andyy121

I want to create a single login page for Admin and user.When admin lo it should go to seperate page and when user log it shoult go to seperate page.I want sample code for this.Thanks..

the password in database doesnt changet when i press change button in the page i show me the pass is change but when i log in with the new pass it show incorrect pass help please <?php session_start (); $user = @$_SESSION['username']; if ($user) { //user is logged in if …

i got following error in my php program error: Parse error: syntax error, unexpected $end in D:\xampp\htdocs\php\project\sample\upload.php on line 85 <?php /** * Upload an image to mysql database. * * * * @author Mr. Vivek Chaurasia * @copyright www.techshri.com * @license LGPL */ // Check for post data. if …

im traying to create a contact form (email). im using xampp. so i need something like HmailServer or something like this to send email to myself. can everybody told me how i can configure HmailServer.

hi all i cant see my erron in the code i have followed the tutorial but i cant see it i wont some help. here is my code. $to = 'andi_138el@hotmail.com'; $subject = 'Contact from submitted'; $body = $contact_name."\n".$contact_text; $headers = 'From: '.$contact_email; if (mail($to,$subject, $body, $headers)){ echo 'Thanks for …

i wont to know why we have to equal this variabel what is theri meaning $contact_name = $_POST['contact_name']; $contact_email = $_POST['contact_email']; $contact_text = $_POST['contact_text'];

i have wrote some html and php my form doesnt work so fine i cant find my error if somebody doesnt mind here is the code.

<form action="index.php" method="post" enctype="multipart/form-data"> File: <input type="file" name="image"><input type="submit" value="Upload"> </form> <?php //connect to db mysql_connect("localhost","root","")or die(mysql_error()); mysql_select_db ("databaseimage")or die (mysql_error()); @$file =$_FILES['image']['tmp_name'];//kjo eshte e lidhur me <input type="file" name="image"> if (!isset($file)) echo ("Please select an image"); else { $image= addslashes(file_get_contents ($_FILES['image'] ['tmp_name'])); $image_name = addslashes($_FILES['image']['name']); $image_size = getimagesize ($_FILES['image'] …

<form action="index.php" method="post" enctype="multipart/form-data"> File: <input type="file" name="image"><input type="submit" value="Upload"> </form> <?php //connect to db mysql_connect("localhost","root","")or die(mysql_error()); mysql_select_db ("databaseimage")or die (mysql_error()); @$file =$_FILES['image']['tmp_name'];//kjo eshte e lidhur me <input type="file" name="image"> if (!isset($file)) echo ("Please select an image"); else { $image= addslashes(file_get_contents ($_FILES['image'] ['tmp_name'])); $image_name = addslashes($_FILES['image']['name']); $image_size = getimagesize ($_FILES['image'] …

$construct .="keywords LIKE '%$search_each%'"; $run = mysql_query($construct); $foundnum = mysql_num_rows($run); i wont to know what display $construct?? i wont to know what is the function of mysql_query and what mean this ($construct); i wont to know what is the function of mysql_num_rows what can it do

<?php if (isset($_POST['generate'])) { $charset = 'absdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWYZ0123456789%$#-()*&.,'; $generated_password = substr (str_shufle($charset),0, 12); } ?> <form action ="password generator" method="post"> <input type="submit" name="generator" value="Generate"><input type="text" value="<?php if (isset($generated_password)) {echo $generated_password; } ?>"/> </form> when i click in generate button it show me this: Object not found! The requested URL was not …

foreach($search_exploded as $search_each) { @$x++; if($x==1) @$construct .="keywords LIKE '%$search_each%'"; else $construct .="AND keywords LIKE '%$search_each%'"; } i just wont to understand this foreach ($search_exploded as $search_each) what is the meaning

$name = $_FILES['myfile'] ['name']; $tmp_name = $_FILES ['myfile'] ['tmp_name']; i wont to know why the variable $name, is begin with name can somebody explain me this variables and parameters.

when i search something in the database it doesn't show me what is in the databese how to connect into it or it doesnt show what is in it????

i have create a html page with log in,log out,register ect... i wont to create a admin page for this,for example if i wont to change images colors border color add image or something else can somebody suggest me what i have to do,give me some ideas.

