netpumber

Hallo ! I have 2 files one called admin.php and the other edit.php So.. in admin file i have this html code : <a href="edit.php" id="hmtext" class="nav">Home Text</a> and in edit.php this code : <?php ini_set ('display_errors',1); error_reporting (E_ALL); include 'dbConf.php'; $connect = mysql_connect($host,$username,$password) or die("Error connecting to Database! " . mysql_error()); mysql_select_db($database , $connect) or die("Cannot select database! " . mysql_error()); $query = "SELECT title,text FROM fronttext"; $q = mysql_query($query); $row = mysql_fetch_array($q); print " <div id=\"frontTexForm\"> <form method= \"post\" action=\"homeTextForm.php\"> <label><font color=\"#99FF00\" >HOME TEXT</font></label> <br /><br /><br /> <label><font color=\"#99FF00\" >Title : </font></label> <br /> <input type=\"text\" size=\"70\" maxlength=\"100\" name=\"title\" value=\"".htmlentities($row['title'])."\" /> <br /><br /> <label><font color=\"#99FF00\" >Text : </font></label> <br /> <textarea cols=\"53\" rows=\"15\" name=\"text\">".htmlentities($row['text'])."</textarea> <br/><br/> <input type=\"submit\" value=\"Αποθήκευση\" name=\"save\"> </form> </div> "; mysql_close(); ?> So i click the link from admin.php and then the browser returns to me this error : Why this happens ? Have you any idea ? The query was executed but i can't print out the results... Thanks a lot!

You mean e.g in a table , do not use width height variables ?

Hi!! I know, my question is very noobish and so am i in web design but if you can... tell me your idea.. I realize that if i see my site from different monitors / different computers / different browsers its size (width mostly ) changes... Is there any way to make it look same for every different computer ? Thanks a lot...

hi!! i have wrote this script.. as a port scanner: <?php ini_set("display_errors", 1); $host = "www.site.com"; for($i=50;$i<6669;$i++) { $fp = @fsockopen($host,$i,$errno,$errstr,10); if($fp) { echo "port " . $i . " open on " . $host . "\n"; @fclose($fp); } flush(); } //end for ?> It scans just for open ports in the hostname. I run it through console and it prints results but to slow.. Also if i runt through web browser doesnt seems to work.. Have you any idea on how to make it faster and work through browser ? Thanks for any of your answer

Any other idea ?

I don't think that this works.. I ll put it in the fuction under window.open(page); but nothing happened..

Ok.. I have one more question. Is it possible to make this script when it is opens the new tab close the old one ? e.g Lets say that we have these sites: http://www.google.com http://www.bing.com http://www.yahoo.com First it opens the google and after some time it opens the bing. I want when it opens the bing tab close the tab with google..and so on.. when it opens the yahoo tab close the bings tab. Is this possible or not ? Thanks in advanced..

hmm i follow finally this tutorial.. and it works Great!! Thanks all of you !! http://www.phpfreaks.com/tutorial/basic-pagination

You had an error here : array_push($pagination, "<a href=\"".$_SERVER['PHP_SELF']."?page=$pagno\">$pageno</a>"); with variable pagno and pageno They are different but it must be the same so now it works little better. The prob is that in the first page doesnt prints the numbers of the other pages.. And in the third page as i said before it prints links with pages 1 and 0 . when you visit 1 it works when you visit 0 it returns the above error.. Actually i think that 0 occurs because of this : $start = ($page - 1) * $perpage; Actually here we multiply (1 -1) * 3 = 0 * 3 = 0 This is my opinion. So something must be change here Mabe the $page = isset($_GET['page']) ? $_GET['page'] : 1; So the $page not to start from 1 .. I dont know

hmm that so so works... In main page it prints the first 3 entries but it doesnt have links for the second and third page. If i add this in url ?page= with 2 or 3 in variable page it sows me the next entries..and then appeared links 1 and 0 and when i visit them returns me this error : Could not retrive the data brcause:mysql_error The query was $query. Anyway.. this is a start...

Yes i found some tutorials but i want also and community's opinion.. Thanks anyway i ll try it out..

Halo my friends! Im little new in php coding and so i need your useful help I have this php code that prints data from mysql.. <?php //[sTART] Retrieving DATA from MySql ini_set ('display_errors',1); error_reporting (E_ALL & ~E_NOTICE); if ($dbc = @mysql_connect ('localhost','user','pass')){ if (!@mysql_select_db ('web_site')){ die('<p>Could not select the database because:<b>'. mysql_error .'</b></p>'); } }else{ die('<p>Could not connect to MYSQL because:<b>' . mysql_error() . '</b></p>'); } $query = "SELECT * FROM site_entries ORDER BY date_entered DESC"; if ($r = mysql_query($query)){ while ($row = mysql_fetch_array($r)){ print "<tr> <td valign=\"top\"> <table width=\"100%\" border=\"0\" cellspacing=\"0\" cellpadding=\"15\" bgcolor=\"#f2f2f2\"> <tr><td> <h3>{$row['title']}</h3><br> {$row['entry']} </td></tr> </table> </td> </tr>"; } }else{ .....blah blah blah.... So.. i want this script print in the first page only 3 entries and the others in other pages.. Have you any idea on how to achieve this ? Thanks in advanced!

