PravinS

your php code should be like this <?php // Always try to connect and select the DB before anything else $con = mysql_connect("localhost", "jingleko_reload", "*******") or die("Couldnt Connect to DB - ".mysql_error()); mysql_select_db("jingleko_reloader", $con) or die("Couldnt Select a DB - ".mysql_error()); // Set post var $Epost = trim(addslashes(strip_tags($_POST['Epost']))); if (isset($_POST['Epost'])) { // Look for it in DB $query = "SELECT Epost FROM newsletter WHERE Epost='".$Epost."'"; $result = mysql_query($query); //If found, do next thing if (mysql_num_rows($result) > 0) { mysql_query("DELETE FROM newsletter WHERE Epost='$Epost'") or die(mysql_error()); echo "<div align=\"center\"><img src=\"Pics/Vlaamse Leeuw.jpg\" width=\"114\" height=\"127\" border=\"0\"></div>"; echo "<p align=\"center\"><b>Thank you, you are now removed from the list.</b></p><br>"; echo "<p align=\"center\"><a href=\"index.htm\"><img src=\"Pics/begin.gif\" width=\"95\" height=\"30\" border=\"0\"></a></p>"; } else { echo "<div align=\"center\"><b><font color=\"red\">This address does not exist</font></b></div><br>"; echo "<div align=\"center\"><a href=\"eruit.htm\"><img src=\"Pics/herbegin.gif\" width=\"95\" height=\"30\" border=\"0\"></a>"; echo "<a href=\"index.htm\"><img src=\"Pics/begin.gif\" width=\"95\" height=\"30\" border=\"0\"></a></div>"; } } mysql_close($con); ?>

you can use isset() function to check if $_FILES['image'] is set of not like if (isset($_FILES['image'])) { // your code here var_dump($_FILES['image']); // your code here } may this will help you

you can check it like this if (is_array($_POST['interest']) && count($_POST['interest']) > 0) { // code here }

do you know URL rewriting? check : http://httpd.apache.org/docs/2.0/misc/rewriteguide.html

i think your code should be like this $event = array("Flash Flood Watch", "Flash Flood Warning", "Severe Thunderstorm Watch", "Severe Thunderstorm Warning", "Torndao Watch", "Torndao Warning"); $searchvalue = "Flash Flood Watch"; //as example if (in_array($searchvalue, $event)) { echo "Match found"; } also go through in_array() function properly http://www.php.net/in_array

use php explode() function http://www.php.net/manual/en/function.explode.php

give name as array for each field, as <?php foreach ($items as $item) { echo'<div class="items-row"> <label for="name_'.$item['item_id'].'">'.$item['item_name'].'</label> <div class="item_input"><input type="text" name="quantity[]" id="'.$item['item_name'].'" value="0"></div> <input type="hidden" name="item_ids[]" value="'.$item['item_id'].'"> <input type="hidden" name="user_ids[]" value="'.$moving_basics_id.'"> </div>'; } ?>

your function should return the value like <?php function test() { $response = "hello"; return $response; } ?> and then you can call the function like this <input type="textbox" name="xx" value="<?php echo test(); ?>" />

can you show us the actual code

it means you haven't created database on your local machine or the database connection in database.php is not correct

at this line add mysql_error() like this and check what is the issue $data=mysql_query("SELECT OT.nostaf, OT.otl_tkh_m, OT.otl_tkh_y ,OT.otl_jum_jam, OT.otl_jabatan FROM ot_line OT") or die(mysql_error());

mysql_select_db("ahamdabad_vpn_ip$", $dbc); $sql = "select main_system from ahamdabad_vpn_ip$"; "ahamdabad_vpn_ip$" is you database or table name? also it is not good practice to use "$" in database of table name

use urlencode() and urldecode() functions

echo the SQL query and execute it in phpmyadmin also write mysql_close($con); at the end the bottom of the code

check you SQL query, remove space between FROM and table name from SQL query $query ="SELECT * From Section order by sectCode ASC";

your form method is post, so use POST while getting variable data $destinationCode=$_POST['destinationcode']; $numberofadults=$_POST['numberofadults']; $numberofchildren=$_POST['numberofchildren']; $startdate=$_POST['stardate']; $enddate=$_POST['enddate'];

you can check this example http://www.w3schools.com/php/php_ajax_database.asp

try uncommenting the if condition <?php //$ab=1; $ab = ''; if(isset($_GET['id'])) $ab= $_GET['id']; echo $ab; ?>

check http://www.phpclasses.org/package/5927-PHP-Display-an-hierarchic-menu-stored-in-a-MySQL-table.html

You can also use TCPDF http://www.tcpdf.org/

check the syntax of INSERT query http://dev.mysql.com/doc/refman/5.6/en/insert.html

input to "mysql_real_escape_string()" function should be string and you have passed array to it, so you are getting this error if possible paste your code here

try using this SQL query SELECT received, SUM(amount) as amt FROM TABLE_NAME GROUP BY received;

if you want to use server side valiation then you can use "filter_var" function to check the email or else you can use javascript email validation http://www.php.net/manual/en/function.filter-var.php

try replacing this code <form action="add_bk_insert.php"> Name <input type="text" name="name"/><br /> E-mail <input type="text" name="email" /><br /> Contact <input type="text" name="contact" /><br /> Requirement <input type="text" name="requirement" /><br /> <input type="submit" value="Submit" name="btnSubmit"/> </form> <?php mysql_connect("localhost","root",""); mysql_select_db("test"); if (isset($_POST['btnSubmit'])) { $name = isset($_POST[name]) ? $_POST[name] : ''; $email_id = isset($_POST[email_id]) ? $_POST[email_id] : ''; $contact = isset($_POST[contact]) ? $_POST[contact] : ''; $requirement = isset($_POST[requirement]) ? $_POST[requirement] : ''; $order = "INSERT INTO sal2 (name,email_id, contact,requirement) VALUES('".$name."','".$email_id."','".$contact."','".$requirement."')"; $result = mysql_query($order); if($result) { echo "Inserted"; //include_once("testimonials.php"); } else{ echo("<br>Input data is fail"); } } ?>