andyy121

[off-topic hijack removed]

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 …

<?php $submit = @$_POST['submit']; //form data $fullname = strip_tags (@$_POST['fullname']); $username = strip_tags (@$_POST['username']); $password= strip_tags(@$_POST['password']); $repeatpassword = strip_tags(@$_POST['repeatpassword']); $date =("Y-m-d"); if ($submit); { // check for existance if($fullname&&$username&&$password&&$repeatpassword) { //encrypt password $password = md5 ($password); $repeatpassword =md5 ($repeatpassword); if ($password==$repeatpassword) { // check char length of username and fullname …

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 = '[email protected]'; $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.

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 …

<?php $submit = @$_POST['submit']; //form data $fullname = strip_tags (@$_POST['fullname']); $username = strip_tags (@$_POST['username']); $password= strip_tags(@$_POST['password']); $repeatpassword = strip_tags(@$_POST['repeatpassword']); $date =date("Y-m-d"); if ($submit) { // check for existance if($fullname&&$username&&$password&&$repeatpassword) { if($password==$repeatpassword) { //chechk char length of username and fullname if (strlen($username)>25||strlen ($fullname)>25) { echo ("Length of username or fullname is …

<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>

<?php echo "<h1>Register</h1>"; $submit = @$_POST['submit']; $fullname = strip_tags (@$_POST['fullname']); $username = strip_tags (@$_POST["username"]); $password = strip_tags (@$_POST['password']); $repeatpassword = strip_tags (@$_POST['repeatpassword']); $date = ("Y-m-d"); if ($submit) { //check for existence if ($fullname&&$username&&$password&&$repeatpassword) { if ($password==$repeatpassword) { //check char length of username and fullname if (strlen($username)>25||strlen ($fullname)>25) { echo ("Length …

<?php $submit = @$_POST['submit']; //form data $fullname = strip_tags (@$_POST['fullname']); $username = strip_tags (@$_POST['username']); $password= strip_tags(@$_POST['password']); $repeatpassword = strip_tags(@$_POST['repeatpassword']); $date =("Y-m-d"); if ($submit); { // check for existance if($fullname&&$username&&$password&&$repeatpassword) { //encrypt password $password = md5 ($password); $repeatpassword =md5 ($repeatpassword); if ($password==$repeatpassword) { // check char length of username and fullname …

Use of undefined constant password - can someone help me