("#^[a-z0-9_]+$#i",$page) ){ $filename = $path.$page.$extension; include($filename); } if (!$page) include 'pages/main.txt'; ?> An example URL I am using is www.mywebsite.com/index.php?$page=pages/tester What am I doing wrong? |
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