Can someone please show me where is the error in this code because i'm trying to selct data from a database i've created in phpmyadmin but the output is:Notice: Undefined variable: POST in C:\xampp\htdocs\php\browsestudents.php on line 25. The database works correctly.This is my code:
<html>
<head>
<title>Online Jobs
</title>
</head>
<body>
<form action="browsestudents.php" method="POST">
<?php
print("Kerko studentet ne varesi te pozicionit te aplikuar te punes:
<form><select name='position'>
<option value='it'>IT</option>
<option value='a'>a</option>
<option value='b'>b</option>
</select>
<input type='submit' value='Search'/>
</form>");
if( !($database=mysql_connect("localhost","root",""))||!(mysql_select_db("st_login",$database)) )
print("Could not connect");
if(isset($_POST['position'])&&!empty($_POST['position'])){
$position=$POST['position'];
if($position=='a'||$position=='b'||$position=='it')
{
$query="SELECT `firstname`,`lastname`,`cv`,`position` /*shto listene notave ne fund*/
FROM `login`
WHERE `position`='$position' /*AND food LIKE 'P%' */
ORDER BY `id`";
if($result=mysql_query($query,$database))
{
if(mysql_num_rows($result)==NULL){//nr i rreshtave te query qe kemi kerkuar esht NULL
echo"no result found";
}
else
while($query_row=mysql_fetch_assoc($result))//krijon nje vektor
//query_row te cilit i asocion te dhenat e query
{
$firstname=$query_row["firstname"];//query_row esht vektor associativ qe i vendoset vlera e food
$lastname=$query_row["lastname"];
$cv=$query_row["cv"];
$position=$query_row["position"];
echo $firstname."<br/> Here is his cv".$cv."and position".$position;
}
}
else
echo mysql_error();
}
}
?>
</body>
</html>