<?php require_once("start.php"); require_once("myconfig.php"); require_once("db.php"); require_once("kontrollo.php"); require_once("mylib.php"); require_once("datavalidation.php"); if (!isset($_REQUEST["id"])) { exit("which category??"); } $category=$_REQUEST["id"]; ?> <html> <head> <title>GarantSig Artikull i ri</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> <link href="../css/travel.css" rel="stylesheet" type="text/css"> <link href="../css/style.css" rel="stylesheet" type="text/css"> <script language="JavaScript"> function validate() { if(document.myform.title.value==""){alert('Verini nje titull');return false;} if(document.myform.subtitle.value==""){alert('Verini nje Nentitull');return false;} if(document.myform.rendi.value==""){ alert('Verini nje rendesi');return …

i m traying this simples code but nothing is displayed what type of problem showld be. <?php function yourname($name) { echo "Your name is" .$name; } yourname("andi"); ?> another code <?php $age= 18; if ($age>=21) { echo 'alan'; }else{ echo 'bob'; } ?>

hi everybody i just need to know what is the function of preg_split and that which is inside of brackets: $keywords = preg_split ('/[\s]+/',$keywords);

if($username&&$password) { $connect = mysql_connect("localhost","root","") or ("Couldn't connect!"); mysql_select_db("phplogin") or die ("Couldn't find db"); $query = mysql_query("Select * FROM users WHERE username='$username'"); $numrow= mysql_num_rows($query); i just wont to know the IF statement when it stars what is his function in the code.

Just noticed that I was no longer logged in in IE9 after a period of doing nothing, but having the website open? Did something change? I don't store cookies or history or whatever, but this never happened before.

this is the index page <?php if (isset ($_POST['keywords'])){ $keywords =mysql_real_escape_string(htmlspecialchars(trim($_POST['keywords']))); $keywords =$_POST['keywords']; $errors = array (); if (empty ($keywords)){ $errors[] = ('Please enter a serch terms'); }else if (strlen($keywords)<3){ $errors[] = ('Your search terms must be three or more characters'); }else if (search_results ($keywords) == false){ $errors []= ('Your …

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <?php include 'func.inc.php'; ?> <title>SEARCH</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> </head> <body> <h2>SEARCH</h2> <form action="" method="post"> <p> <input type="text" name="keywords"/> <input type="submit" value="Search"/> </p> </form> <?php if (isset ($_POST['keywords'])){ $suffix =""; $keywords =mysql_real_escape_string(htmlspecialchars(trim($_POST['keywords']))); $keywords =$_POST['keywords']; $errors = array (); if (empty …

when i click in the search button it must display whatever i type in

hi i wont to change the activatet value of user log in in database example:when i log in in the page the value must change from 0 to 1 any idea

Warning: include(funck.inc.php): failed to open stream: No such file or directory in C:\xampp\htdocs\search\index.php on line 2 Warning: include(): Failed opening 'funck.inc.php' for inclusion (include_path='.;C:\xampp\php\PEAR') in C:\xampp\htdocs\search\index.php on line 2 help plz what i have to change

it show me this when i click the previw button to show the page how i can fix this? Warning: include(funck.inc.php): failed to open stream: No such file or directory in C:\xampp\htdocs\search\index.php on line 1 Warning: include(): Failed opening 'funck.inc.php' for inclusion (include_path='.;C:\xampp\php\PEAR') in C:\xampp\htdocs\search\index.php on line 1

in database i have added the random and activated with boolean type and the defaul was set to As defined with 0 value.when i click log in it must desplay this: your accoun si not activate.Please check your email. but i show you are in click here to click here …

i have added the $random in line 53 and '$random' ,'0' in 57 line inside the (),in the database i have added random and activated with boolean value which is 0 or 1,and when i click the previu/debug in browser button it show me this: Warning: mysql_num_rows() expects parameter 1 …

I am new in php and I have created a php login page where it will retrive data from a table and check a print "login successful" or "login failed". Now after execution of the codeI am giving correct data but all the time it is replying "login failed" plese …

<form action="login.php" method="post"> Username: <input type='text' name ='username'><br> Password: <input type="password" name='password'><br> <input type="submit" value="Log in"> </form> </body> </html>