galafura
-
Posts
18 -
Joined
-
Last visited
Never
galafura's Achievements
Newbie (1/5)
0
Reputation
-
Script to call a PHP code from an HTML programm
galafura replied to galafura's topic in PHP Coding Help
Yes ... please ... because I don't know how to do it !!! Tks -
Hi, I need a script in order to allow me run a PHP programm (default.php) within a HTML page (index.html). Thanks very much ...
-
Hi ... Thanks, So are you telling me that my /webroot is the directory C:\Program Files\Apache Software Foundation\Apache2.2\htdocs ?????
-
Hi, How can I know which is my webroot. I'm working in a PC with Windows 7, Apache 2.2, PHP 5 and MySQL 5.0 I want to install cakePHP in my webroot ... but I'm having some troubles to do it. Thanks
-
Yes ... I have read, but I'm not getting the results I was expected ... I think there is something missing in my PC !!!!!! Thanks
-
Hi, Can someone tell me how to install CakePHP in my machine. My Pc has installed the W 7, Apache 2.2 (C:\Program Files\Apache Software Foundation\Apache2.2), MySQL 5.0 (C:\Program Files\MySQL\MySQL Server 5.0) and PHP 5.3.0 (C:\Program Files\EasyPHP5.3.0). Thanks ....
-
Apache error using PHPAdmin
galafura replied to galafura's topic in PHP Installation and Configuration
:'( Tks ... but that I already know !!! How can I see what are the applications that are running and using that port number - 80 ????? Thanks -
:'( Hi, I have just installed PHPAdmin 5.3.0 ... and when I start PHPAdmin (apache2.jpg), MySQL starts and gives no error, but the Apache doesn't start and gives the error (Apache.jpg) What should I do to put Apache running. Thanks [attachment deleted by admin]
-
Hi, I'm running this script in my PC: <? //========================================= // Enter on the system //========================================= include "conecta.php"; //Connect to the database MySQL if(($name_login) AND ($pass_login)) { $sql = mysql_query("SELECT * FROM tb_user WHERE apelido='$name_login' AND senha='$pass_login'") or die("SQL error :".mysql_error()); if(mysql_num_rows($sql) > 0) { $id_imasters = mysql_result($sql,0,"id_user"); $apelido_imasters = mysql_result($sql,0,"apelido"); $senha_imasters = mysql_result($sql,0,"senha"); session_start("imasters"); session_register("id_imasters","apelido_imasters","senha_imasters"); header("Location:options.php"); } } ?> <html> <head> <title>iMasters Test Application</title> </head> <body bgcolor="#FFFFFF" text="#000000"> <form name="frm_login" method="post" action="<?echo $PHP_SELF;?>"> <table width="40%" border="0" cellspacing="0" cellpadding="0"> <tr> <td colspan="2"><b><font face="Arial" size="3">IDENTIFICATION</font></b></td> </tr> <tr> <td width="33%" height="25"><font face="Arial" size="2">Name:</font></td> <td width="67%" height="25"><font face="Arial" size="2"> <input type="text" name="name_login"></font></td> </tr> <tr> <td width="33%" height="25"><font face="Arial" size="2">Password:</font></td> <td width="67%" height="25"><font face="Arial" size="2"><input type="password" name="pass_login"></font></td> </tr> <tr> <td colspan="2"><input type="submit" name="enter" value="Enter >>"></td> </tr> </table> </form> </body> </html> And I'm geting the attached picture ... Why is not running as expected ... Tks. [attachment deleted by admin]
-
Hi, Tks, I have installed: Apache 2.2.3 MySQL 5.0.45 PHP 5.3.1 I still get the same error ... Help me please.
-
Hi, How can Icheck my PH intalled version ? Tks.
-
Hi I'm running this php file in my PC: <?php $link = mysql_connect('localhost', 'root', 'sa'); if (!$link) { die('Could not connect: ' . mysql_error()); } echo 'Connected successfully'; mysql_close($link); ?> But I'm getting this error: mysql_connect(): Client does not support authentication protocol requested by server. How can I solve this issue. Tks
-
Can you send me an example ... if you have ... tks
-
Hi again ... See the picture in attach. Tks [attachment deleted by admin]
-
Ok ... The problem is like this: On my web site I have several text fields where the user enters the name of several items related with personal information. He can enter e.g his address in a text box with 40 char, but if he wants enter more information for the address, he will have a button (+ sign) when he can clicks on it and another same text box will be open to him ... and he will be able to enter another address ... Thanks