another help please.

[asherinho]

Iam a beginer to php thats why i frequently ask questions.

I have a table in the database havin 3 fields:- firstname,lastname,age.i want to fetch the age basing on the firstname selected from the form's select menu which has all the names present in the firstname field.the form is saved as php since i had to first fetch the names to be available in the select menu.here are the codes and the warning i got

 

<?php

$db=mysql_connect("localhost","root");

mysql_select_db("my_db",$db);

$query="SELECT age FROM person WHERE firstname=$firstname";

$result=mysql_query($query);

$record=mysql_fetch_assoc($result);

list($fieldname,$fieldvalue)=each($record);

echo $fieldvalue;

?>

 

Warning: Variable passed to each() is not an array or object in C:\Program Files\xampp\webalizer\age.php on line 15

 

[trq]

Assuming your form uses the POST method....

 

<?php

  $db = mysql_connect("localhost","root");
  mysql_select_db("my_db",$db);
  $firstname = mysql_real_escape_string($_POST['firstname']);
  $query="SELECT age FROM person WHERE firstname='$firstname'";
  if ($result = mysql_query($query)) {
    if (mysql_num_rows($result)) {
      $record = mysql_fetch_assoc($result);
      list($fieldname,$fieldvalue) = each($record);
      echo $fieldvalue;
    } else {
      echo "No results found for $firstname<br />";
    }
  } else {
    echo "Query failed<br />" . mysql_error() . "<br />$query";
  }

?>