[There is a mistake somewhere....]

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);
?>

[Notice: Undefined index: image]

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

[is_array]

you can check it like this

if (is_array($_POST['interest']) && count($_POST['interest']) > 0)
{
       // code here
}

[Generate PHP profile page with PHP data]

do you know URL rewriting?

 

check : http://httpd.apache.org/docs/2.0/misc/rewriteguide.html

[Problem with Array]

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

[Convert comma seperated strings into array]

use php explode() function

 

http://www.php.net/manual/en/function.explode.php

[Adding Values from looped input fields]

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>'; 
      }
?>

[Output to textbox from within a function]

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(); ?>" />

[PHP Arrays]

can you show us the actual code

[Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given]

thanks mate. its solve. it work perfectly on live server but its appearing "no database found" when running it offline. the database.php also include in the folder.

 

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

[Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given]

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());

[problem in my program]

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

[& problem in URL]

use urlencode() and urldecode() functions

[Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\Exer10\prtSect.php on line 20]

echo the SQL query and execute it in phpmyadmin

 

also write  mysql_close($con); at the end the bottom of the code

[Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\Exer10\prtSect.php on line 20]

check you SQL query, remove space between FROM and table name from SQL query

$query ="SELECT * From Section order by sectCode ASC";

[get variables not working]

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'];

[ajax with php]

you can check this example

 

http://www.w3schools.com/php/php_ajax_database.asp

[Query string error: Undefined index: id]

try uncommenting the if condition

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

[Dropdown php]

check

 

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

[Generate pdf report in php]

You can also use TCPDF

http://www.tcpdf.org/

[Help me i am makeing a login]

check the syntax of INSERT query

 

http://dev.mysql.com/doc/refman/5.6/en/insert.html

[Unfamiliar Warning Message]

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

[MYSQL HELP ... i find very difficult... need help...]

try using this SQL query

SELECT received, SUM(amount) as amt FROM TABLE_NAME GROUP BY received;

[Email validation check]

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

[Notice: Undefined index: name in C:\xampp\htdocs]

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");

	}
}
?>