donglv7097 2016-01-14 23:21
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尝试使用php [duplicate]从phpmyadmin中存储的数据库中选择数据时出错

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>

    
  

</div>
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2条回答 默认 最新

  • dourong4031 2016-01-14 23:31
    关注

    On line 25 you have $POST instead of $_POST which would be the correct way.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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