Oh!! Thanks a lot my friend..

So.. i said to try first without setTimeout and run this but didnt worked too var myArray = [ 'http://www.google.com', http://www.bing.com', 'http://www.yahoo.com']; for(var i=0;i<=3;i++) { window.open('+myArray[i]+'); }

Is this : <script type='text/javascript'> var myArray = [ 'http://www.google.com', http://www.bing.com', 'http://www.yahoo.gom']; for(var i=0;i<=3;i++) { setTimeout("openPage('+myArray+')",60000); } function openPage(page) { window.open(page); } </script> And in for loop the myArray.length is just a number as i put it here or something like this myArray.3 ? I run the above script but didnt worked..:s

Thanks for your answer my friend optikalefx but i am little new in javascript coding.. So here is what i have : <script type='text/javascript'> var counter = 0; var myArray = [ 'http://www.google.com', http://www.bing.com', 'http://www.yahoo.gom']; //function creation function openPage() { window.open(myArray[counter]); counter ++; } </script> I didn understand what exactly to put in the for loop and that you say "then use javascript to setTimeout('openPage()',60000);" Can you be more specific to understand better ? Thanks a lot my friend and sorry for my english !

Halo out there! So listen to my story.. I want to make a web page that will do these simple things . It only for personal use so i have to tell you that i use Firefox if that info is useful. Ok.. lets continue. It wil have an array like below var myArray = [ 'http://www.google.com', http://www.bing.com', 'http://www.yahoo.gom']; now i want these sites that are in array page open them automatically in a new tabs after one minute each of them. E.g when i open this page with firefox , after a minute a new tab will be created and http://www.google.com loaded . After one minute do the same with bing.com and so on... This project as i can imagine needs a timer but for now i need your help on how to make it open in new tabs the array's items.. Thanks for any answer!

And have you any idea on how to make it ?

Hi!! I have this css code.. #menu { background: #333; list-style: none; margin: 0; padding: 0; width: 12em; } #menu li { font: 67.5% "Lucida Sans Unicode", "Bitstream Vera Sans", "Trebuchet Unicode MS", "Lucida Grande", Verdana, Helvetica, sans-serif; margin: 0; padding: 0; } #menu a { background: #333; border-bottom: 1px solid #393939; color: #ccc; display: block; margin: 0; padding: 8px 12px; text-decoration: none; font-weight:normal; } #menu a:hover { background: #2580a2 url("images/hover.gif") left center no-repeat; color: #fff; padding-bottom: 8px; } #menur { background: #333; list-style: none; margin: 0; padding: 0; width: 6em; } #menur li { font: 67.5% "Lucida Sans Unicode", "Bitstream Vera Sans", "Trebuchet Unicode MS", "Lucida Grande", Verdana, Helvetica, sans-serif; margin: 0; padding: 0; } #menur a { background: #333; border-bottom: 1px solid #393939; color: #ccc; display: block; margin: 0; padding: 8px 12px; text-decoration: none; font-weight:normal; } #menur a:hover { background: #2580a2 url("images/hover.gif") left center no-repeat; color: #fff; padding-bottom: 8px; } And here is my html code with the menu: <ul id="menu"> <li><a href="" target="_self">Main Menu</a> <ul> <li><a href="#">Subitem 1</a></li> <li><a href="#">Subitem 2</a></li> <li><a href="#">Subitem 3</a></li> </ul> </li> </ul> It is vertical menu and its like here in screen : Main Menu - Subitem 1 - Subitem 2 - Subitem 3 So.. my question is if i can make subitems to appear when you click on the Main Menu.. Thanks in advanced!

yes i know. the server that will run this script has the hdd that i share.. So these two things are in the same pc.

Halo.. So this is what i want to create. I have a folder that i share and it has many movies.. So i will setup a local web browser and i want to make a .php file that after you login it will saw you the files that are in this share(movies names). How i can make php print out a list of files in that specific directory ? Thanks..!

Thanks a lot at seanlim...His help me a lot...!!

One more question... Here is the textarea code... <textarea style=\"background:#B0D2D7\" rows=\"10\" name=\"entry\" cols=\"48\" >".htmlentities($row['entry'])."</textarea> Lets say i add this in the edit form : This will create a link but its not correct and this is how it likes in the source code..: <a href=\"http://www.google.com\">Google</a>. Why this occurs ?

Hmm thanks a lot my friend... I really mean it http://validator.w3.org helped me alot I forgot to close a form that uses get method

I don't understand how this one has something wrong and not the other one...when i have copy and paste the same code... Its crazy dude..